Electricity

Class 11 Physics, unit 11, book pages 197–218

Practise 81 MCQs Take the unit quiz Review 51 cards

Electricity is about charges in motion. This unit starts from the slow drift of electrons that makes a current, then treats potential, emf and internal resistance, resistivity and its change with temperature, and electric power and energy. It then gives Kirchhoff's two laws, uses them to find the combined resistance of resistors in series and in parallel and to solve circuits, and ends with the null methods (Wheatstone bridge and potentiometer) and with sensors such as the LDR, the thermistor and the conducting carbon fibres in concrete bridges.

Learning outcomes (18)

Use, for a current-carrying conductor, the expression $I = A n v q$ [where $n$ is the number of charge carriers per unit volume].
State and use $V = W / Q$ .
State and use $P = I V$ , $P = I^{2} R$ and $P = V^{2} / R$ .
State and use $R = ρ L / A$ .
State that the resistance of a light-dependent resistor (LDR) decreases as the light intensity increases.
Define and use the electromotive force (e.m.f.) [of a source as energy transferred per unit charge in driving charge around a complete circuit].
Distinguish between e.m.f. and potential difference (p.d.) in terms of energy considerations.
Explain the effects of the internal resistance of a source of e.m.f. on the terminal potential difference.
State Kirchhoff's first law and describe that it is a consequence of conservation of charge.
State Kirchhoff's second law and describe that it is a consequence of conservation of energy.
Derive, using Kirchhoff's laws, a formula for the combined resistance of two or more resistors in series.
Derive and apply a formula for the combined resistance of two or more resistors in parallel.
Use Kirchhoff's laws to solve simple circuit problems.
State and use the principle of the potentiometer as a means of comparing potential differences.
Explain the use of a galvanometer in null methods.
Explain the use of thermistors and light-dependent resistors in potential dividers [to provide a potential difference that is dependent on temperature and light intensity].
Explain the internal resistance of sources and its consequences for external circuits.
Explain how inspectors can easily check the reliability of a concrete bridge with carbon fibres as the fibres conduct electricity.

11.1Drift velocityOften examined

Electricity, conductors and insulators

Electricity is the branch of physics in which we deal with the dynamic state of charges. There are two categories: static electricity and current electricity. Static electricity is the discharging of an electrical pulse that appears because of an imbalance of positive and negative charges. Current electricity is the movement of charge carriers when a potential difference is provided across a conductor.

Materials fall into two categories according to whether they let charge flow through them:

Good and bad conductors.
Type of material	What it does	Examples (from the book)
Good conductor	Allows electric charge to flow through it	Aluminium, gold, steel, brass, copper
Bad conductor	Does not allow charge to flow	Wool, rubber, plastic, wood

Drift velocity of electrons

Drift velocity: The average velocity of the free electrons by which they drift towards the high potential terminal of a conductor in the presence of an electric field.

In a conductor billions of free electrons are moving at random, colliding with their neighbours and passing energy to them on the way. This random motion alone does not make a current. When a potential difference is applied across the conductor, an electric field is set up in it. The field pushes all the electrons in the same sense, from the low potential end towards the high potential end, on top of their random motion. The resulting slow, steady shift is the drift, and it produces the net current. The drift velocity of electrons in a metal is only about $1 0^{- 3} m s^{- 1}$ .

Figure 11.1 (a): Motion of electrons in the absence of an electric field. The electron wanders along a random zigzag path and has no net displacement along the wire.

The concept of drift velocity cannot be described without an external electric field. In a metal the Fermi velocity of the electrons, their random speed, is very high, about $1 0^{6} m s^{- 1}$ , and it does not depend on the current or the applied voltage. The drift velocity, in contrast, is only about $1 0^{- 3} m s^{- 1}$ . In a metal the electrons gain energy and lose it again in collisions, which is why a metal becomes hot when a current passes. For the flow of charge to continue, an external electric field must keep helping the electrons to drift.

When an emf source is attached across the ends of the conducting material, all the electrons become aligned in a well-organised way in one direction. Emf is the work done per unit charge in converting non-electrical energy to electrical energy, and the potential difference is the effect of that emf. Both are measured in the same unit (volt). The external electric field in the conductor can be measured as the potential difference across the conductor per unit length.

Figure 11.1 (b): Motion of electrons in the presence of an electric field. The electrons drift towards the positive terminal, so the conventional current is in the opposite sense.

Drift velocity and mobility

Mobility compares the drift velocity with the field that causes it. Table 11.1 sets the two quantities side by side.

Table 11.1: Comparison of drift velocity and mobility.
	Drift velocity	Mobility
Meaning	Average velocity acquired by the charge carriers within the conducting material in the presence of an external electric field	Magnitude of the drift velocity of the charge carriers per unit electric field
Formula	$v = - μ E$	$μ = \frac{v}{E}$
Unit	metre per second ( $m s^{- 1}$ )	$m^{2} s^{- 1} V^{- 1}$

Drift velocity factors

Derivation Current in terms of drift velocity: $I = n e A v$

Aim. Find the current $I$ in a metal wire of length $L$ and uniform cross-section area $A$ , when an emf source of potential difference $V$ is attached across it and the free electrons drift with velocity $v$ .

Assumptions.

The wire has uniform cross-section area $A$ and length $L$ .
Every free electron drifts with the same average speed $v$ .

The potential difference $V$ across the length $L$ produces a uniform electric field in the wire.
$E = \frac{V}{L}$
Let $N$ be the number of free electrons. The volume of the wire is $A L$ , so the free electron density (number per unit volume) is $n = \frac{N}{A L}$ . Hence the number of free electrons in the wire is
$N = n A L$
One electron carries charge $e$ , so the charge of all the free electrons in the wire is
$Q = N e = n A L e$
The electrons drift with speed $v$ , so the whole of this charge passes a given cross-section of the wire in the time needed to cover the length $L$ .
$t = \frac{L}{v}$
Current is charge per unit time.
$I = \frac{Q}{t} = \frac{n A L e}{L / v} = n e A v$

Result

I = n e A v (11.1)

Rearranging gives the drift velocity $v = \frac{I}{n A e}$ . The current per unit cross-section area, the current density, is $J = \frac{I}{A} = n e v$ . For carriers of charge $q$ the same result reads $I = A n v q$ , the form given in the unit's learning outcomes. Current and drift velocity are directly related: a larger drift velocity means a larger current.

Example 11.1: Drift velocity in a copper conductor

A current of

3 A

is flowing in a copper conductor with a cross-section of

1 m m^{2}

. Find the drift velocity of the electrons. (For copper,

n = 8.5 \times 1 0^{28}

per

m^{3}

Given

$I = 3 A$ current
$A = 1 m m^{2} = 1 \times 1 0^{- 6} m^{2}$ cross-section
$n = 8.5 \times 1 0^{28} m^{- 3}$ free electron density of copper
$e = 1.6 \times 1 0^{- 19} C$ charge of an electron

To find

drift velocity $v$

Use the relation between current and drift velocity.
$I = n A v e$
Make $v$ the subject.
$v = \frac{I}{n A e}$
Put in the values.
$v = \frac{3}{( 8.5 \times 1 0 ^{28} ) ( 1 \times 1 0 ^{- 6} ) ( 1.6 \times 1 0 ^{- 19} )} = \frac{3}{1.36 \times 1 0 ^{4}}$
Divide.
$v \approx 2.2 \times 1 0^{- 4} m s^{- 1}$

v \approx 2.2 \times 1 0^{- 4} m s^{- 1}

Assignment 11.1

A wire with diameter

0.02 m

contains

1 0^{28}

free electrons per

m^{3}

. Find the drift velocity of the free electrons in the wire with an electric current of

100 A

Given

$d = 0.02 m$ diameter
$n = 1 0^{28} m^{- 3}$ free electron density
$I = 100 A$ current
$e = 1.6 \times 1 0^{- 19} C$

To find

drift velocity $v$

The cross-section of the wire is a circle of diameter $d$ .
$A = \frac{π d ^{2}}{4} = \frac{π ( 0.02 ) ^{2}}{4} = 3.14 \times 1 0^{- 4} m^{2}$
Use $v = \frac{I}{n A e}$ .
$v = \frac{100}{( 1 0 ^{28} ) ( 3.14 \times 1 0 ^{- 4} ) ( 1.6 \times 1 0 ^{- 19} )} = \frac{100}{5.03 \times 1 0 ^{5}}$
Divide.
$v \approx 2.0 \times 1 0^{- 4} m s^{- 1}$

v \approx 2.0 \times 1 0^{- 4} m s^{- 1}

11.2Electric potential

Electric potential energy

A charged particle placed in an electric field experiences a force, $F = E q$ . So if a charge carrier is to be moved in a constant electric field, work is required, and this work changes the electric potential energy of the charge carrier. It is like moving a massive object in a uniform gravitational field: the work changes the potential energy of the object. When an object moves naturally (along the force) no external force and no expenditure of energy is needed, but when it moves against the natural force we must supply external energy. The same idea is used in electricity.

Electric potential energy: The energy required by a charge carrier to move against the electric field. Equivalently, the work done to place a charged particle at a point of an electric field, from any point at infinity.

Consider a charge carrier $q$ in a constant electric field $E$ . A positive charge moves from the left plate to the right plate, that is, from the positively charged plate to the negatively charged plate (Fig. 11.2 (a)). If the distance covered by the charge in the direction of the field is $S$ (Fig. 11.2 (b)), the work done is

W = F S = q E S

Figure 11.2 (a): A charge

q

experiences a force in an electric field. Field lines run from the positive plate to the negative plate.

Figure 11.2 (b): The charge covers a distance

s

in the direction of the field; the force on it is

F = q E

The electrostatic force is conservative in nature, so the work done on the charge $q$ does not depend on the path followed. A charged particle possesses electric potential energy when it is placed in an electric field.

Relation between electric potential energy and electric potential

As the electric potential energy is $q E S$ , the electric potential is the amount of electric potential energy per unit charge needed to move a charged object from any point at infinity into the uniform or non-uniform electric field region, against the direction of the field lines.

Electric potential: The electric potential energy per unit charge, or the work done per unit charge against the electric field: $V = \frac{U}{q} = \frac{W}{Q}$ .

V = \frac{U}{q} so U = V q

$V$ : electric potential; $U$ : electric potential energy; $q$ : charge under observation. Electric potential is measured in volt, $1 V = 1 J C^{- 1}$ .

Derivation Potential in a uniform field: $V = - E r$

Aim. Express the electric potential at a point a distance $r$ from the reference point in a uniform electric field $E$ .

Assumptions.

The field $E$ is uniform (constant in size and direction).
$θ = 18 0^{\circ}$ represents work done against the electric field lines.

Electric potential is work per unit charge.
$V = \frac{W}{Q}$
The force on the charge $Q$ in the field is $F = E Q$ . For a displacement $r$ at an angle $θ$ to this force the work is $F r cos θ$ .
$V = \frac{F r}{Q} = \frac{F r cos θ}{Q} = \frac{E Q r cos θ}{Q}$
Cancel $Q$ .
$V = E r cos θ$
For $θ = 18 0^{\circ}$ the displacement is against the electric field lines, that is, the work is done against the field. Put $cos 18 0^{\circ} = - 1$ .
$V = - E r$

Result

V = - E r

The minus sign records that the displacement is against the field. The magnitude of the potential difference between two points a distance $r$ apart along a uniform field is $E r$ .

Potential due to point charges

Let any point be at a distance $r$ from a point charge $q$ taken as the reference. The electric potential at that point is

V = k \frac{q}{r}

$k = 9 \times 1 0^{9} N m^{2} C^{- 2}$ is Coulomb's constant; $q$ : the point charge; $r$ : distance from the charge.

Electric field intensity and electric potential are both field quantities. For more than one charge the electric potential is the sum of the potentials of the separate charges:

V = k i = 1 \sum N \frac{q _{i}}{r _{i}}

Worked example: potential at a point due to two charges

A point P lies

0.30 m

from a charge

q_{1} = + 4.0 nC

and

0.20 m

from a charge

q_{2} = - 2.0 nC

(Fig. below). (a) Find the electric potential at P. (b) How much potential energy does a charge of

+ 5.0 nC

have when placed at P?

The point P lies between

q_{1} = + 4 nC

and

q_{2} = - 2 nC

, with

r_{1} = 0.30 m

and

r_{2} = 0.20 m

. Each charge makes its own potential at P; the two are added as numbers (signs included).

Given

$q_{1} = + 4.0 nC = 4.0 \times 1 0^{- 9} C$
$r_{1} = 0.30 m$
$q_{2} = - 2.0 nC = - 2.0 \times 1 0^{- 9} C$
$r_{2} = 0.20 m$
$k = 9 \times 1 0^{9} N m^{2} C^{- 2}$
$q = + 5.0 nC$

To find

(a) potential $V$ at P
(b) potential energy $U$ of $q$ at P

Potential is a scalar, so the potentials of the separate charges are added as ordinary numbers, with the sign of each charge kept: $V = k \sum q_{i} / r_{i}$ .
$V = k (\frac{q _{1}}{r _{1}} + \frac{q _{2}}{r _{2}})$
Potential due to the positive charge $q_{1}$ alone (positive).
$V_{1} = \frac{( 9 \times 1 0 ^{9} ) ( 4.0 \times 1 0 ^{- 9} )}{0.30} = + 120 V$
Potential due to the negative charge $q_{2}$ alone (negative, since $q_{2}$ is negative).
$V_{2} = \frac{( 9 \times 1 0 ^{9} ) ( - 2.0 \times 1 0 ^{- 9} )}{0.20} = - 90 V$
Add the two values.
$V = V_{1} + V_{2} = 120 - 90 = + 30 V$
(b) The potential energy of a charge placed in a potential is $U = V q$ .
$U = V q = (30) (5.0 \times 1 0^{- 9}) = 1.5 \times 1 0^{- 7} J$

V = + 30 V, U = 1.5 \times 1 0^{- 7} J

The electron-volt

In atomic and nuclear physics the charged objects of interest are electrons and protons, each with charge of magnitude $q_{e} = 1.6 \times 1 0^{- 19} C$ . The electrostatic potential energy of such objects is expressed in the unit electron-volt.

1 eV = 1.6 \times 1 0^{- 19} J

Electric potential is very similar to gravitational potential, because both are relative quantities. Electric potential cannot be measured without measuring the difference in potential, for example across the terminals of a battery. In any electric circuit, the potential difference tells us about the energy conversion from electrical energy to any other form of energy, with respect to every point charge that moves between two defined points in the circuit.

Example 11.2: Potential difference from energy

What is the potential difference between two points in an electric field if it takes

600 J

of energy to move a charge of

2 C

between these two points?

Given

$Δ U = 600 J$ energy needed
$Q = 2 C$ charge moved

To find

potential difference $Δ V$

Potential difference is energy per unit charge.
$Δ V = \frac{Δ U}{Q}$
Put in the values.
$Δ V = \frac{600}{2} = 300 V$

Δ V = 300 V

Assignment 11.2

Calculate the current passing through a conducting material of cross-section area

1 c m^{2}

, if after applying a constant potential the electrons acquire a drift velocity of

1.5 \times 1 0^{- 2} m s^{- 1}

and the free electron density is

1 0^{24}

electrons per cubic metre.

Given

$A = 1 c m^{2} = 1 \times 1 0^{- 4} m^{2}$
$v = 1.5 \times 1 0^{- 2} m s^{- 1}$ drift velocity
$n = 1 0^{24} m^{- 3}$ free electron density
$e = 1.6 \times 1 0^{- 19} C$

To find

current $I$

Use Eq. (11.1), $I = n e A v$ .
$I = n A e v$
Put in the values.
$I = (1 0^{24}) (1 \times 1 0^{- 4}) (1.6 \times 1 0^{- 19}) (1.5 \times 1 0^{- 2})$
Multiply the numbers and add the powers of ten.
$I = 0.24 A$

I = 0.24 A

Table 11.2: Conversion factors for energy. Read across: 1 unit of the row equals the listed number of the column unit.
Energy unit	Joule	Kilojoule per mole	Electron-volt
Joule	1	$6.66 \times 1 0^{20}$	$6.24 \times 1 0^{18}$
Kilojoule per mole	$1.66 \times 1 0^{- 21}$	1	0.01
Electron-volt	$1.6 \times 1 0^{- 19}$	96.5	1

11.3Electromotive force and maximum power outputOften examined

To make charge carriers move and so provide a potential difference across a conducting material, we need an emf source. An emf source such as a cell or battery converts chemical energy into electrical energy. The term electromotive force is not a mechanical force at all: it represents work per unit charge.

