Alternating current

Class 12 Physics, unit 20, book pages 106–123

Alternating current (AC) keeps reversing its direction; its value follows $x = x_{0} sin ω t$ . This unit defines the quantities that describe it (period, frequency, peak and r.m.s. values, mean power), shows how diodes rectify AC into DC and how a capacitor smooths the output, and explains self and mutual induction. It then follows AC through a resistor, an inductor and a capacitor, where phase, reactance and phasor diagrams lead to the impedance of RL, RC and RLC series circuits.

Learning outcomes (13)

Use the terms period, frequency and peak value as applied to an alternating current or voltage.
Use equations of the form $x = x_{0} sin ω t$ representing a sinusoidally alternating current or voltage.
Use the fact that the mean power in a resistive load is half the maximum power for a sinusoidal alternating current.
Distinguish between root-mean-square (r.m.s.) and peak values, including stating and using $I_{rms} = I_{0} / 2$ and $V_{rms} = V_{0} / 2$ for a sinusoidal alternating current.
Distinguish graphically between half-wave and full-wave rectification.
Explain the use of a single diode for the half-wave rectification of an alternating current.
Explain the use of four diodes (bridge rectifier) for full-wave rectification of an alternating current.
Analyse the effect of a single capacitor in smoothing current flow, including the effect of the values of capacitance and the load resistance.
Define mutual inductance ( $M$ ) and self-inductance ( $L$ ), and their unit henry.
Describe the phase of AC and how phase lags and leads in AC circuits.
Identify inductors as important components of AC circuits termed chokes (devices which present a high resistance to alternating current).
Calculate the reactances of capacitors and inductors.
Describe impedance as the vector summation of resistances and reactances.

20.1Alternating current and voltageOften examined

A cell or a battery gives direct current (DC): charge flows round the circuit in one direction only. The mains supply in your home gives alternating current (AC): the current flows first in one direction, then in the opposite direction, and keeps changing its direction (polarity) at regular intervals of time. The direction of an alternating current at any instant depends on the polarity of the voltage at that instant.

AC powers our homes and industries because its voltage can be stepped up or down easily with a transformer. Electrical power is sent over long distances at a very high voltage and a small current, so very little energy is wasted as heat ( $I^{2} R$ ) in the transmission lines. Near our homes it is stepped down to 220 V for safe use.

Equation of an alternating current or voltage

An AC generator is a coil rotating with a constant angular velocity $ω$ in a uniform magnetic field. The emf it produces, and the current it drives, vary with time like a sine function:

x = x_{0} sin (ω t)

$x$ : the instantaneous value of the current or voltage, that is, its value at time $t$ .
$x_{0}$ : the peak value (maximum value) of the current or voltage.
$ω$ : the angular frequency of the generator, and hence of the current and voltage, in $rad s^{- 1}$ .
$ω t$ : the phase angle in radians. It tells how far the cycle has advanced at time $t$ .

I = I_{0} sin (ω t)

(20.1 a)

alternating current, peak value $I_{0}$

V = V_{0} sin (ω t)

(20.1 b)

alternating voltage, peak value $V_{0}$

One full turn of the coil ( $2 π$ radians) produces one complete cycle. So the angular frequency is linked to the frequency $f$ and the time period $T$ by

ω = 2 π f = \frac{2 π}{T}

A sinusoidal alternating quantity

\frac{π}{2}

π

\frac{3 π}{2}

2 π

x_{0}

- x_{0}

ω t

x

T (one cycle)

peak

zero crossing

x = x_{0} sin ω t

Fig 20.1 redrawn. The value rises from zero to the peak

x_{0}

, falls through zero to

- x_{0}

and returns to zero in one period

T

. The curve is steepest (changing fastest) at the zero crossings and flat (changing slowest) at the peaks.

The graph of voltage or current against time is a sine wave (Fig 20.1). It shows that a sinusoidal voltage or current:

changes its polarity (direction) after regular intervals;
changes its magnitude continuously;
changes smoothly from one polarity to the other;
changes most rapidly at the zero (crossover) point and most slowly at its peak.

Reading an AC equation

An alternating voltage is

V = 20 sin (157 t)

volt. Find its peak value, angular frequency, frequency and time period, and its value at

t = 0.01 s

Compare with $V = V_{0} sin ω t$ . The number in front of the sine is the peak value and the number multiplying $t$ is $ω$ .
$V_{0} = 20 V, ω = 157 rad s^{- 1}$
Frequency from $ω = 2 π f$ .
$f = \frac{ω}{2 π} = \frac{157}{2 \times 3.14} = 25 Hz$
Time period is the reciprocal of frequency.
$T = \frac{1}{f} = \frac{1}{25} = 0.04 s$
At $t = 0.01 s$ the phase angle is $ω t = 157 \times 0.01 = 1.57 rad = π /2$ (a quarter of a cycle). Use radian mode on the calculator.
$V = 20 sin (1.57) = 20 \times 1 = 20 V$

V_{0} = 20 V, ω = 157 rad s^{- 1}, f = 25 Hz, T = 0.04 s, V (0.01 s) = 20 V

20.1AC terms: cycle, period, frequency, peak, average and r.m.s. valuesOften examined

A few terms describe any alternating quantity. They apply equally to alternating current and alternating voltage, so $x$ below stands for either.

Instantaneous value: The value of the alternating quantity at a particular instant of time, $x = x_{0} sin ω t$ .
Cycle: One complete set of positive and negative values of an alternating quantity. Fig 20.2 shows one cycle of an alternating voltage.
Time period $T$: The time taken to complete one cycle of an alternating quantity. SI unit: second (s).
Frequency $f$: The number of cycles that occur in one second. It is measured in cycles per second, called hertz (Hz): $f = 1/ T$ .
Peak value $x_{0}$: The maximum value of the alternating quantity (current or voltage). It is also called the amplitude.
Average value: The average of all the values of the waveform over a stated period of time: all the values added up over the time interval, divided by that time.
Root-mean-square (r.m.s.) value: The steady current (DC) which, when flowing through a resistor, produces the same amount of heat as that produced by the alternating current flowing through the same resistance for the same time.

f = \frac{1}{T}

Average value

From a graph, the average value is found with

Average value = \frac{total (net) area under the curve for time T}{time T}

The area above the time axis is taken as positive and the area below it as negative. For a sine wave the positive half-cycle and the negative half-cycle enclose equal areas (the shaded areas in the next graph). So the average value over one complete cycle is zero. That is why the average value is not used to describe a sinusoidal voltage or current, and it cannot be used to calculate power.

Peak value and r.m.s. value

The peak value $x_{0}$ is the largest value of the wave. It occurs twice in each cycle: once as the positive maximum $+ x_{0}$ and once as the negative maximum $- x_{0}$ .

Peak and r.m.s. values

\frac{π}{2}

π

\frac{3 π}{2}

2 π

x_{0}

x_{rms}

- x_{rms}

- x_{0}

ω t

x

+

-

x = x_{0} sin ω t

Fig 20.2 redrawn. The peak value

x_{0}

is the highest point of the wave. The r.m.s. value sits