Electromotive force (e.m.f.): The energy transferred per unit charge by a source in driving charge around a complete circuit: \varepsilon=\dfrac{W}{Q}. It is measured in volt.

The difference between e.m.f. and potential difference in terms of energy.
	e.m.f. (ε)	Potential difference (p.d.)
Energy change	Non-electrical energy (for example chemical) is changed into electrical energy	Electrical energy is changed into other forms (heat, light and so on)
Where	The source, for the whole circuit including the inside of the source	Between two points of a circuit, for example across one component
Defined as	Energy supplied per unit charge by the source	Energy changed per unit charge between the two points
Unit	Volt	Volt

Terminal potential difference and internal resistance

The source itself has some resistance to the flow of charge, called its internal resistance r. The potential difference across the terminals of a cell that is delivering current is therefore less than its emf (Fig. 11.3).

Figure 11.3: The emf source \varepsilon (with internal resistance r) converts chemical energy into electrical energy and drives current I through the load R. The terminal potential difference is V=\varepsilon-Ir.

Derivation Terminal potential difference and the current in the circuit

Aim. Find the potential difference V across a cell of emf \varepsilon and internal resistance r that sends a current I through a load R, and then find I.

Assumptions.

The load is a resistor R connected across the terminals of the cell.
The current is steady.

In a time t the charge that passes is It, so the source supplies the energy \varepsilon It. Part of it is wasted as heat in the internal resistance, I^{2}rt, and the rest, VIt, is delivered to the outside circuit.
\varepsilon It=VIt+I^{2}rt
Divide by It.
\varepsilon=V+Ir
The potential difference across the cell terminals is therefore
V=\varepsilon-Ir
The same V appears across the load, and for the load V=IR. Substitute.
IR=\varepsilon-Ir
Collect the terms in I.
\varepsilon=I(R+r)

ResultI=\dfrac{\varepsilon}{R+r}\qquad (11.2)

The internal resistance acts like an extra resistor in series with the load.

The effects of the internal resistance on the terminal potential difference follow from V=\varepsilon-Ir:

If no current is drawn (I=0) there is no drop inside the source, so V=\varepsilon.
The more current the source delivers, the larger the drop Ir and the smaller the terminal potential difference.
A source with a large internal resistance has a terminal potential difference that falls sharply when current is drawn; for a source with r very small, V\approx\varepsilon even for a large current.

Worked example: finding the emf and internal resistance of a cell

A student connects a cell to a variable load and a voltmeter. When the current is 0.20\,\mathrm{A} the terminal potential difference is 1.40\,\mathrm{V}, and when the current is 0.50\,\mathrm{A} it is 1.25\,\mathrm{V}. Find the emf and the internal resistance of the cell, and the current if the terminals are joined by a thick wire of negligible resistance.

Given

I_1 = 0.20\,\mathrm{A}
V_1 = 1.40\,\mathrm{V}
I_2 = 0.50\,\mathrm{A}
V_2 = 1.25\,\mathrm{V}

To find

emf \varepsilon
internal resistance r
current with R\approx0

For each reading the terminal potential difference is the emf minus the drop inside the cell, V=\varepsilon-Ir. Write it twice.
1.40=\varepsilon-0.20\,r\qquad(1)\\1.25=\varepsilon-0.50\,r\qquad(2)
Subtract (2) from (1) so that \varepsilon cancels.
0.15=0.30\,r\;\Rightarrow\;r=0.50\,\Omega
Put r back into (1).
\varepsilon=1.40+0.20(0.50)=1.50\,\mathrm{V}
A wire of negligible resistance gives R\approx0, so the current is limited only by the internal resistance, I=\varepsilon/(R+r).
I=\dfrac{1.50}{0+0.50}=3.0\,\mathrm{A}
On the graph this is the point where the line meets the I axis (V=0). The intercept on the V axis is \varepsilon and the magnitude of the slope is r.

Terminal potential difference against current

V=\varepsilon-Ir is a straight line. It cuts the V axis at the emf \varepsilon=1.5\,\mathrm{V} and its slope is -r=-0.50\,\Omega; the two measured readings lie on it.

\varepsilon=1.5\,\mathrm{V},\quad r=0.50\,\Omega,\quad I_{\text{short}}=3.0\,\mathrm{A}

Maximum power transfer

In many electronic circuits it is important to have the maximum transfer of power from the source to the load. In radio and TV transmission we want maximum power transfer from the transmitting medium to the antenna system, and from an amplifier to the speaker system. This is accomplished by a proper matching of the load resistance R and the source resistance r.

Derivation Maximum power transfer theorem

Aim. Show that the power delivered to a load R by a source of emf \varepsilon and internal resistance r is greatest when R=r, and find the maximum power.

If V is the potential difference across R, the loss of potential energy per second in R, the power delivered to R by the current I, is
P_{\mathrm{out}}=I^{2}R
Put I=\dfrac{\varepsilon}{R+r} from Eq. (11.2).
P_{\mathrm{out}}=\dfrac{\varepsilon^{2}R}{(R+r)^{2}}
Expand the denominator.
P_{\mathrm{out}}=\dfrac{\varepsilon^{2}R}{R^{2}+r^{2}+2Rr}
Write 2Rr=-2Rr+4Rr, which turns the denominator into a perfect square plus 4Rr.
P_{\mathrm{out}}=\dfrac{\varepsilon^{2}R}{R^{2}+r^{2}-2Rr+4Rr}=\dfrac{\varepsilon^{2}R}{(R-r)^{2}+4Rr}
Divide the top and the bottom by R to see what changes with R.
P_{\mathrm{out}}=\dfrac{\varepsilon^{2}}{\dfrac{(R-r)^{2}}{R}+4r}
For a given source, \varepsilon and r are fixed. The term \dfrac{(R-r)^{2}}{R} is never negative and is zero only when R=r. So the denominator is smallest, and P_{\mathrm{out}} largest, when R=r.
Put R=r in the last expression.
\left(P_{\mathrm{out}}\right)_{\max}=\dfrac{\varepsilon^{2}}{4r}

Result\left(P_{\mathrm{out}}\right)_{\max}=\dfrac{\varepsilon^{2}}{4r}\quad\text{or}\quad\dfrac{\varepsilon^{2}}{4R}\qquad (11.3)

Maximum power is delivered to a load R when the internal resistance of the source of emf is equal to the load resistance. This statement is called the maximum power transfer theorem. If the load resistance is less or greater than the source resistance, the power delivered to the load is low. (Since R=r at the maximum, \dfrac{\varepsilon^{2}}{4r} and \dfrac{\varepsilon^{2}}{4R} are the same.)

Power delivered to the load

The output power P_{\mathrm{out}} (as a fraction of its maximum) against the load resistance R (as a multiple of r). The peak is at R=r; a smaller or larger load receives less power.

11.4Variation of resistance with temperature

Temperature and resistance

How the temperature changes the resistance of a conducting material is given by the equation

R_t=R_0\left[1+\alpha\,(T-T_0)\right]

(11.4)

R_t: resistance at temperature T; R_0: resistance at the reference temperature T_0; \alpha: temperature coefficient of resistance of the material, in \mathrm{K^{-1}} (or ^\circ\mathrm{C}^{-1}).

The sign of \alpha decides what happens to the resistance as the temperature rises. Table 11.3 lists it for different electrical materials.

Table 11.3: Electrical material versus temperature coefficient of resistance (α).
Material	Temperature coefficient of resistance, α
Metals	Greater than zero; resistance increases with temperature.
Solid non-metals	Equal to zero; resistance is independent of temperature.
Semiconductors	Less than zero; resistance decreases with increase in temperature.
Alloys	Has a small value greater than zero.
Superconductors	At low temperature, the resistance of certain substances becomes exactly zero.

Resistance, length and cross-section area

The resistance of a conducting material depends upon physical factors such as the length and the thickness (cross-section area) of the conductor, and upon a non-physical factor like the temperature. The resistance R is directly proportional to the length L of the conductor and inversely proportional to its cross-section area A:

R\propto L\quad\text{and}\quad R\propto\dfrac{1}{A}\qquad\text{so}\qquad R\propto\dfrac{L}{A}\quad\text{or}\quad R=\text{constant}\times\dfrac{L}{A}

The constant is a property of the material, called its specific resistance or resistivity \rho. So

R=\rho\dfrac{L}{A}

\rho: resistivity, in ohm metre (\Omega\,\mathrm{m}); L: length; A: cross-section area.

Figure 11.4 (a): A cylindrical shaped conductor of length L, cross-section area A and resistivity \rho has resistance R=\rho L/A.

Consider a cylindrical conductor of length l and cross-section area a with resistance R_1=\dfrac{\rho l}{a} (Fig. 11.4 (b)). Now change its dimensions:

Double the length. If the length is changed from l to 2l with the area unchanged, the resistance rises from R_1 to R_2=\dfrac{\rho\,2l}{a}=2R_1 (Fig. 11.4 (c)).
Double the area. If the length stays the same but the wire is replaced by another wire of the same length and double the cross-section area (2a), the resistance becomes R_3=\dfrac{\rho l}{2a}=\dfrac{R_1}{2} (Fig. 11.4 (d)).

Figure 11.4 (b), (c), (d): Conductor 1 (length l, area a), conductor 2 (length 2l) and conductor 3 (area 2a). Doubling the length doubles R; doubling the area halves it.

It means that a thicker wire resists the flow of charge less than a thinner wire. The resistance of copper is much less than that of rubber, and that is why copper is used for the conduction of charge carriers while rubber is used for insulating.

Conductivity

Conductivity: The reciprocal of the resistivity, \sigma=\rho^{-1}=\dfrac{1}{\rho}. Its SI unit is (\Omega\,\mathrm{m})^{-1}, also written as siemens per metre.

Resistivity and resistance

Resistivity is the physical property of a conducting material that shows the ability of the atoms of that material to impede the flow of electrons. The higher the resistivity, the more strongly the electrons are attracted to the atoms and the more difficulty they have in moving through the material.

Resistivity and resistance compared, as in the book.
	Resistivity	Resistance
Kind of property	Intrinsic: a property of the material, a composition function	Extrinsic: depends on the physical dimensions of the conductor and on its temperature
Depends on the shape?	No	Yes (L and A)

Table 11.4: Resistivity.
Material	Resistivity (Ω m)
Aluminium	2.7\times10^{-8}
Copper	1.7\times10^{-7} (as printed)
Graphite	8\times10^{-6}
Quartz	5\times10^{16}
Silicon	2.3\times10^{3}

Example 11.3: Resistance of a wire

Calculate the resistance of a wire 10\,\mathrm{m} long that has a diameter of 2\,\mathrm{mm} and resistivity of 2.63\times10^{-2}\,\Omega\,\mathrm{m}.

Given

L = 10\,\mathrm{m} length
r = \dfrac{2\,\mathrm{mm}}{2}=1\,\mathrm{mm}=1\times10^{-3}\,\mathrm{m} radius
\rho = 2.63\times10^{-2}\,\Omega\,\mathrm{m} resistivity

To find

resistance R

The cross-section is a circle, A=\pi r^{2}, so
R=\dfrac{\rho L}{A}=\dfrac{\rho L}{\pi r^{2}}
Put in the values (taking \pi=3.14).
R=\dfrac{(2.63\times10^{-2})(10)}{(3.14)(1\times10^{-3})^{2}}=\dfrac{0.263}{3.14\times10^{-6}}
Divide.
R=0.83758\times10^{5}\,\Omega\approx83\,758\,\Omega

R\approx8.38\times10^{4}\,\Omega=83\,758\,\Omega

Assignment 11.3

A wire has a length of 10\,\mathrm{m} and a resistance of 100\,\Omega. If the wire is stretched to 3 times its original length, how much will the resistance of the wire increase?

Given

L_1 = 10\,\mathrm{m} original length
R_1 = 100\,\Omega original resistance
L_2 = 3L_1 stretched length

To find

new resistance R_2 and the increase R_2-R_1

The wire is only stretched, so its volume stays the same and its resistivity does not change. When the length becomes three times, the cross-section area falls to one third.
L_2=3L_1\quad\Rightarrow\quad A_2=\dfrac{A_1}{3}\qquad(\text{because }A_1L_1=A_2L_2)
Use R=\dfrac{\rho L}{A} for the new wire.
R_2=\dfrac{\rho L_2}{A_2}=\dfrac{\rho\,(3L_1)}{A_1/3}=9\,\dfrac{\rho L_1}{A_1}=9R_1
Put in the number.
R_2=9\times100\,\Omega=900\,\Omega
The increase is the difference.
R_2-R_1=900\,\Omega-100\,\Omega=800\,\Omega

R_2=900\,\Omega\ \text{(an increase of }800\,\Omega\text{)}

11.5Light-dependent resistors (LDRs)

A light-dependent resistor (LDR), also called a photo-resistor, is a device used to detect light levels, for example in a security system based on light detection. The basic mechanism behind its working is the mutual variation between the resistance of the material and the intensity of the light falling on it: when the intensity of light increases, the resistance of the material decreases.

Light-dependent resistor (LDR): A special type of resistor that works on the principle of photoconductivity: its resistance decreases as the intensity of the light falling on it increases (and increases as the light is reduced).

Figure 11.5 (a): An LDR. A track of cadmium sulphide, shown here as a serpentine pattern, joins the two leads. On the right is the circuit symbol.

The basic principle of the LDR is photoconductivity. Its conducting behaviour is better explained with the energy band theory: the material reduces its conductance when the incident light intensity reduces, and raises it when the light gets stronger. Fig. 11.5 (b) shows this inverse behaviour of the resistance at different luminance levels, from the darkness of a night to bright sunlight.

Resistance of an LDR against light intensity

Figure 11.5 (b): Inverse behaviour of resistance. Both scales are logarithmic: the resistance falls from the order of a megohm in the dark to a few hundred ohms in bright sunlight. The curve is a sketch of the book's figure.

Commonly two types of LDR exist, the intrinsic and the extrinsic photo-resistors. Intrinsic LDRs are made of pure semiconducting material such as germanium or silicon. Extrinsic LDRs are based on modified semiconducting materials, made by a doping process. Most of these LDRs cannot be used for shorter wavelengths of light and are effective only for longer wavelengths. LDRs have a lot of practical applications, such as electronic-eye security systems, street-light controlling systems, and light switching for sunset or sunrise conditions.

An LDR is a part of automation systems, which help to reduce human effort in many fields; in power saving the loss of electricity should be minimal. Its important characteristics include the maximum power it can dissipate within a given range of temperature. As the LDR is a sensitive device, its sensitivity changes with the wavelength of the light; it can detect in the range from 4\times10^{3}\,\text{\AA} to 10^{4}\,\text{\AA} (that is, 400\,\mathrm{nm} to 1000\,\mathrm{nm}).

In a dark environment the resistance of an LDR reaches its maximum value, of the order of megohms; in a lighted environment it falls to a few hundred ohms.
Photo-resistors have lower sensitivity at high temperatures, but higher sensitivity at lower temperatures.
LDRs have a sharp resistance recovery rate.

LDR in a potential divider

Potential divider: A voltage divider is a circuit that takes a larger voltage and divides it down by a fixed ratio, according to the electronic components used, to give a smaller output voltage. A thermistor or an LDR can be used within a potential divider circuit.

An LDR is used in a potential divider because its resistance can be increased or decreased by changing the intensity of the light that falls on it (Fig. 11.5 (c)). By increasing or decreasing the light the resistance in the circuit changes, and this changes the potential drop that appears across the LDR. This is one of the best practical applications of the LDR.

Figure 11.5 (c): LDR-based potential divider circuit. The supply V is shared between the fixed resistor R_1 and the LDR; the output V_2 is taken across the LDR.

How the output changes with light

The fixed resistor R_1 and the LDR are in series, so the same current flows through both.
I=\dfrac{V}{R_1+R_{\mathrm{LDR}}}
The output V_2 is the potential difference across the LDR.
V_2=IR_{\mathrm{LDR}}=V\,\dfrac{R_{\mathrm{LDR}}}{R_1+R_{\mathrm{LDR}}}
In bright light R_{\mathrm{LDR}} is small, so only a small share of V appears across it and V_2 is small. In the dark R_{\mathrm{LDR}} is large and V_2 is close to V.

11.6Electric power and electric energy

Electric power: The rate at which energy is transferred; the electrical energy consumed in a circuit per unit time.

Consider an electric circuit that consists of a resistor of load resistance R connected with an emf source, as shown in Fig. 11.6. The electric power consumed in the circuit can be measured as follows.

Figure 11.6: An electric circuit. The voltage V across the resistor drives the current I and the power P=VI is dissipated in R as heat.

Derivation Electric power: P=VI=I^{2}R=V^{2}/R

Aim. Express the electric power consumed in a resistor R in terms of the voltage V across it and the current I through it.

Power is the energy consumed by the circuit divided by time.
P=\dfrac{\text{energy}}{\text{time}}
Potential difference is energy per unit charge, so the energy used in moving a charge Q through a potential difference V is VQ. Put this in.
P=\dfrac{VQ}{t}
The charge per unit time is the current, \dfrac{Q}{t}=I.
P=VI
For a resistor, Ohm's law gives V=IR. Substitute it for V, or use I=\dfrac{V}{R} for I.
P=I^{2}R=\dfrac{V^{2}}{R}

ResultP=\dfrac{\text{energy}}{\text{time}}=\dfrac{VQ}{t}=VI=I^{2}R=\dfrac{V^{2}}{R}

The power loss in a resistor is energy lost as heat. The SI unit of electric power is the watt: 1\,\mathrm{W}=1\,\mathrm{V}\times1\,\mathrm{A}.

1\,\text{watt}=1\,\text{volt}\times1\,\text{ampere}

Electric energy

The energy required or acquired by charge carriers to move from one position to another is called electrical energy. When electrons move through a resistor under a potential difference applied across its ends, the energy is dissipated as heat. In other words, one coulomb of charge per unit time (Q/t) falls through the potential difference V, so the rate of energy dissipation is VI. Multiplying the power by the time gives the energy:

\text{Energy dissipation}=VQ=VIt=I^{2}Rt=\dfrac{V^{2}t}{R}

The commercial unit of electrical energy is the kilowatt-hour (kW h), the unit on an electricity bill.

1\,\text{kilowatt-hour}=1000\,\text{watt}\times3600\,\text{seconds}=3.6\times10^{6}\,\text{joule}

\text{number of units}=\dfrac{\text{power (watt)}\times\text{time of use (hours)}}{1000}

One unit of electricity is one kilowatt-hour.

Example 11.4: Cost of electricity

Calculate the cost of using a 50\,\mathrm{W} energy saver for 8 hours daily in your study room for one month. Assume that the price of a unit is Rs. 20.

Given

P = 50\,\mathrm{W} power
t = 8\times30=240\,\mathrm{h} use in 30 days
\text{cost per unit} = \text{Rs. }20

To find

cost of electricity

Find the number of units (kilowatt-hours).
\text{units}=\dfrac{50\times240}{1000}=12
Multiply by the cost of one unit.
\text{cost}=12\times20=\text{Rs. }240

\text{Cost}=\text{Rs. }240

Assignment 11.4

Estimate the cost of electricity consumed for a month if the following devices are used as specified: (i) 10 bulbs of 40\,\mathrm{W} for 10 hours; (ii) 10 fridges of 250\,\mathrm{W} for 24 hours; (iii) 40 tube lights of 25\,\mathrm{W} for 10 hours. (The rate of electricity is Rs. 50 per unit.)

Given

\text{rate} = \text{Rs. }50\text{ per unit}
\text{period} = 30\ \text{days} each device is used for the stated hours every day

To find

the monthly cost

Daily energy of each group, power (kW) \times number of devices \times hours. (i) bulbs:
E_1=\dfrac{40\times10\times10}{1000}=4\,\mathrm{kW\,h}
(ii) fridges:
E_2=\dfrac{250\times10\times24}{1000}=60\,\mathrm{kW\,h}
(iii) tube lights:
E_3=\dfrac{25\times40\times10}{1000}=10\,\mathrm{kW\,h}
Units for the month are 30 times the daily energy.
4\times30=120,\qquad60\times30=1800,\qquad10\times30=300\ \text{units}
Cost of each group at Rs. 50 per unit.
120\times50=6000,\qquad1800\times50=90\,000,\qquad300\times50=15\,000
Add the three costs.
6000+90\,000+15\,000=111\,000

\text{(i) Rs. }6000,\ \text{(ii) Rs. }90\,000,\ \text{(iii) Rs. }15\,000;\ \text{total Rs. }111\,000

11.7Kirchhoff's lawsOften examined

Gustav Robert Kirchhoff introduced laws relating to two electrical physical quantities, current and voltage, as a modified form of Ohm's work. Kirchhoff formulated a pair of laws as practical applications of the laws of conservation of charge and of energy. These laws help in the calculation of current in complex combinations of resistances with more than one emf source.

For example, the circuit in Fig. 11.7 is a multi-loop circuit. It has junctions (also known as nodes), and resistors and emf sources are attached in several branches to produce current. As the circuit is complex, the equivalent resistance cannot be found by simple series and parallel rules, so the circuit cannot be solved by applying Ohm's law alone. Kirchhoff's rules can be used instead of Ohm's law.

Figure 11.7: A multi-loop circuit. Several emf sources and resistors are joined at junctions, so series and parallel rules alone cannot reduce it.

Junction (node): A point of a circuit where three or more conductors meet.
Loop: Any closed path in a circuit.

Kirchhoff's first law is the junction rule (current law) and the second is the loop rule (voltage law). They follow in the next sections.

11.7.1Kirchhoff's first law or Kirchhoff's current law (KCL)Often examined

Kirchhoff's first law (current law, KCL): The currents entering a node (junction) always equal the currents leaving the node.

Consider five wires connected at a junction point, carrying the currents I_1, I_2, I_3, I_4 and I_5 (Fig. 11.8). I_1, I_2 and I_3 show the flow of charges entering the node and are taken positive, while I_4 and I_5 show the flow of charges leaving the node and are taken negative.

Figure 11.8: Kirchhoff's first law. I_1, I_2, I_3 flow into the node and I_4, I_5 flow out of it.

I_1+I_2+I_3-I_4-I_5=0\qquad\text{hence}\qquad\sum I=0

Currents entering the node are positive and currents leaving it are negative. Equivalently, I_1+I_2+I_3=I_4+I_5.

Derivation KCL as a consequence of conservation of charge

Aim. Explain why the current law is a statement of the conservation of charge.

Charge can neither be created nor destroyed, and a junction is only a point: it cannot store charge.
So the charge that has entered the junction as current in any time must leave it in equal amount in that same time.
\text{charge in}=\text{charge out}
Divide the charges by the time. Charge per unit time is current.
I_{\mathrm{in}}=I_{\mathrm{out}}\quad\text{or}\quad\sum I=0

Result\sum I=0

Kirchhoff's current law is just a specific way of stating the law of conservation of charge.

11.7.2Kirchhoff's second law or Kirchhoff's voltage law (KVL)Often examined

Kirchhoff's second law (voltage law, KVL): The algebraic sum of all the changes in potential around any closed path (loop) must be zero: \sum V=0.

According to Kirchhoff's law, the algebraic sum of the potential differences, including the voltage supplied by the voltage sources (emf) and the voltage drops in resistive elements (IR), in any loop must be equal to zero. The convention is that a rise in potential is taken as positive and a drop in potential as negative, as shown in Fig. 11.9 (a).

\sum V=0

Figure 11.9 (a): Sign convention for Kirchhoff's law. \Delta V is the change in potential along the direction in which the loop is travelled (blue arrow).

For example, consider a simple loop ABCDA with no junctions, having two emf sources \varepsilon_1 and \varepsilon_2 with internal resistances r_1 and r_2 and a resistor R_1 (Fig. 11.9 (b)). Assume a clockwise current I, and travel around the circuit in the direction of the assumed current, starting from A.

Figure 11.9 (b): A simple loop ABCDA. The longer plate of each cell is its positive terminal; the loop is travelled clockwise.

Applying KVL to the loop ABCDA

At \varepsilon_1 the direction is from the negative to the positive terminal of the battery, so there is a rise in potential, taken as positive.
\Delta V_{\varepsilon_1}=+\varepsilon_1
At the internal resistor r_1 there is a drop Ir_1, taken as negative.
\Delta V_{r_1}=-Ir_1
At R_1 there is a drop.
\Delta V_{R_1}=-IR_1
At \varepsilon_2 the direction is from the positive to the negative terminal, so there is a drop in potential, taken as negative.
\Delta V_{\varepsilon_2}=-\varepsilon_2
Through the internal resistor r_2 there is again a drop.
\Delta V_{r_2}=-Ir_2
Kirchhoff's voltage law says the sum of all the changes is zero.
\Delta V_{\varepsilon_1}+\Delta V_{r_1}+\Delta V_{R_1}+\Delta V_{\varepsilon_2}+\Delta V_{r_2}=0
Write in the values.
\begin{aligned}+\varepsilon_1+(-Ir_1)+(-IR_1)+(-\varepsilon_2)+(-Ir_2)&=0\\ \Rightarrow\ \varepsilon_1-Ir_1-IR_1-\varepsilon_2-Ir_2&=0\end{aligned}
Solving for the current gives
I=\dfrac{\varepsilon_1-\varepsilon_2}{r_1+r_2+R_1}

Derivation KVL as a consequence of conservation of energy

Aim. Explain why the voltage law is a statement of the conservation of energy.

Voltage is defined as energy per unit charge.
V=\dfrac{\Delta U}{q_0}
As a charge goes once around a closed loop it returns to its starting point, so its potential energy has not changed overall. The energy it gains in the sources must equal the energy it loses in the resistors and elsewhere in the loop.
Dividing the energies by the charge, the total amount of energy gained per unit charge must equal the amount lost per unit charge. With rises positive and drops negative this is
\sum V=0

Result\sum V=0

As energy and charge are both conserved, Kirchhoff's voltage law is just a specific way of stating the law of conservation of energy.

11.7.3Applying Kirchhoff's lawsOften examined

To solve problems by using Kirchhoff's laws, consider the circuit of Fig. 11.10 and take two closed loops:

Loop 1 (ABCFA), assuming that a current I_1 flows through it.
Loop 2 (CDEFC), assuming that a current I_2 flows through it.

Figure 11.10: Complex circuit. Loop 1 (ABCFA) carries I_1 clockwise and loop 2 (CDEFC) carries I_2 anticlockwise; the resistor R_2 is shared by both loops. The longer plate of each cell is its positive terminal.

Go around loop 1 in the direction of the assumed current I_1, starting from point A. The battery \varepsilon_1 is taken as positive, because we go from its negative to its positive terminal in the branch from A to B. At R_1 the drop in potential is negative, V_{R_1}=-I_1R_1. At R_2 the current is I_1+I_2 (both loop currents pass through it in the same direction, C to F), so the drop is V_{R_2}=-(I_1+I_2)R_2. Kirchhoff's loop rule gives

\varepsilon_1+V_{R_1}+V_{R_2}=0\quad\Rightarrow\quad\varepsilon_1-I_1R_1-(I_1+I_2)R_2=0

(11.5)

Go around loop 2 in the direction of I_2 (C, F, E, D, C), starting from point C. At R_2 the drop is again V_{R_2}=-(I_1+I_2)R_2. At R_3 the drop is V_{R_3}=-I_2R_3. The battery \varepsilon_2 is positive, because again we go from its negative to its positive terminal, in the branch from E to D. Hence

V_{R_2}+\varepsilon_2+V_{R_3}=0\quad\Rightarrow\quad-(I_1+I_2)R_2+\varepsilon_2-I_2R_3=0

(11.6)

By solving the equations (11.5) and (11.6) simultaneously, the branch currents in the circuit can be determined.

Worked example: a two-loop circuit where one current comes out negative

In the circuit of the figure \varepsilon_1=10\,\mathrm{V}, \varepsilon_2=6\,\mathrm{V}, R_1=2\,\Omega, R_2=4\,\Omega and R_3=3\,\Omega; the cells have negligible internal resistance. Use Kirchhoff's laws to find the current in each resistor and say what a negative answer means.

The two-loop circuit with the assumed currents. I_1 goes up the left branch, I_2 goes up the right branch and both pass downwards through R_2 as I_1+I_2. The cells have their positive plate (the longer line) at the top.

Given

\varepsilon_1 = 10\,\mathrm{V}
\varepsilon_2 = 6\,\mathrm{V}
R_1 = 2\,\Omega
R_2 = 4\,\Omega
R_3 = 3\,\Omega

To find

currents in R_1, R_2, R_3

Assume I_1 up the left branch and I_2 up the right branch. KCL at the top junction: both flow into it, so R_2 carries I_1+I_2 downwards.
I_{R_2}=I_1+I_2
KVL for loop 1 (left branch and middle branch), going round in the direction of I_1. The cell \varepsilon_1 is crossed from its negative to its positive plate (a rise), and both resistors are crossed along the current (drops).
\varepsilon_1-I_1R_1-(I_1+I_2)R_2=0\\10-2I_1-4(I_1+I_2)=0\;\Rightarrow\;3I_1+2I_2=5\qquad(1)
KVL for loop 2 (right branch and middle branch), going round in the direction of I_2 in the same way.
\varepsilon_2-I_2R_3-(I_1+I_2)R_2=0\\6-3I_2-4(I_1+I_2)=0\;\Rightarrow\;4I_1+7I_2=6\qquad(2)
Eliminate I_2: multiply (1) by 7 and (2) by 2, then subtract.
21I_1+14I_2=35,\quad 8I_1+14I_2=12\;\Rightarrow\;13I_1=23\;\Rightarrow\;I_1=1.77\,\mathrm{A}
Put I_1 into (1).
I_2=\dfrac{5-3(1.769)}{2}=-0.154\,\mathrm{A}
The current in the middle resistor.
I_{R_2}=I_1+I_2=1.769-0.154=1.62\,\mathrm{A}
Check with equation (2): 4(1.769)+7(-0.154)=7.08-1.08=6.0, which is correct.

I_{R_1}=1.77\,\mathrm{A},\quad I_{R_2}=1.62\,\mathrm{A},\quad I_{R_3}=0.15\,\mathrm{A}\ (\text{opposite to the arrow})

11.7.3Deriving the equation for resistors in seriesOften examined

Derivation Equivalent resistance of resistors in series

Aim. Using Kirchhoff's laws, find the single resistor R equivalent to two resistors R_1 and R_2 connected in series, and extend the result to several resistors.

Assumptions.

Two resistors R_1 and R_2 are joined end to end and a single resistor R is equivalent to them (the same total current and the same total potential difference V).

Figure 11.11: Resistors connected in series. The same current I flows through each; the potential differences add to the total V. The single resistor R on the right is the equivalent.

By Kirchhoff's first law the current I through each resistor is the same, since they are connected in series (there are no junctions between them).
By Kirchhoff's second law the total potential difference in the closed loop equals the sum of the potential differences across both resistors.
V=V_1+V_2\qquad (11.7)
By Ohm's law the potential difference is the product of current and resistance, V_1=IR_1 and V_2=IR_2, and for the equivalent resistor V=IR.
IR=IR_1+IR_2\qquad (11.8)
The current I is the same for all the resistors, so divide Eq. (11.8) by I.
R=R_1+R_2

ResultR=R_1+R_2+R_3+\cdots\qquad (11.9)

The equivalent resistor R of several resistors connected in series is the sum of the separate resistances, so it is larger than the largest of them.

11.7.3Deriving the equation for resistors in parallelOften examined

Derivation Equivalent resistance of resistors in parallel

Aim. Using Kirchhoff's laws, find the single resistor R equivalent to two resistors R_1 and R_2 connected in parallel, and extend the result to several resistors.

Assumptions.

Two resistors R_1 and R_2 are joined between the same two junctions, and a single resistor R is equivalent to them.

Figure 11.12: Resistors connected in parallel. The current I splits into I_1 and I_2 while each branch has the same potential difference V. The single resistor R on the right is the equivalent.

By Kirchhoff's first law the current through each resistor is different, because the current splits at the junction. The current through the equivalent resistor R is the total current I.
I=I_1+I_2\qquad (11.10)
By Kirchhoff's second law the potential difference across the resistors in different branches must be the same, and the resistor R has the same potential difference.
V=V_1=V_2\qquad (11.11)
By Ohm's law I=\dfrac{V}{R} for each part, so Eq. (11.10) becomes
\dfrac{V}{R}=\dfrac{V}{R_1}+\dfrac{V}{R_2}\qquad (11.12)
The potential difference V is the same for all, so divide Eq. (11.12) by V.
\dfrac{1}{R}=\dfrac{1}{R_1}+\dfrac{1}{R_2}\qquad (11.13)

Result\dfrac{1}{R}=\dfrac{1}{R_1}+\dfrac{1}{R_2}+\dfrac{1}{R_3}+\cdots\qquad (11.14)

The reciprocal of the equivalent resistance of resistors in parallel is the sum of the reciprocals of the separate resistances, so the equivalent resistance is smaller than the smallest of them.

Example 11.5: Currents in a two-loop circuit

Find the current flowing through each resistor, using Kirchhoff's law, in the circuit of the figure.

The circuit of Example 11.5. I_1 is assumed in loop BAFEB and I_2 in loop DCBED, both anticlockwise. The longer plate of each cell is its positive terminal.

Given

\varepsilon_1 = 3\,\mathrm{V}
\varepsilon_2 = 4.5\,\mathrm{V}
R_1 = 3\,\Omega
R_2 = 2\,\Omega
R_3 = 5\,\Omega
R_4 = 6\,\Omega shared by both loops
R_5 = 4\,\Omega

To find

current in each resistor

Loop BAFEB. Travel in the direction of I_1. The cell \varepsilon_1 gives a rise. R_1 and R_2 carry I_1. R_4 carries I_1 (upwards, E to B) and I_2 (downwards, B to E), so the net current in the direction of travel is (I_1-I_2).
\varepsilon_1-I_1R_1-I_1R_2-(I_1-I_2)R_4=0
Put in the values.
3-3I_1-2I_1-6(I_1-I_2)=0
Expand and collect the terms.
3-5I_1-6I_1+6I_2=0\quad\Rightarrow\quad11I_1-6I_2=3\qquad(1)
Loop DCBED. Travel in the direction of I_2. The cell \varepsilon_2 gives a rise. R_5 and R_3 carry I_2. R_4 carries I_2 (B to E) and I_1 (against), so the net current is (I_2-I_1).
\varepsilon_2-I_2R_5-(I_2-I_1)R_4-I_2R_3=0
Put in the values.
4.5-4I_2-6(I_2-I_1)-5I_2=0
Expand and collect the terms.
4.5-9I_2-6I_2+6I_1=0\quad\Rightarrow\quad-6I_1+15I_2=4.5\qquad(2)
Eliminate I_2: multiply Eq. (1) by 5 and Eq. (2) by 2.
55I_1-30I_2=15\qquad\text{and}\qquad-12I_1+30I_2=9
Add the two equations.
43I_1=24\quad\Rightarrow\quad I_1=0.56\,\mathrm{A}
Put I_1 in Eq. (1).
11(0.558)-6I_2=3\quad\Rightarrow\quad I_2=\dfrac{6.14-3}{6}=0.52\,\mathrm{A}
The current in R_4 is the difference of the two loop currents.
I_1-I_2=0.56-0.52\approx0.03\,\mathrm{A}\ \text{(upwards, E to B)}

Both I_1 and I_2 come out positive, so the assumed directions are their actual directions. R_1 and R_2 carry I_1; R_5 and R_3 carry I_2; R_4 carries I_1-I_2 upwards.

I_{R_1}=I_{R_2}=0.56\,\mathrm{A},\quad I_{R_3}=I_{R_5}=0.52\,\mathrm{A},\quad I_{R_4}\approx0.03\,\mathrm{A}\ \text{(upwards)}

Assignment 11.5

Find the current flowing through each resistor in the circuit having two loops, as shown in the figure.

The circuit of Assignment 11.5: a 10\,\mathrm{V} cell in the left branch, a 2\,\Omega resistor in the middle branch, and a 6\,\mathrm{V} cell with a 1\,\Omega resistor on the right. The longer plate of each cell (here the upper one) is its positive terminal; the loop currents are taken anticlockwise as in the book's figure.

Given

\varepsilon_1 = 10\,\mathrm{V} left branch, positive terminal up
\varepsilon_2 = 6\,\mathrm{V} right branch, positive terminal up
R_a = 2\,\Omega middle branch
R_b = 1\,\Omega bottom right

To find

current in every branch

Take I_1 anticlockwise in loop 1 and I_2 anticlockwise in loop 2, as drawn. In the middle branch I_1 runs upwards and I_2 runs downwards.
Loop 1: travel down the left branch, along the bottom and up the middle branch. Down through the 10\,\mathrm{V} cell is from its positive to its negative terminal, a drop. In the 2\,\Omega the current along the direction of travel (upwards) is I_1-I_2.
-10-2(I_1-I_2)=0\quad\Rightarrow\quad I_2-I_1=5\qquad(1)
Loop 2: travel down the middle branch, along the bottom through the 1\,\Omega and up through the 6\,\mathrm{V} cell (negative to positive terminal, a rise). In the 2\,\Omega the current along the direction of travel (downwards) is I_2-I_1.
-2(I_2-I_1)-1\cdot I_2+6=0\quad\Rightarrow\quad2I_1-3I_2+6=0\qquad(2)
Put I_2=I_1+5 from Eq. (1) into Eq. (2).
2I_1-3(I_1+5)+6=0\quad\Rightarrow\quad-I_1-9=0\quad\Rightarrow\quad I_1=-9\,\mathrm{A}
Then
I_2=I_1+5=-4\,\mathrm{A}
The current in the 2\,\Omega resistor, downwards, is
I_2-I_1=-4-(-9)=5\,\mathrm{A}

The negative signs mean that both assumed directions were wrong. The 10\,\mathrm{V} cell sends 9\,\mathrm{A} clockwise, upwards through itself; 5\,\mathrm{A} goes down through the 2\,\Omega resistor; and 4\,\mathrm{A} goes down the right branch, through the 1\,\Omega resistor and against the emf of the 6\,\mathrm{V} cell.

I_{10\,\mathrm{V}}=9\,\mathrm{A},\quad I_{2\,\Omega}=5\,\mathrm{A},\quad I_{1\,\Omega}=4\,\mathrm{A}

Worked example: a mixed series and parallel circuit

A 4\,\Omega resistor is joined in series with a parallel pair of 12\,\Omega and 6\,\Omega resistors, and the combination is connected across a 16\,\mathrm{V} battery of negligible internal resistance (Fig. below). Find the equivalent resistance, the battery current, the potential difference across the parallel pair and the current in each of its branches.

A 4\,\Omega resistor in series with a parallel pair (12\,\Omega and 6\,\Omega) across a 16\,\mathrm{V} battery. The two parallel branches share the same potential difference.

Given

R_1 = 4\,\Omega
R_2 = 12\,\Omega
R_3 = 6\,\Omega
V = 16\,\mathrm{V}

To find

R_{eq}
battery current I
p.d. across the parallel pair
I_2 and I_3

Replace the parallel pair by one equivalent resistor, using \dfrac{1}{R_p}=\dfrac{1}{R_2}+\dfrac{1}{R_3}.
\dfrac{1}{R_p}=\dfrac{1}{12}+\dfrac{1}{6}=\dfrac{1+2}{12}=\dfrac{1}{4}\;\Rightarrow\;R_p=4\,\Omega
The 4\,\Omega resistor and R_p are in series, so the resistances add.
R_{eq}=R_1+R_p=4+4=8\,\Omega
The battery current is the same current that flows through R_1 (series).
I=\dfrac{V}{R_{eq}}=\dfrac{16}{8}=2.0\,\mathrm{A}
Potential difference across R_1, then what is left for the parallel pair (the p.d.s in series add up to the battery voltage).
V_1=IR_1=2.0\times4=8\,\mathrm{V}\;\Rightarrow\;V_p=16-8=8\,\mathrm{V}
Each parallel branch has the full V_p across it, so the currents follow from Ohm's law.
I_2=\dfrac{8}{12}=0.67\,\mathrm{A},\qquad I_3=\dfrac{8}{6}=1.33\,\mathrm{A}
Check with KCL: the branch currents add up to the supply current.
I_2+I_3=0.67+1.33=2.0\,\mathrm{A}=I

R_{eq}=8\,\Omega,\quad I=2.0\,\mathrm{A},\quad V_p=8\,\mathrm{V},\quad I_2=0.67\,\mathrm{A},\quad I_3=1.33\,\mathrm{A}

11.8Null method of measurement

In the null method a galvanometer shows a null (zero) reading when two quantities are balanced, so a known quantity can be compared with an unknown one. The method is used to find the value of an unknown resistance in the circuit called the Wheatstone bridge. The potentiometer is another electrical device that uses the null method, for finding potentials.

Why a null reading is useful: the galvanometer is only used to tell whether any current flows. When it reads zero, the unknown is exactly balanced against a known quantity, so the galvanometer scale itself need not be accurate or calibrated, and no current is drawn from the circuit being measured at the balance point.

11.8.1Wheatstone bridgeOften examined

A Wheatstone bridge consists of four resistors (three known resistors and one unknown resistor) and a galvanometer connected between two points, say b and d (Fig. 11.13). The bridge becomes balanced when no deflection is seen on the galvanometer. This null reading can then be used to calculate the value of the unknown resistance.

Figure 11.13: Wheatstone bridge circuit. P, Q and S are known resistors, R is the unknown, and G is the galvanometer between b and d.

Derivation Balance condition of the Wheatstone bridge

Aim. Find the unknown resistance R of the bridge in terms of the three known resistances P, Q and S when the bridge is balanced.

Assumptions.

The bridge is balanced: the galvanometer shows no deflection.

Let I_1 and I_2 be the currents through P and R respectively when the bridge is balanced. There is no current through the galvanometer, so the current in Q is also I_1 and the current in S is also I_2.
The galvanometer reads zero, so the points b and d are at the same potential. Hence the potential drop across ab equals the potential drop across ad.
I_1P=I_2R\qquad (11.15)
Similarly the potential drop across bc equals the potential drop across dc.
I_1Q=I_2S\qquad (11.16)
Divide Eq. (11.15) by Eq. (11.16). The currents I_1 and I_2 cancel.
\dfrac{P}{Q}=\dfrac{R}{S}

ResultR=\dfrac{PS}{Q}\qquad (11.17)

If the three resistances on the right-hand side are known, the fourth resistance R can be calculated.

11.8.2PotentiometerOften examined

A potentiometer is a null-type resistance-network device for measuring potential differences. Its principle of action is that an unknown emf or p.d. is measured by balancing it, wholly or in part, against a known potential difference.

The simplest potentiometer consists of a wire AB of uniform cross-section, stretched alongside a scale and connected across a battery of potential V (Fig. 11.14). A standard cell of known emf \varepsilon_1 is connected between A and terminal 1 of a two-way switch S.

Figure 11.14: Potentiometer. The slide wire AB is connected across the battery V (with a rheostat Rh and key S). The cells \varepsilon_1 (known) and \varepsilon_2 (unknown) are joined in turn through the galvanometer G to the slider N.

Derivation Comparing two emfs with the potentiometer

Aim. Find the unknown emf \varepsilon_2 in terms of the standard emf \varepsilon_1 and the two balance lengths.

Assumptions.

The wire AB is of uniform cross-section and the current in it is steady, so the potential fall is the same for each unit length.
The emf of the driving battery is larger than \varepsilon_1 and \varepsilon_2.

Press the slider N momentarily against the wire AB with the standard cell \varepsilon_1 in the circuit (switch at 1), and adjust its position until the galvanometer deflection is zero. Let l_1 be the corresponding distance between A and N. The fall of potential over the length l_1 of the wire is then the same as the emf \varepsilon_1. The wire is uniform and carries a steady current, so the potential fall is proportional to the length.
\varepsilon_1\propto l_1\qquad (11.18)
Move the switch to 2, replacing the standard cell by the cell whose emf \varepsilon_2 is to be measured. Adjust the slider N again to give zero deflection on G. If l_2 is the new distance between A and N, then
\varepsilon_2\propto l_2\qquad (11.19)
Divide Eq. (11.19) by Eq. (11.18). The constant of proportionality (the potential fall per unit length of the wire) cancels.
\dfrac{\varepsilon_2}{\varepsilon_1}=\dfrac{l_2}{l_1}

Result\varepsilon_2=\left(\dfrac{l_2}{l_1}\right)\varepsilon_1\qquad (11.20)

The unknown emf is found from the standard emf and two lengths measured on a scale.

Worked example: comparing two emfs with a potentiometer

A potentiometer wire AB is 100\,\mathrm{cm} long and the driver battery puts 2.0\,\mathrm{V} across it. A standard cell of emf \varepsilon_1=1.018\,\mathrm{V} balances at a length l_1=50.9\,\mathrm{cm} from A. An unknown cell \varepsilon_2 balances at l_2=75.0\,\mathrm{cm}. Find \varepsilon_2 and explain why the galvanometer reads zero at balance.

Given

\varepsilon_1 = 1.018\,\mathrm{V}
l_1 = 50.9\,\mathrm{cm}
l_2 = 75.0\,\mathrm{cm}
V_{AB} = 2.0\,\mathrm{V}
L = 100\,\mathrm{cm}

To find

\varepsilon_2
potential gradient along the wire

The wire is uniform, so the potential difference per unit length is the same everywhere. This is the potential gradient.
k=\dfrac{V_{AB}}{L}=\dfrac{2.0}{100}=0.020\,\mathrm{V\,cm^{-1}}
At balance the galvanometer shows no current, so the cell's emf is exactly equal to the potential difference across the length of wire it is balanced against. For the standard cell this must be \varepsilon_1=kl_1.
kl_1=0.020\times50.9=1.018\,\mathrm{V}\;\checkmark
The unknown cell balances against l_2 on the same wire, so \varepsilon_2=kl_2. Dividing the two balance conditions removes k.
\dfrac{\varepsilon_2}{\varepsilon_1}=\dfrac{l_2}{l_1}\;\Rightarrow\;\varepsilon_2=\dfrac{l_2}{l_1}\,\varepsilon_1
Substitute.
\varepsilon_2=\dfrac{75.0}{50.9}\times1.018=1.50\,\mathrm{V}
Check from the gradient: kl_2=0.020\times75.0=1.50\,\mathrm{V}, the same answer.

Potential difference along the slide wire

The potential difference between the end A and a point on the wire grows in proportion to the length l (0.020\,\mathrm{V} for each centimetre here). The balance lengths 50.9\,\mathrm{cm} and 75.0\,\mathrm{cm} therefore pick out 1.018\,\mathrm{V} and 1.50\,\mathrm{V} on the line.

\varepsilon_2=1.50\,\mathrm{V}

The basic purposes of this electrical instrument are:

determination of electromotive force;
measurement of internal resistance;
comparison of the potentials of two emf sources;
a better output in measuring potential difference as compared with a voltmeter.

11.9Thermistor

Thermistor: Short for thermal resistor: a heat-sensitive resistor, in which a change of temperature changes the resistance. In the book's words, a special kind of thermometer that shows an inverse relation between temperature and resistance.

Samuel Ruben was the first to invent a thermistor, which was later used for commercial purposes. Thermistors are usually made of a semiconductor material (semiconductor oxides of iron, nickel and cobalt). A pair of platinum leads is attached at the two ends for electrical connections. The arrangement is enclosed in a very small glass bulb and sealed. They are generally in the form of discs, rods, beads and so on (Fig. 11.15).

Thermistors are classified on the basis of the sign of the temperature coefficient of resistance as negative temperature coefficient (NTC) and positive temperature coefficient (PTC) thermistors. The difference between them is the inverse and the direct relation between temperature and resistance, respectively. For a PTC thermistor the resistance increases with the rise in temperature; for an NTC thermistor the resistance decreases with the rise in temperature.

There are many practical applications of a thermistor, such as temperature compensation and temperature control.

The special characteristic of this electrical device is that it can be used as a potential divider. As the thermistor's resistance changes with temperature, if it is connected in a potential divider circuit with a known value of resistance, the desired voltage can be measured on the basis of the temperature variation.

Figure 11.16: Temperature-based potential divider circuit. The output voltage across the thermistor increases or decreases with the temperature, depending on whether the thermistor is NTC or PTC.

In a potential divider circuit the thermistor can be used because its resistance can be increased or decreased by changing the temperature. By increasing the temperature the resistance of the thermistor falls (an NTC thermistor), and by decreasing the temperature it rises, depending upon which type of thermistor is connected, NTC or PTC.

How the output depends on the temperature

The fixed resistor R_1 and the thermistor R_{\mathrm{th}} are in series, with the same current I=V/(R_1+R_{\mathrm{th}}). The output across the thermistor is
V_{\mathrm{out}}=V\,\dfrac{R_{\mathrm{th}}}{R_1+R_{\mathrm{th}}}
With an NTC thermistor, a higher temperature makes R_{\mathrm{th}} smaller, so a smaller share of V appears across it and V_{\mathrm{out}} falls. A lower temperature makes V_{\mathrm{out}} rise.
With a PTC thermistor the changes are the other way round.

So if the temperature of the (NTC) thermistor is high, the potential across it drops, because its resistance is decreased, and vice versa.

11.10Carbon fibers in concrete bridge

Carbon-fibre-reinforced concrete has been used in construction projects because of its ability to improve the compressive strength, tensile strength, elasticity modulus, chemical stability, abrasion resistance and corrosion resistance of the concrete. It also reduces shrinkage cracks and brittleness, and it is light in weight and has good thermal conductivity. Carbon fibre was introduced into concrete bridges (as in Fig. 11.18) to improve the flexural strength and reduce the crack width.

Carbon fibres contain mainly carbon atoms and are commonly used in civil engineering works. The optimum amount of carbon fibre is 0.3 % by volume of the concrete.

Checking the reliability of the bridge

An inspector can easily check the reliability of a concrete bridge made with carbon fibres, because the fibre conducts electricity. A sensor measures the electrical resistance of the fibres in the concrete. If the sensor shows that the electric resistance is increasing over time, the fibres are separating because of cracks.

The resistance of the fibres through the concrete shows the state of the bridge: a continuous fibre path has a low resistance; cracks separate the fibres and the measured resistance rises.

Carbon is a very good conducting material because its molecular structure is very similar to that of graphite. There are many electrical applications of carbon fibres, for example carbon electrodes for electrolysis, electromagnetic-interference shielding, sensing and electrical switching.

Formula sheet

Current and drift velocity

I=neAv

n: free electron density; e: electron charge; A: cross-section area; v: drift velocity (Eq. 11.1; for any carrier I=Anvq)

Unit: \mathrm{A}

Drift velocity

v=\dfrac{I}{nAe}

from Eq. (11.1)

Unit: \mathrm{m\,s^{-1}}

Current density

J=nev

current per unit cross-section area

Unit: \mathrm{A\,m^{-2}}

Mobility

\mu=\dfrac{v}{E}

v: drift velocity; E: electric field (the book writes v=-\mu E for electrons)

Unit: \mathrm{m^{2}\,s^{-1}\,V^{-1}}

Electric field in a wire

E=\dfrac{V}{L}

V: potential difference across the wire; L: length

Unit: \mathrm{V\,m^{-1}}

Work in a uniform field

W=qES

q: charge; E: field; S: distance along the field

Unit: \mathrm{J}

Electric potential

V=\dfrac{U}{q}=\dfrac{W}{Q}

U: potential energy; q, Q: charge

Unit: \mathrm{V}=\mathrm{J\,C^{-1}}

Potential in a uniform field

V=-Er

E: field; r: displacement against the field

Unit: \mathrm{V}

Potential of a point charge

V=k\dfrac{q}{r}

k=9\times10^{9}\,\mathrm{N\,m^{2}\,C^{-2}}; q: charge; r: distance

Unit: \mathrm{V}

Potential of several charges

V=k\sum_{i=1}^{N}\dfrac{q_i}{r_i}

sum over all the charges

Unit: \mathrm{V}

Electron-volt

1\,\mathrm{eV}=1.6\times10^{-19}\,\mathrm{J}

energy gained by an electron through 1\,\mathrm{V}

Unit: \mathrm{J}

Terminal potential difference

V=\varepsilon-Ir

\varepsilon: emf; I: current; r: internal resistance

Unit: \mathrm{V}

Current in a circuit

I=\dfrac{\varepsilon}{R+r}

R: load resistance (Eq. 11.2)

Unit: \mathrm{A}

Power delivered to the load

P_{\mathrm{out}}=\dfrac{\varepsilon^{2}R}{(R+r)^{2}}

I^{2}R with I=\varepsilon/(R+r)

Unit: \mathrm{W}

Maximum output power

\left(P_{\mathrm{out}}\right)_{\max}=\dfrac{\varepsilon^{2}}{4r}

occurs when R=r (Eq. 11.3)

Unit: \mathrm{W}

Resistance and temperature

R_t=R_0\left[1+\alpha(T-T_0)\right]

R_0: resistance at T_0; \alpha: temperature coefficient of resistance (Eq. 11.4)

Unit: \Omega

Resistance of a conductor

R=\rho\dfrac{L}{A}

\rho: resistivity; L: length; A: cross-section area

Unit: \Omega

Conductivity

\sigma=\dfrac{1}{\rho}

reciprocal of the resistivity

Unit: (\Omega\,\mathrm{m})^{-1}

Electric power

P=VI=I^{2}R=\dfrac{V^{2}}{R}

V: potential difference; I: current; R: resistance

Unit: \mathrm{W}

Electric energy

E=VQ=VIt=I^{2}Rt=\dfrac{V^{2}t}{R}

t: time

Unit: \mathrm{J}

Kilowatt-hour

1\,\mathrm{kW\,h}=3.6\times10^{6}\,\mathrm{J}

1000\,\mathrm{W}\times3600\,\mathrm{s}

Unit: \mathrm{J}

Number of units

\text{units}=\dfrac{P(\mathrm{W})\times t(\mathrm{h})}{1000}

one unit is one kilowatt-hour

Unit: \mathrm{kW\,h}

Kirchhoff's first law

\sum I=0

currents into a node are positive, out of it negative

Unit: \mathrm{A}

Kirchhoff's second law

\sum V=0

around a closed loop: rises positive, drops negative

Unit: \mathrm{V}

Resistors in series

R=R_1+R_2+R_3+\cdots

Eq. (11.9)

Unit: \Omega

Resistors in parallel

\dfrac{1}{R}=\dfrac{1}{R_1}+\dfrac{1}{R_2}+\dfrac{1}{R_3}+\cdots

Eq. (11.14)

Unit: \Omega

Wheatstone bridge

R=\dfrac{PS}{Q}

at balance; P, Q, S known (Eq. 11.17)

Unit: \Omega

Potentiometer

\varepsilon_2=\left(\dfrac{l_2}{l_1}\right)\varepsilon_1

l_1, l_2: balance lengths for the standard cell \varepsilon_1 and the unknown \varepsilon_2 (Eq. 11.20)

Unit: \mathrm{V}

Key terms

Electricity: The branch of physics in which we deal with the dynamic state of charges; it has two categories, static electricity and current electricity.
Drift velocity: The average velocity of the free electrons by which they drift towards the high potential terminal of a conductor in the presence of an electric field.
Mobility: The magnitude of the drift velocity of the charge carriers per unit electric field, \mu=v/E; its unit is \mathrm{m^{2}\,s^{-1}\,V^{-1}}.
Electric potential energy: The energy required by a charge carrier to move against the electric field, or the work done to place a charged particle in an electric field from infinity.
Electric potential: The electric potential energy per unit charge, or the work done per unit charge against the electric field, V=U/q; its unit is the volt, \mathrm{J\,C^{-1}}.
Potential gradient: The slope of the graph of potential against distance.
Electromotive force (e.m.f.): The energy transferred per unit charge by a source in driving charge around a complete circuit; the work per unit charge in converting non-electrical energy into electrical energy.
Internal resistance: The resistance of the source of emf itself; it lowers the terminal potential difference to V=\varepsilon-Ir.
Maximum power transfer theorem: Maximum power is delivered to a load R when the internal resistance of the source of emf is equal to the load resistance.
Temperature coefficient of resistance: The constant \alpha in R_t=R_0[1+\alpha(T-T_0)]; it is positive for metals, zero for solid non-metals, negative for semiconductors, small and positive for alloys.
Resistance: A measure of the opposition to the flow of charge carriers in an electrical circuit; its unit is the ohm.
Resistivity: A fundamental, specific property of a material that measures its electrical resistance; \rho=RA/L, in ohm metre.
Conductivity: The reciprocal of resistivity, \sigma=1/\rho.
Light-dependent resistor (LDR): A special type of resistor that works on the principle of photoconductivity: its resistance decreases as the light intensity increases.
Electric power: The rate of transfer of energy; P=VI=I^{2}R=V^{2}/R, in watt.
Kilowatt-hour: The commercial unit of electrical energy, 1\,\mathrm{kW\,h}=3.6\times10^{6}\,\mathrm{J}.
Kirchhoff's first law (KCL): The currents entering a node always equal the currents leaving the node; it expresses the conservation of charge.
Kirchhoff's second law (KVL): The algebraic sum of all the changes in potential around any closed loop is zero; it expresses the conservation of energy.
Null measurement method: A measuring method in which the quantity to be measured is balanced by an opposing known quantity that is varied until the resultant of the two is zero, seen as a null reading on a galvanometer.
Wheatstone bridge: A circuit of four resistors and a galvanometer that finds an unknown resistance from the balance condition R=PS/Q.
Potentiometer: A null-type device for measuring potential differences, by balancing an unknown emf or p.d. against a known potential difference along a uniform wire.
Thermistor: A heat-sensitive (thermal) resistor whose resistance changes with temperature; NTC and PTC are its two types.
Potential divider: A circuit that takes a larger voltage and divides it down by a fixed ratio to give a smaller output voltage; a thermistor or an LDR in it makes the output depend on temperature or light.

Multiple choice questionsp. 215–217

On which of the following factor the drift velocity does not depend:

Alength of the wireAnswer
Bcross-section of the wire
Cnumber of free electrons
Dmagnitude of the electric field

Why

The drift velocity is v=\dfrac{I}{neA} (equation 11.1). It depends on the current I, the electron density n and the cross-section A, and on the field through v=\mu E. The length L does not appear in v=I/(neA), so the drift velocity does not depend on the length of the wire.

B, C and D are wrong: A is in the denominator, n is in the denominator, and v rises with the field strength E, so each one does change the drift velocity.

Note: if the potential difference is kept fixed instead of the current, E=V/L and the length does matter. The book's relation I=neAv is the basis of this answer.

Drift velocity of electrons is independent of ____.

Atime consumedAnswer
Bthe length of the wire
Cthe number of free electrons
Dthe strength of the electric field

Why

Drift velocity is the steady average velocity of the free electrons, given by v=\dfrac{I}{neA} or v=\mu E. Neither expression contains time, so the time consumed has no effect on the drift velocity.

C and D are wrong because n appears in v=I/(neA) and v grows with E. B is a weaker option: L is not in v=I/(neA), but with a fixed potential difference E=V/L changes with L, so the length can change v. Time never does.

A 10\,\Omega resistor is connected across the terminals of a 10\,\mathrm{V} battery. The power dissipation of the resistor is:

A100\,\mathrm{W}
B50\,\mathrm{W}
C25\,\mathrm{W}
D10\,\mathrm{W}Answer

Why

The resistor is across the battery, so the potential difference across it is V=10\,\mathrm{V}. The power is P=\dfrac{V^{2}}{R}=\dfrac{(10)^{2}}{10}=10\,\mathrm{W}. The same answer comes from I=V/R=1\,\mathrm{A} and P=I^{2}R=1^{2}\times10=10\,\mathrm{W}.

A (100\,\mathrm{W}) is V^{2} itself, found by forgetting to divide by R.

A heavy duty refrigerator is used about 1000\,\mathrm{W}, if it is allowed to run continuously for 10 hours daily, how many kilowatt-hours of energy does it consume in 10 days?

A100\,\mathrm{kWh}Answer
B200\,\mathrm{kWh}
C300\,\mathrm{kWh}
D350\,\mathrm{kWh}

Why

Running time =10\,\mathrm{h/day}\times10\,\mathrm{days}=100\,\mathrm{h}. Power =1000\,\mathrm{W}=1\,\mathrm{kW}. Energy =P\,t=1\,\mathrm{kW}\times100\,\mathrm{h}=100\,\mathrm{kWh}.

B corresponds to 20 days of running, so the number of days was doubled.

What will be the value of emf if 3\,\mathrm{A} current within 5\,\mathrm{mins} perform work about 900\,\mathrm{J}?

A10\,\mathrm{V}
B3\,\mathrm{V}
C1\,\mathrm{V}Answer
D5\,\mathrm{V}

Why

Time t=5\,\mathrm{min}=300\,\mathrm{s}, so the charge is Q=It=3\times300=900\,\mathrm{C}. The emf is the work per unit charge: \varepsilon=\dfrac{W}{Q}=\dfrac{900\,\mathrm{J}}{900\,\mathrm{C}}=1\,\mathrm{V}.

B (3\,\mathrm{V}) is simply the current in amperes read as a voltage; the time was not used.

Which one of the following electrical devices helps us to measure the accurate value of emf?

Ameter bridge
Bvoltmeter
Cmultimeter
DpotentiometerAnswer

Why

A potentiometer balances the unknown emf against a known potential difference. At balance no current is drawn from the source, so the reading is the true emf.

A voltmeter draws a current, so it reads the terminal potential difference V=\varepsilon-Ir, which is less than the emf. A meter bridge is used to find resistances, not an emf.

The unit of dynamic charge is:

Acoulomb
Bohm
CampereAnswer
Dvolt

Why

Charge in motion is current, I=\dfrac{Q}{t}. Its SI unit is the ampere, 1\,\mathrm{A}=1\,\mathrm{C\,s^{-1}}, which is a rate of flow of charge.

A (coulomb) is the unit of charge itself, not of its rate of flow. The phrase dynamic charge is not a book term; the answer follows from reading it as charge in motion.

The resistance of a steel wire of particular length and thickness is 100\,\Omega. If the length and the diameter of the wire both are doubled, then the resistivity in \Omega\,\mathrm{m} will be:

A1\times10^{-7}\,\Omega\,\mathrm{m}
B2\times10^{-7}\,\Omega\,\mathrm{m}Answer
C4\times10^{-7}\,\Omega\,\mathrm{m}
D5\times10^{-7}\,\Omega\,\mathrm{m}

Why

Resistivity is a property of the material, \rho=\dfrac{RA}{L}. Doubling L and d changes the resistance but not \rho. The new resistance is R'=\rho\dfrac{2L}{\pi(2d)^{2}/4}=\dfrac{R}{2}=50\,\Omega, while \rho stays the same.

The given values fix the resistance but not the resistivity, because L and d are not given. The key is therefore the option closest to the resistivity of steel, about 2\times10^{-7}\,\Omega\,\mathrm{m}.

Note. The book prints no key, and the data do not fix \rho (no L or d is given). The key B is the closest option to steel and is unchanged by doubling L and d.

How much power dissipation will be increased if current passing through the circuit is increased 100\,\% (let keeping temperature constant)?

A100\,\%
B200\,\%
C300\,\%Answer
D500\,\%

Why

For a resistor at constant temperature R is constant, so P=I^{2}R. A 100\,\% increase makes the current 2I, so the power becomes (2I)^{2}R=4P. The power has risen by 300\,\%.

A and B would be right only if power were proportional to current, which is not the case for P=I^{2}R.

There are 2 resistors of having values 2\,\Omega and 4\,\Omega respectively are connected in series to a 6\,\mathrm{V} battery. The heat dissipated by the 4\,\Omega resistor in 5\,\mathrm{s} will be:

A5\,\mathrm{J}
B10\,\mathrm{J}
C20\,\mathrm{J}Answer
D30\,\mathrm{J}

Why

In series the total resistance is 2+4=6\,\Omega, so the current is I=\dfrac{6\,\mathrm{V}}{6\,\Omega}=1\,\mathrm{A}. The same current flows in the 4\,\Omega resistor, and the heat is H=I^{2}Rt=1^{2}\times4\times5=20\,\mathrm{J}.

B (10\,\mathrm{J}) is the heat in the 2\,\Omega resistor, not in the 4\,\Omega one.

The algebraic sign of an IR drop primarily depends upon the

Adirection of flow of current.Answer
Bbattery connections.
Cmagnitude of current flowing through it.
Dvalue of resistance.

Why

In the sign convention for Kirchhoff's voltage law, an IR drop is negative when the loop is traversed in the direction of the current, and positive when it is traversed against it. The sign therefore depends on the direction of flow of the current.

C and D give only the size of the drop, and B is wrong because battery connections fix the sign of an emf, not of an IR drop.

What will be the nearest value of protected resistance, if LDR having resistance 20\,\Omega is operated at specific intensity of light is to be protect through a series resistance in such a way that up to 12\,\mathrm{mA} current in the provision of 10\,\mathrm{V} potential across the circuit as shown in following figure?

A681\,\Omega
B773\,\Omega
C813\,\OmegaAnswer
D973\,\Omega

Why

The total resistance must be R_{\text{tot}}=\dfrac{V}{I}=\dfrac{10\,\mathrm{V}}{12\times10^{-3}\,\mathrm{A}}=833.3\,\Omega. The protective resistor is in series with the LDR, so R=R_{\text{tot}}-R_{\text{LDR}}=833.3-20=813\,\Omega.

A, B and D do not come from this subtraction; only C follows from V/I with the LDR removed.

Which of the following electrical instrument can be used as a null detector in the Wheatstone bridge?

AgalvanometerAnswer
Bammeter
Cvoltmeter
Dmultimeter

Why

A null detector must show zero when the bridge is balanced. The galvanometer shows no deflection at balance, and this zero reading is what the null method uses.

An ammeter or voltmeter gives a reading, not a zero, and so cannot show the balance point clearly.

142026 Annual I, Set B, MCQ 11

Potentiometer can be used for ____.

Acomparing two voltages.Answer
Bcomparing two currents.
Cmeasuring a voltage.
Dmeasuring a current.

Why

A potentiometer is a null device that compares potential differences. The book lists the comparison of the potential of two emf sources as one of its basic purposes.

B and D are wrong because a potentiometer compares potential differences, not currents. C is less exact because a voltage is found only by comparison with a known potential difference.

152025 Annual I, Set B, MCQ 12

In order to achieve high accuracy, the slide wire of a potentiometer should be:

Aas long as possible.Answer
Bas short as possible.
Cneither too small not too large.
Dvery thick.

Why

A longer wire spreads the same potential difference over a greater length, so the potential gradient is smaller. The balance point can then be set more precisely, which gives higher accuracy.

A short wire gives a larger potential gradient and a coarser balance point (B). Thickness (D) does not improve the sensitivity of the balance point.

There are ____ types of thermistor.

Aone
BtwoAnswer
Cthree
Dfour

Why

Thermistors are classified by the sign of their temperature coefficient: NTC (negative temperature coefficient), whose resistance falls as the temperature rises, and PTC (positive temperature coefficient), whose resistance rises with temperature.

So there are two types.

Thermistor is special kind of resistor named as:

Alaser resistor
Bphoto resistor
Cthermal resistorAnswer
Delectric resistor

Why

A thermistor is short for thermal resistor, a resistor that is sensitive to heat (the book's definition).

B (photo resistor) is an LDR, which is sensitive to light and not to temperature.

Short questionsp. 217

11.13 marks

Why it is not possible to measure the drift speed for electron by timing their travel along a conductor?

Solution

A single electron does not travel steadily along the wire. Its thermal motion is random, with speed of the order of 10^{6}\,\mathrm{m\,s^{-1}}, and it changes direction at every collision with the atoms and other electrons.

The net progress along the wire, the drift velocity, is only about 10^{-3}\,\mathrm{m\,s^{-1}}. This small drift is hidden inside the much larger random motion, so the path of one electron cannot be followed or timed.

Electrons are identical and billions of them move at once, so the drift velocity is found only as an average from the current, v=\dfrac{I}{neA} (equation 11.1), and not by timing one electron.

11.23 marks

What is the difference between e.m.f. and potential difference?

Solution

The emf of a source is the work done by the source per unit charge in driving charge round the complete circuit, \varepsilon=\dfrac{W}{Q}. It is the cause of the current. The potential difference is the work per unit charge between two points of the external circuit, V=\dfrac{W}{Q}, and it is the effect of that emf.

Feature	emf	Potential difference
Energy	Non-electrical energy converted into electrical energy in the source	Electrical energy converted into other forms in a component
Relation	V=\varepsilon-Ir for a source with internal resistance r	Equal to \varepsilon when no current flows
Measured	By a potentiometer (no current drawn at balance)	By a voltmeter across the component

Both are measured in volts, so the units do not distinguish them; the difference lies in the energy transfer.

11.33 marks

Why we use Kirchhoff's law for circuit problems solution?

Solution

Many circuits have several loops and several emf sources. Such a circuit cannot be reduced to a single equivalent resistor by simply combining series and parallel resistors, and Ohm's law alone is not enough to find the branch currents.

Kirchhoff's two laws give one equation for each junction and each loop. Solving these equations together gives all the unknown currents, so Kirchhoff's rules are used instead of Ohm's law for complex circuits.

11.43 marks

The Kirchhoff's current rule is based on conservation of charge. Explain how?

Solution

Charge can be neither created nor destroyed. Consider a junction: in any time interval the charge entering the junction must equal the charge leaving it, so the charge does not accumulate at the junction.

With currents entering taken as positive and currents leaving as negative, this gives Kirchhoff's first rule:

\sum I=I_1+I_2+I_3-I_4-I_5=0

This is the same statement as conservation of charge, written as a rule for currents.

11.53 marks

While analyzing a circuit, the internal resistance of e.m.f. sources is ignored. Why?

Solution

The internal resistance r of a source is usually very small compared with the external resistances, so the potential drop Ir inside the source is small. The terminal potential difference V=\varepsilon-Ir is then almost equal to the emf.

For example, the book gives about 0.15\,\Omega for a fresh alkaline AA cell at room temperature, which is very small compared with the resistors in most circuit problems. Ignoring r therefore makes the working simpler with little error.

11.63 marks

Why rise in temperature of a conductor, with positive temperature coefficient, is accompanied by a rise in the resistance?

Solution

The resistance of a conductor changes with temperature by R_t=R_0[1+\alpha(T-T_0)]. For a metal the temperature coefficient \alpha is greater than zero, so a rise in temperature T-T_0 makes R_t greater than R_0.

Physically, as the temperature rises the atoms vibrate more strongly, so the free electrons collide with them more often and drift less easily. The electrons lose more energy in these collisions, so the same potential difference produces a smaller current and the resistance rises.

11.72025 Annual I, Set D, Q.2 (xiv)3 marks

Under what circumstances can the terminal potential difference of a battery exceed its e.m.f.?

Solution

The terminal potential difference is V=\varepsilon-Ir. It can exceed the emf only if I is negative, that is, if the current is forced through the battery against its emf.

This happens when the battery is being charged by another source of larger emf. The current then enters the positive terminal, and the relation becomes V=\varepsilon+Ir, which is greater than \varepsilon.

11.83 marks

What is the working principle of Galvanometer?

Solution

A galvanometer is a sensitive instrument for detecting a small current. The current flows through a coil placed in a magnetic field, and the magnetic torque on the coil turns it. The pointer deflects in proportion to the current.

In a null method the galvanometer is used only to detect zero current. When it shows no deflection, no current passes through it, and this is the balance point used in the Wheatstone bridge and the potentiometer.

11.93 marks

How inspectors can easily check the reliability of a concrete bridge with carbon fibres?

Solution

Carbon fibres conduct electricity, and the optimum amount of carbon fibre in concrete is 0.3 % by volume. Because the fibres conduct, the concrete can be tested electrically.

An inspector connects electrodes to the concrete and measures its electric resistance over time. If the resistance increases over time, the fibres are separating because of cracks, so the bridge is no longer reliable. A steady resistance shows that the fibre network is still continuous.

11.103 marks

Could electronic devices charge themselves without being plugged into an electricity source?

Solution

No. Charging a device means moving charge through it against the electric field, which needs energy. Energy is supplied by an emf source, which converts non-electrical (for example chemical) energy into electrical energy, measured as work per unit charge.

A device cannot draw this energy from nothing. Without an external source of energy it cannot gain electrical energy, so it cannot charge itself.

Note. The book prints the number 11.10 twice. Both questions are kept as printed; this one is the question on electronic devices, and the next one (short-q11) is the second 11.10.

11.103 marks

Does a source of electricity ever run out of electrons?

Solution

No. The source does not supply electrons; it supplies energy. The free electrons are already present in the conductor, with a density of about n=8.5\times10^{28}\,\mathrm{m^{-3}} in copper, and they simply circulate round the circuit.

The emf source converts chemical energy into electrical energy, which drives the electrons round the circuit. Charge is conserved, so the same electrons go round the loop again and again and none is used up.

Note. The second question printed as 11.10 (the first is short-q10).

11.113 marks

Two wires have equal length, one is made of copper and the other of manganin and they have the same resistance. Which wire here will be thicker between these two given wires?

Solution

For wires of the same length and the same resistance, R=\rho\dfrac{L}{A} gives A=\dfrac{\rho L}{R}. The cross-section is therefore proportional to the resistivity: the wire with the larger \rho must be the thicker one.

Manganin has a much higher resistivity than copper, about 30 times as large. So the manganin wire is the thicker of the two, with about 30 times the cross-sectional area of the copper wire.

Comprehensive questionsp. 217–218

11.12026 Annual I, Set B, Q.2 (ix)2026 Annual I, Set D, Q.2 (xii)

What is LDR? Discuss in detail.

Solution

What an LDR is

A light dependent resistor (LDR), also called a photo-resistor, is a resistor whose resistance depends on the intensity of the light falling on it. In light the resistance is low, and in darkness it is high. Its resistance decreases as the light intensity increases.

In the dark the resistance of an LDR reaches its maximum, which is in megaohms (\mathrm{M\Omega}). In a lighted environment it falls to a few hundred ohms.

Working principle

The principle is photoconductivity: the conductance of the material increases when light falls on it. The conducting behaviour is explained by energy band theory. When the incident light intensity falls, the material reduces its conductance, so the resistance rises.

The inverse behaviour is shown in the graph below. Resistance is high at low illumination (dark) and falls steadily as the illumination increases.

Figure 11.5 (b): Resistance falls as the light intensity rises (schematic of the inverse behaviour).

Types of LDR

Intrinsic LDR: made of pure semiconducting material such as germanium or silicon.
Extrinsic LDR: made from semiconductors modified by doping. It is mostly used for longer wavelengths and cannot be used for light of shorter wavelength.

Uses and characteristics

An LDR is sensitive to light in the wavelength range of about 4\times10^{3} to 10^{4} angstrom. Because of its sharp resistance recovery rate and small size, it is used in security systems based on light detection, street light control, light switching for sunset and sunrise, and light meters.

LDRs are also part of automation systems. They reduce human effort and save electricity, because the light is switched off automatically when it is not needed.

11.2

Explain the use of thermistors and light dependent resistors in potential dividers.

Solution

Potential divider

A potential divider takes a voltage across two resistors in series and gives a fraction of it as the output. If one of the resistors is a thermistor or an LDR, its resistance changes with temperature or light, so the output voltage changes with them.

LDR in a potential divider

Figure 11.5 (c): LDR based potential divider circuit.

The LDR is the lower arm and the output is taken across it. In light its resistance falls, so its share of the total voltage falls and the output drops. In darkness the resistance rises and the output rises. The circuit therefore gives a potential difference that depends on the light intensity.

Thermistor in a potential divider

A thermistor is a heat sensitive resistor. The book's negative temperature coefficient (NTC) thermistor has a resistance that falls as its temperature rises. If it is the lower arm of the divider, the output voltage falls when it is heated, and rises when it cools.

A positive temperature coefficient (PTC) thermistor works the other way round, because its resistance rises with temperature. Choosing the right type gives the required change of voltage with temperature.

Why this is useful

The output is a potential difference that depends on temperature or on light intensity. It can be used for temperature compensation, control of circuits and automatic switching, which are the practical applications the book gives.

11.3

Explain the following terms. (A) Drift velocity, (B) emf, (C) internal resistance.

Solution

(A) Drift velocity

The drift velocity is the average velocity with which the free electrons drift towards the positive terminal in the presence of an electric field. Without a field the electrons have random motion and no net drift. The drift velocity is very small, about 10^{-3}\,\mathrm{m\,s^{-1}}, while the random speed is about 10^{6}\,\mathrm{m\,s^{-1}}.

It is related to the current by:

I=neAv

where n is the number of free electrons per unit volume, e the charge of one electron and A the cross-section. The drift velocity is v=\mu E with the mobility \mu, and the electrons drift against the field.

(B) emf

The electromotive force (emf) of a source is the work done per unit charge in converting non-electrical energy into electrical energy and driving charge round a complete circuit:

\varepsilon=\dfrac{W}{Q}

Its unit is the volt. It is not a force, although it is measured as work per unit charge; the source is the device that maintains the potential difference across the circuit.

(C) Internal resistance

The internal resistance r is the resistance of the source itself. When a current I flows through the source, a potential drop Ir occurs inside it, so the terminal potential difference is:

V=\varepsilon-Ir

The terminal potential difference therefore falls as the current increases. A source with a large internal resistance gives a low terminal potential difference when it is loaded.

11.4

What is potential divider? Explain.

Solution

Definition

A potential divider is a circuit in which a larger voltage is divided by a fixed ratio across two or more resistors in series, to give a smaller output voltage. The output is taken across one of the resistors.

Output voltage of a potential divider

Derivation Output voltage of a two resistor potential divider

Aim. Find the output voltage V_{\text{out}} across R_2 when the input voltage is V across R_1 and R_2 in series.

Assumptions.

The two resistors are in series, so the same current I flows through both.
The source has no internal resistance, so the full voltage V appears across the two resistors.

The output is taken across R2.

Both resistors are in series, so the total resistance is R_1+R_2 and Ohm's law gives the current.
I=\dfrac{V}{R_1+R_2}
The output is the potential difference across R_2, given by Ohm's law.
V_{\text{out}}=IR_2
Substitute the current from step 1.
V_{\text{out}}=\dfrac{V}{R_1+R_2}\,R_2
Simplify.
V_{\text{out}}=V\,\dfrac{R_2}{R_1+R_2}

ResultV_{\text{out}}=V\,\dfrac{R_2}{R_1+R_2}

The fraction R_2/(R_1+R_2) is less than one, so the output is always smaller than the input. If one resistor is an LDR or a thermistor, changing its resistance changes the output voltage.

Use

A thermistor or an LDR in place of R_2 (or R_1) gives an output voltage that depends on temperature or light intensity, as described for the thermistor and the LDR in the book.

11.5

What is Wheatstone bridge? Explain in detail.

Solution

Definition

A Wheatstone bridge is a circuit of four resistors and a galvanometer, with a battery across two opposite corners. The galvanometer is connected between the other two corners. It is used with the null method to find an unknown resistance.

Balance condition

The bridge is balanced when the galvanometer shows no deflection. No current then flows through the galvanometer, and the two corners it joins are at the same potential. The known resistors P and Q and the variable resistor S are adjusted until this happens; the unknown resistance is then R.

Figure 11.13: Wheatstone bridge circuit.

Derivation Balance condition of the Wheatstone bridge

Aim. Find the relation between P, Q, R and S when the bridge is balanced.

Assumptions.

Balanced: no current flows through the galvanometer, so b and d are at the same potential.
The current through P is I_1 and through R is I_2. Because there is no current in the galvanometer, the current through Q is also I_1 and through S is also I_2.

The potential drop from a to b equals the drop from a to d, because b and d are at the same potential.
I_1P=I_2R\qquad(11.15)
Similarly the drop from b to c equals the drop from d to c.
I_1Q=I_2S\qquad(11.16)
Divide equation (11.15) by equation (11.16). The currents cancel.
\dfrac{I_1P}{I_1Q}=\dfrac{I_2R}{I_2S}\quad\Rightarrow\quad\dfrac{P}{Q}=\dfrac{R}{S}
Make R the subject.
R=\dfrac{PS}{Q}\qquad(11.17)

ResultR=\dfrac{PS}{Q}

If three resistances are known, the fourth can be calculated from the balance. The method compares the unknown with known resistances, so the galvanometer needs only to show zero.

11.6

What is a potentiometer? Explain.

Solution

Definition and principle

A potentiometer is a null type resistance network for measuring potential differences. Its principle is that an unknown emf or potential difference is measured by balancing it, wholly or in part, against a known potential difference.

A uniform wire AB is connected across a driver battery of potential V. Because the wire is uniform and carries the same current, the potential falls at a constant rate along it. The potential drop over a length \ell of the wire is therefore proportional to \ell.

Figure 11.14: Potentiometer (simplified, one comparison cell shown).

Comparing two emfs

Derivation Ratio of two emfs with a potentiometer

Aim. Compare an unknown emf \varepsilon_2 with a standard emf \varepsilon_1 and find \varepsilon_2.

Assumptions.

The wire is uniform, so the potential drop per unit length, k, is constant.
At balance the galvanometer shows zero, so no current flows in the comparison cell branch.

The standard cell is balanced at length \ell_1, so its emf equals the potential drop over \ell_1.
\varepsilon_1=k\,\ell_1\qquad(11.18)
Replace the standard cell with the unknown cell and find the new balance length \ell_2.
\varepsilon_2=k\,\ell_2\qquad(11.19)
Divide equation (11.19) by equation (11.18). The gradient k cancels.
\dfrac{\varepsilon_2}{\varepsilon_1}=\dfrac{\ell_2}{\ell_1}
Make \varepsilon_2 the subject.
\varepsilon_2=\left(\dfrac{\ell_2}{\ell_1}\right)\varepsilon_1\qquad(11.20)

Result\varepsilon_2=\left(\dfrac{\ell_2}{\ell_1}\right)\varepsilon_1

The constant gradient cancels in the ratio, so the emf of the unknown cell is found from two lengths and the known emf of the standard cell.

Uses and advantage

Determination of electromotive force.
Measurement of internal resistance.
Comparison of the potential of two emf sources.
Better output in measuring potential difference than a voltmeter.

A potentiometer gives a more accurate reading than a voltmeter, because at balance it draws no current from the source. A voltmeter draws a current and so reads a potential difference less than the emf.

11.7

Explain Kirchhoff's first law and describe that it is a consequence of conservation of charge.

Solution

Statement

Kirchhoff's first law (current law, KCL): the currents entering a node are always equal to the currents leaving the node. Currents entering are taken as positive and currents leaving as negative, so the algebraic sum of the currents at a node is zero:

\sum I=0

Figure 11.8: Kirchhoff's first law.

Consequence of conservation of charge

Derivation KCL from conservation of charge

Aim. Show that the sum of the currents at a junction is zero.

Assumptions.

Charge cannot be created or destroyed at the junction.
The junction is a point, so no charge is stored there in a steady state.

In a time \Delta t the charge entering the junction is the sum of the entering currents times \Delta t.
\Delta Q_{\text{in}}=(I_1+I_2)\,\Delta t
In the same time the charge leaving is the sum of the leaving currents times \Delta t.
\Delta Q_{\text{out}}=(I_3+I_4)\,\Delta t
Conservation of charge: the charge in equals the charge out.
\Delta Q_{\text{in}}=\Delta Q_{\text{out}}
Divide by \Delta t and bring all terms to one side.
I_1+I_2-I_3-I_4=0

Result\sum I=0

The charge that enters a junction must leave it in the same time. So KCL is a specific way of stating the law of conservation of charge.

11.8

Explain Kirchhoff's second law and describe that it is a consequence of conservation of energy.

Solution

Statement

Kirchhoff's second law (voltage law, KVL): the algebraic sum of all the potential changes around any closed loop is zero. Going round the loop, a rise in potential is positive and a drop is negative. For a loop with emf sources and resistors:

\sum V=0

Figure 11.9 (b): A simple loop.

Consequence of conservation of energy

Derivation KVL from conservation of energy

Aim. Find the current in the loop and show that the energy balance gives \sum V=0.

Assumptions.

The loop has no junctions, so the same current I flows everywhere in it.
Voltage is energy per unit charge, V=\Delta U/q.

Go round the loop clockwise, starting at \varepsilon_1. Passing through \varepsilon_1 from its negative to its positive terminal is a rise. Passing through \varepsilon_2 is a drop, because it opposes the current.
+\varepsilon_1-\varepsilon_2
Each resistance, including the internal ones, causes a drop IR in the direction of the current.
-I\,r_1-I\,R-I\,r_2
Add the terms and set the sum to zero.
\varepsilon_1-\varepsilon_2-I\,(r_1+R+r_2)=0
Solve for the current.
I=\dfrac{\varepsilon_1-\varepsilon_2}{R+r_1+r_2}

Result\varepsilon_1-\varepsilon_2-I\,(r_1+R+r_2)=0

The energy per unit charge gained from the sources, \varepsilon_1-\varepsilon_2, equals the energy per unit charge lost in the resistances, I(r_1+R+r_2). This is conservation of energy, so KVL is a specific way of stating it.

11.9

Using Kirchhoff's laws, derive a formula for the combined resistance of two or more resistors in series.

Solution

Result

For resistors in series the same current flows through each, and the total potential difference is shared among them. The equivalent resistance is the sum of the individual resistances.

Derivation Combined resistance of resistors in series

Aim. Find the equivalent resistance R of R_1 and R_2 in series.

Assumptions.

There are no junctions between R_1 and R_2, so the same current I flows through both (Kirchhoff's first law).
The potential difference V is across the whole combination.

Figure 11.11: Resistors connected in series.

Kirchhoff's second law: the total potential difference across the closed loop equals the sum of the potential differences across the two resistors.
V=V_1+V_2\qquad(11.7)
Ohm's law for each resistor.
IR=IR_1+IR_2\qquad(11.8)
Divide by the current I, which is common to all the resistors.
R=R_1+R_2
For several resistors connected in series the same method gives the sum of all of them.
R=R_1+R_2+R_3+\cdots\qquad(11.9)

ResultR=R_1+R_2+R_3+\cdots

The series combination is always larger than the largest individual resistor.

11.10

Using Kirchhoff's laws, derive a formula for the combined resistance of two or more resistors in parallel.

Solution

Result

For resistors in parallel the total current divides between the branches, and each branch has the same potential difference. The reciprocal of the equivalent resistance is the sum of the reciprocals.

Derivation Combined resistance of resistors in parallel

Aim. Find the equivalent resistance R of R_1 and R_2 in parallel.

Assumptions.

The current from the source splits at the junctions, so the total current I is shared by the two branches.
Each branch is connected between the same two junctions, so both have the same potential difference V.

Figure 11.12: Resistors connected in parallel.

Kirchhoff's first law at the junction: the total current is the sum of the branch currents.
I=I_1+I_2\qquad(11.10)
Kirchhoff's second law: the potential difference across each branch is the same, and equal to the potential difference across R.
V=V_1=V_2\qquad(11.11)
Ohm's law for the total and for each branch.
\dfrac{V}{R}=\dfrac{V}{R_1}+\dfrac{V}{R_2}\qquad(11.12)
Cancel the common potential difference V.
\dfrac{1}{R}=\dfrac{1}{R_1}+\dfrac{1}{R_2}\qquad(11.13)
For several resistors in parallel the same method gives the sum of the reciprocals.
\dfrac{1}{R}=\dfrac{1}{R_1}+\dfrac{1}{R_2}+\dfrac{1}{R_3}+\cdots\qquad(11.14)

Result\dfrac{1}{R}=\dfrac{1}{R_1}+\dfrac{1}{R_2}+\dfrac{1}{R_3}+\cdots

The equivalent resistance is smaller than the smallest individual resistance, because each extra branch gives the current another path.

11.11

Explain the electronic current in a metallic wire due to the drift of free electrons in the wire.

Solution

Free electrons in a metal

A metal has a very large number of free electrons per unit volume, n\approx10^{28}\,\mathrm{m^{-3}}. Without an external field they move randomly at high speed, so there is no net current. When a potential difference is applied across the wire, the electric field makes the electrons drift steadily in one direction on top of this random motion.

Figure 11.1 (b): Motion of electrons in the presence of an electric field.

Current from the drift of electrons

Derivation Current from the drift velocity

Aim. Show that the current in the wire is I=neAv.

Assumptions.

The wire has uniform cross-section A and free electron density n.
The drift velocity v is the same for all electrons.

In time t every electron within a distance vt of a section crosses that section. These electrons fill a cylinder of length vt.
V_{\text{cyl}}=A\,v\,t
The number of free electrons in the cylinder is the density times the volume.
N=n\,A\,v\,t
Each electron carries charge e, so the charge crossing the section is Q=Ne.
Q=n\,e\,A\,v\,t
The current is the charge per unit time.
I=\dfrac{Q}{t}=n\,e\,A\,v

ResultI=neAv

The drift velocity is v=I/(neA). The electrons move against the electric field, so the conventional current, which is taken in the direction of positive charge, flows in the opposite direction to the electrons.

Numerical problemsp. 218

11.1

If 1\,\mathrm{A} current flows through a copper wire having 1\,\mathrm{cm^{2}} area of cross-section and 10\,\mathrm{km} length; calculate the time required by charge carrier to travel from one end to other end of the conductor (free electron density of copper is 8.5\times10^{28} per \mathrm{m^{3}}).

Solution

Given

I = 1\,\mathrm{A}
A = 1\,\mathrm{cm^{2}}=1\times10^{-4}\,\mathrm{m^{2}}
L = 10\,\mathrm{km}=1\times10^{4}\,\mathrm{m}
n = 8.5\times10^{28}\,\mathrm{m^{-3}}
e = 1.6\times10^{-19}\,\mathrm{C}

To find

t, the time to travel the length L

The drift velocity follows from I=neAv.
v=\dfrac{I}{neA}
Substitute the values.
v=\dfrac{1}{8.5\times10^{28}\times1\times10^{-4}\times1.6\times10^{-19}}=\dfrac{1}{1.36\times10^{6}}=7.35\times10^{-7}\,\mathrm{m\,s^{-1}}
The time is the length divided by the drift velocity.
t=\dfrac{L}{v}=\dfrac{1\times10^{4}}{7.35\times10^{-7}}=1.36\times10^{10}\,\mathrm{s}
Convert to years, using 1\,\mathrm{yr}=3.156\times10^{7}\,\mathrm{s}.
t=\dfrac{1.36\times10^{10}}{3.156\times10^{7}}\approx431\,\mathrm{years}

t\approx1.36\times10^{10}\,\mathrm{s}\approx431\,\mathrm{years}

Book answer. 431 years

Note. The drift velocity is very small, so one electron would need about four centuries to cross a 10 km wire. The book's answer agrees.

11.2

Calculate the flow of charges per unit time inside the emf source having potential 2\,\mathrm{V} and terminals are connected with each other having internal resistance 0.02\,\Omega.

Solution

Given

\varepsilon = 2\,\mathrm{V}
r = 0.02\,\Omega
R = 0\,\Omega\ (\text{terminals joined})

To find

I, the flow of charge per unit time

The flow of charge per unit time is the current. With the terminals joined the external resistance is zero, so only the internal resistance limits the current.
I=\dfrac{\varepsilon}{R+r}=\dfrac{\varepsilon}{r}
Substitute.
I=\dfrac{2\,\mathrm{V}}{0.02\,\Omega}=100\,\mathrm{A}

I=100\,\mathrm{A}

Book answer. 100 A

Note. The flow of charge is 100\,\mathrm{C} per second.

11.3

Calculate the terminal potential of the emf source if internal resistance of the battery is 10\,\Omega and current and internal potential are 1\,\mathrm{A} and 10\,\mathrm{V} respectively?

Solution

Given

r = 10\,\Omega
I = 1\,\mathrm{A}
\varepsilon = 10\,\mathrm{V}\ (\text{the internal potential, read as the emf})

To find

V, the terminal potential difference

The terminal potential difference of a source is the emf less the drop across its internal resistance.
V=\varepsilon-Ir
Substitute. The drop across r is Ir=1\times10=10\,\mathrm{V}.
V=10\,\mathrm{V}-1\,\mathrm{A}\times10\,\Omega=0\,\mathrm{V}

V=0\,\mathrm{V}

Book answer. 0 V

Note. The question says the internal potential is 10\,\mathrm{V}. This is read as the emf, which is the only reading that gives the book's answer of 0\,\mathrm{V}. The drop across the internal resistance is Ir=10\,\mathrm{V} in any case, so the terminal potential difference is zero.

11.42026 Annual I, Set B, Q.5 OR

There is a copper coil having 2000 number of turns with 0.8\,\mathrm{mm^{2}} cross-section are, the length per turn is 80\,\mathrm{cm}. Calculate the resistance of the coil? (Resistivity of copper is 0.02\,\mu\Omega\,\mathrm{m})?

Solution

Given

N = 2000
A = 0.8\,\mathrm{mm^{2}}=0.8\times10^{-6}\,\mathrm{m^{2}}
l_{\text{turn}} = 80\,\mathrm{cm}=0.80\,\mathrm{m}
\rho = 0.02\,\mu\Omega\,\mathrm{m}=2\times10^{-8}\,\Omega\,\mathrm{m}

To find

R, the resistance of the coil

The length of the wire is the number of turns times the length per turn.
L=N\,l_{\text{turn}}=2000\times0.80\,\mathrm{m}=1600\,\mathrm{m}
The resistance of a conductor is R=\rho\dfrac{L}{A}.
R=\rho\dfrac{L}{A}
Substitute the values.
R=\dfrac{2\times10^{-8}\times1600}{0.8\times10^{-6}}=\dfrac{3.2\times10^{-5}}{0.8\times10^{-6}}=40\,\Omega

R=40\,\Omega

Book answer. 40 Ω

Note. The resistivity is taken as given in the question, 0.02\,\mu\Omega\,\mathrm{m}. The book's Table 11.4 misprints the copper value (see the report).

11.5

A piece of silver wire has a resistance of 1\,\Omega. What will be the resistance of magnum wire of one third of the length and one third of diameter, if the specific resistance of wire is 30 times that of silver?

Solution

Given

R = 1\,\Omega\ (\text{silver wire, length }L\text{, diameter }d)
L' = \dfrac{L}{3}
d' = \dfrac{d}{3}
\rho' = 30\,\rho

To find

R', the resistance of the magnum wire

The cross-section of the magnum wire is A'=\dfrac{\pi d'^{2}}{4}=\dfrac{\pi(d/3)^{2}}{4}=\dfrac{A}{9}.
A'=\dfrac{A}{9}
Use R'=\rho'\dfrac{L'}{A'} with \rho'=30\rho.
R'=30\rho\,\dfrac{L/3}{A/9}
Simplify, using R=\rho L/A=1\,\Omega.
R'=30\times3\times\dfrac{\rho L}{A}=90\,\dfrac{\rho L}{A}=90\,\Omega

R'=90\,\Omega

Book answer. 90 Ω

11.6

Calculate the current that flows in the 1\,\Omega resistor in the following circuit.

The long plate of each cell is its positive terminal.

Solution

Given

\varepsilon_{AE} = 6\,\mathrm{V}\ (\text{positive terminal at }A)
\varepsilon_{CB} = 9\,\mathrm{V}\ (\text{positive terminal at }B)
R_{EB} = 2\,\Omega
R_{CF} = 3\,\Omega
R_{FD} = 3\,\Omega
R_{EF} = 1\,\Omega

To find

I in the 1\,\Omega resistor

Two loops are used, with currents I_1 round A\to D\to F\to E\to A and I_2 round E\to F\to C\to B\to E. Both are taken as positive in these directions. The current in the 1\,\Omega resistor from F to E is I_1-I_2.
Loop 1 (start at A): the 6\,\mathrm{V} cell is a rise of 6\,\mathrm{V} from E to A. The 3\,\Omega and the 1\,\Omega resistors are drops.
6-3I_1-1\,(I_1-I_2)=0\quad\Rightarrow\quad4I_1-I_2=6
Loop 2 (start at E): from E to F is a rise of I_1-I_2. The 3\,\Omega and 2\,\Omega resistors are drops, and the 9\,\mathrm{V} cell is a rise from C to B.
(I_1-I_2)-3I_2+9-2I_2=0\quad\Rightarrow\quad I_1-6I_2+9=0
Use the second equation to write I_1 in terms of I_2, and put it in the first.
I_1=6I_2-9\quad\Rightarrow\quad4(6I_2-9)-I_2=6\quad\Rightarrow\quad23I_2=42
Solve for the two currents.
I_2=\dfrac{42}{23}=1.83\,\mathrm{A},\qquad I_1=\dfrac{45}{23}=1.96\,\mathrm{A}
The current in the 1\,\Omega resistor is the difference.
I=I_1-I_2=\dfrac{3}{23}=0.13\,\mathrm{A}

I\approx0.13\,\mathrm{A}\ (\text{from }F\text{ to }E)

Book answer. 0.13 A

Note. Both loop currents come out positive, so the current in the 1\,\Omega resistor flows upwards from F to E. The polarities are read from the symbols in the figure (long plate positive). The result agrees with the book.

FBISE 2025 and 2026 questions

Questions from the FBISE HSSC-I 2026 (1st Annual) and FBISE HSSC-I 2025 (1st Annual) papers that belong to this unit, answered the way an examiner expects.

2026 Annual I, Set B, MCQ 111 mark

A potentiometer is used to compare:

AResistance
BCurrent
CPotential differenceAnswer
DPower

Why

A potentiometer is a null device. It balances an unknown potential difference (or emf) against a known one, so it is used to compare potential differences, for example the emfs of two cells, using \varepsilon_2/\varepsilon_1=\ell_2/\ell_1.

Option A is wrong because the comparison of resistances is done with a Wheatstone bridge. Current and power are not compared by a potentiometer.

2025 Annual I, Set B, MCQ 121 mark

In order to achieve high accuracy, the slide wire of a potentiometer should be:

AAs long as possibleAnswer
BAs short as possible
CVery thick
DNeither too small, nor too large

Why

A longer slide wire spreads the same potential difference over a greater length, so the potential gradient is smaller. The balance point can then be located more precisely, which gives higher accuracy. The answer is A.

A short wire (B) gives a larger gradient and a coarser balance point. Thickness (C) does not improve the sensitivity of the balance point, and there is no best 'medium length' (D): the longer the wire, the better.

2026 Annual I, Set B, Q.2 (v)3 marks

Calculate the resistance of wire 10\,\mathrm{m} long that has diameter of 3\,\mathrm{mm} and resistivity of 2.63\times10^{-2}\,\Omega\mathrm{m}.

Solution

Given

L = 10\,\mathrm{m}
d = 3\,\mathrm{mm}=0.003\,\mathrm{m}
\rho = 2.63\times10^{-2}\,\Omega\,\mathrm{m}

To find

Formula (1 mark). The resistance of a wire is R=\rho\dfrac{L}{A}. The cross-section is a circle of diameter d, so A=\dfrac{\pi d^{2}}{4}.
R=\rho\dfrac{L}{A}
Area (1 mark).
A=\dfrac{\pi(0.003)^{2}}{4}=7.07\times10^{-6}\,\mathrm{m^{2}}
Substitute and answer with unit (1 mark).
R=\dfrac{2.63\times10^{-2}\times10}{7.07\times10^{-6}}=3.72\times10^{4}\,\Omega

R\approx3.72\times10^{4}\,\Omega

2025 Annual I, Set B, Q.2 (vi)3 marks

Calculate KE in electron volt and joules gained by an electron moving through a potential difference of 2 volts.

Solution

Given

q = e=1.6\times10^{-19}\,\mathrm{C}
V = 2\,\mathrm{V}

To find

kinetic energy in eV and in J

Energy gained (1 mark). The work done on the charge by the potential difference is qV, and an electron starting from rest gains it as kinetic energy.
K=qV=eV
In electron volts (1 mark). One electron volt is the energy gained by an electron through 1\,\mathrm{V}.
K=e\times2\,\mathrm{V}=2\,\mathrm{eV}
In joules (1 mark). Use 1\,\mathrm{eV}=1.6\times10^{-19}\,\mathrm{J}.
K=2\times1.6\times10^{-19}=3.2\times10^{-19}\,\mathrm{J}

K=2\,\mathrm{eV}=3.2\times10^{-19}\,\mathrm{J}

2025 Annual I, Set B, Q.2 (vii)3 marks

Why does the resistance of LDR decrease with increase in intensity of light on it?

Solution

An LDR works on photoconductivity: the conductance of its semiconductor material increases when light falls on it. Light of greater intensity delivers more energy to the material, so more electrons are freed from their bonds and become available to carry current (this is explained by the energy band theory).

With more charge carriers the conductance rises, and since R=1/G the resistance falls as the intensity of the light increases. In the dark there are few free carriers, so the resistance is high, of the order of megohms.

2026 Annual I, Set B, Q.2 (ix)2+1 = 3 marks

Briefly explain light dependent resistor (LDR). Give one application of LDR.

Solution

LDR (2 marks). A light-dependent resistor (photo-resistor) is a resistor whose resistance depends on the intensity of the light falling on it. It works on the principle of photoconductivity: when the light intensity increases the resistance decreases, and when the light is reduced the resistance increases. In the dark its resistance is of the order of megohms, and in bright light it falls to a few hundred ohms.

Application (1 mark). Street-light control: an LDR in a potential divider switches the lamp on automatically at sunset and off at sunrise. (Other uses are security systems based on light detection and light meters.)

2026 Annual I, Set B, Q.5 OR3+3 = 6 marks

a. State Kirchoff's laws for current and voltage. b. There is a copper coil having 2000 turns with 0.8\,\mathrm{mm^2} cross-section area, the length per turn is 80\,\mathrm{cm}. Calculate the resistance of the coil.

Solution

(a) Kirchhoff's laws (3 marks)

Kirchhoff's first law (current law). The currents entering a node are equal to the currents leaving it. Taking entering currents as positive and leaving currents as negative, the algebraic sum of the currents at a node is zero:

\sum I=0

It is a consequence of the conservation of charge: the charge that enters a junction in a given time must leave it in the same time.

Kirchhoff's second law (voltage law). The algebraic sum of all the potential changes around any closed loop is zero. A rise in potential is taken as positive and a drop as negative:

\sum V=0

It is a consequence of the conservation of energy: the energy per unit charge gained from the sources equals the energy per unit charge lost in the resistances.

(b) Resistance of the coil (3 marks)

Given

N = 2000
A = 0.8\,\mathrm{mm^{2}}=0.8\times10^{-6}\,\mathrm{m^{2}}
l_{\text{turn}} = 80\,\mathrm{cm}=0.80\,\mathrm{m}
\rho = 2\times10^{-8}\,\Omega\,\mathrm{m} copper, as in the book's exercise

To find

The length of the wire is the number of turns times the length per turn.
L=N\,l_{\text{turn}}=2000\times0.80=1600\,\mathrm{m}
Use R=\rho\dfrac{L}{A} and substitute.
R=\dfrac{2\times10^{-8}\times1600}{0.8\times10^{-6}}=40\,\Omega

R=40\,\Omega

2025 Annual I, Set B, Q.6 OR6 marks

Find the current flowing through each resistor of the circuit.

Solution

The cells are taken as ideal (no internal resistance), because none is given. Both cells have their long plate, the positive terminal, at the top. Three branches join the top and bottom wires, so choose a current in each: I_1 upward through the 10\,\mathrm{V} cell, I_2 downward through the 2\,\Omega resistor, and I_3 downward through the 6\,\mathrm{V} cell and the 1\,\Omega resistor.

The circuit of the question with the currents chosen for the working. The long plate of each cell is its positive terminal, so both cells have their positive terminal at the top.

Working

KCL at the top junction. The current entering it is I_1; the currents leaving it are I_2 and I_3.
I_1=I_2+I_3
KVL, left loop (through the 10\,\mathrm{V} cell upward, a rise of 10\,\mathrm{V}, then down through the 2\,\Omega resistor, a drop 2I_2).
10-2I_2=0\;\Rightarrow\;I_2=5\,\mathrm{A}
KVL, right loop (down through the 2\,\Omega resistor, a drop 2I_2; then along the 1\,\Omega resistor against I_3, a rise 1\cdot I_3; then up through the 6\,\mathrm{V} cell, a rise of 6\,\mathrm{V}).
-2I_2+I_3+6=0\;\Rightarrow\;I_3=2(5)-6=4\,\mathrm{A}
**Current in the 10\,\mathrm{V} cell** from KCL.
I_1=5+4=9\,\mathrm{A}

Current in each resistor: 2\,\Omega resistor, 5\,\mathrm{A} (downward); 1\,\Omega resistor, 4\,\mathrm{A} (in the direction of I_3, towards the junction at the bottom of the middle branch). The 10\,\mathrm{V} cell supplies 9\,\mathrm{A}.

I_{2\Omega}=5\,\mathrm{A},\ I_{1\Omega}=4\,\mathrm{A}

2026 Annual I, Set D, MCQ 121 mark

At balance point, current in galvanometer is:

AZeroAnswer
BMaximum
CMinimum
DInfinite

Why

At the balance point the potentials at the two ends of the galvanometer are equal, so there is no potential difference across it and no current flows. The galvanometer reads zero, which is why the method is called the null method.

Option B (maximum) describes the galvanometer when the bridge is far from balance. A reading of minimum or infinite current would not give a sharp balance point.

2025 Annual I, Set D, MCQ 131 mark

A 100 ohm resistor is connected across the terminals of a 10 V battery. The power dissipation of the resistor is:

A1 wattAnswer
B10 watts
C100 watts
D1000 watts

Why

The resistor is across the battery, so V=10\,\mathrm{V} and P=\dfrac{V^{2}}{R}=\dfrac{(10)^{2}}{100}=1\,\mathrm{W}. The same answer comes from I=V/R=0.1\,\mathrm{A} and P=VI=1\,\mathrm{W}.

Option C (100\,\mathrm{W}) is V^{2} itself, found by forgetting to divide by R. Option B (10\,\mathrm{W}) would be the answer for a 10\,\Omega resistor.

2026 Annual I, Set D, Q.2 (v)3 marks

Calculate the resistance of a wire 10\,\mathrm{m} long that has diameter of 2\,\mathrm{mm} and resistivity of 2.63\times10^{-2}\,\Omega\mathrm{m}.

Solution

Given

L = 10\,\mathrm{m}
d = 2\,\mathrm{mm}=0.002\,\mathrm{m}
\rho = 2.63\times10^{-2}\,\Omega\,\mathrm{m}

To find

Formula (1 mark). The resistance of a wire is R=\rho\dfrac{L}{A}. The cross-section is a circle of diameter d, so A=\dfrac{\pi d^{2}}{4}.
R=\rho\dfrac{L}{A}
Area (1 mark).
A=\dfrac{\pi(0.002)^{2}}{4}=3.14\times10^{-6}\,\mathrm{m^{2}}
Substitute and answer with unit (1 mark).
R=\dfrac{2.63\times10^{-2}\times10}{3.14\times10^{-6}}=8.37\times10^{4}\,\Omega

R\approx8.37\times10^{4}\,\Omega

2025 Annual I, Set D, Q.2 (ix)3 marks

What is meant by null method of measurement? Enlist two applications.

Solution

Null method (1 mark). In the null method of measurement the quantity to be measured is balanced against an opposing known quantity, which is varied until the resultant of the two is zero. A galvanometer shows the null (zero) reading at balance. Because only a zero reading has to be recognised, the galvanometer need not be calibrated, and no current is drawn from the circuit being measured at the balance point.

Two applications (2 marks).

Wheatstone bridge: finds an unknown resistance by balancing the bridge until the galvanometer reads zero.
Potentiometer: finds or compares potentials, for example the emfs of two cells, by balancing against the potential drop along a wire.

2026 Annual I, Set D, Q.2 (xii)1+2 = 3 marks

What is a light dependent resistor (LDR)? Why is LDR useful in automatic circuits?

Solution

LDR (1 mark). A light-dependent resistor is a resistor that works on photoconductivity: its resistance decreases as the light falling on it gets brighter, and increases as the light is reduced.

Use in automatic circuits (2 marks). The resistance of an LDR changes with the light level, so in a potential divider the output voltage changes with the light:

V_{2}=V\,\dfrac{R_{\mathrm{LDR}}}{R_{1}+R_{\mathrm{LDR}}}

In bright light R_{\mathrm{LDR}} is small and V_{2} is small; in the dark R_{\mathrm{LDR}} is large and V_{2} is close to V. This change of voltage can switch a circuit on or off with no one present, for example street lights that turn on at dusk and off at dawn. It reduces human effort and saves electricity, because the light is off when it is not needed.

Figure 11.5 (c): LDR-based potential divider circuit. The supply V is shared between the fixed resistor R_1 and the LDR; the output V_2 is taken across the LDR.

2025 Annual I, Set D, Q.2 (xiv)3 marks

Under what conditions can the terminal potential differences of a battery exceed its emf?

Solution

The terminal potential difference of a battery of emf \varepsilon and internal resistance r is V=\varepsilon-Ir. It can exceed the emf only if I is negative, that is, if current is forced through the battery against its emf.

This happens when the battery is being charged by another source of larger emf. The current then enters the positive terminal, and V=\varepsilon+Ir, which is greater than \varepsilon.

Exam focus

Apply Kirchhoff's laws to find the current in each resistor of a circuit with two cells. (long)
2025 Set B Q.6 OR (6 marks); 2026 Set B Q.5 OR (a) asks the statements of both laws; numerical question 6 of the book.
Go to answer
What is an LDR? Why does its resistance fall in light, and how is it used (potential divider, automatic circuits)? (long)
Asked in all three: 2026 Set B Q.2 (ix), 2026 Set D Q.2 (xii) and 2025 Set B Q.2 (vii); comprehensive questions 1 and 2.
Go to answer
Numerical: R=\rho L/A for a wire or a coil. (numerical)
2026 Set B and Set D Q.2 (v) (wire of given diameter) and 2026 Set B Q.5 OR (b) (coil of 2000 turns); numerical question 4 of the book.
Go to answer
Potentiometer: principle, comparison of emfs, and why a long slide wire is more accurate. (long)
2026 Set B MCQ 11 and 2025 Set B MCQ 12; MCQs 14 and 15 and comprehensive question 6 of the book.
Go to answer
Null method of measurement and the Wheatstone bridge (zero galvanometer current at balance). (short)
2025 Set D Q.2 (ix) and 2026 Set D MCQ 12; comprehensive question 5.
Go to answer
When can the terminal potential difference exceed the emf? (short)
2025 Set D Q.2 (xiv); short question 7 of the book.
Go to answer
Kinetic energy gained through a potential difference, in eV and joules. (numerical)
2025 Set B Q.2 (vi); uses 1\,\mathrm{eV}=1.6\times10^{-19}\,\mathrm{J}.
Go to answer
Power in a resistor, P=V^{2}/R. (mcq)
2025 Set D MCQ 13; MCQ 3 of the book is the same type.
Go to answer
Drift velocity, emf and internal resistance, and the series and parallel resistor derivations. (long)
Not in the 2025 or 2026 papers, but comprehensive questions 3, 9, 10 and 11 of the book and the unit SLOs.
Go to answer

11.1Drift velocityOften examined

Electricity, conductors and insulators

Drift velocity of electrons

Drift velocity and mobility

Drift velocity factors

Derivation Current in terms of drift velocity: I=neAv

11.2Electric potential

Electric potential energy

Relation between electric potential energy and electric potential

Derivation Potential in a uniform field: V=−Er

Potential due to point charges

The electron-volt

11.3Electromotive force and maximum power outputOften examined

Terminal potential difference and internal resistance

Derivation Terminal potential difference and the current in the circuit

Maximum power transfer

Derivation Maximum power transfer theorem

11.4Variation of resistance with temperature

Temperature and resistance

Resistance, length and cross-section area

Conductivity

Resistivity and resistance

11.5Light-dependent resistors (LDRs)

LDR in a potential divider

11.6Electric power and electric energy

Derivation Electric power: P=VI=I^{2}R=V^{2}/R

Electric energy

11.7Kirchhoff's lawsOften examined

11.7.1Kirchhoff's first law or Kirchhoff's current law (KCL)Often examined

Derivation KCL as a consequence of conservation of charge

11.7.2Kirchhoff's second law or Kirchhoff's voltage law (KVL)Often examined

Derivation KVL as a consequence of conservation of energy

11.7.3Applying Kirchhoff's lawsOften examined

11.7.3Deriving the equation for resistors in seriesOften examined

Derivation Equivalent resistance of resistors in series

11.7.3Deriving the equation for resistors in parallelOften examined

Derivation Equivalent resistance of resistors in parallel

11.8Null method of measurement

11.8.1Wheatstone bridgeOften examined

Derivation Balance condition of the Wheatstone bridge

11.8.2PotentiometerOften examined

Derivation Comparing two emfs with the potentiometer

11.9Thermistor

11.10Carbon fibers in concrete bridge

Checking the reliability of the bridge

Formula sheet

Key terms

Multiple choice questionsp. 215–217

Short questionsp. 217

Comprehensive questionsp. 217–218

What an LDR is

Working principle

Types of LDR

Uses and characteristics

Potential divider

LDR in a potential divider

Thermistor in a potential divider

Why this is useful

(A) Drift velocity

(B) emf

(C) Internal resistance

Definition

Output voltage of a potential divider

Derivation Output voltage of a two resistor potential divider

Use

Definition

Balance condition

Derivation Balance condition of the Wheatstone bridge

Definition and principle

Comparing two emfs

Derivation Ratio of two emfs with a potentiometer

Uses and advantage

Statement

Consequence of conservation of charge

Derivation KCL from conservation of charge

Statement

Consequence of conservation of energy

Derivation KVL from conservation of energy

Result

Derivation Combined resistance of resistors in series

Derivation Current in terms of drift velocity: $I = n e A v$

Derivation Potential in a uniform field: $V = - E r$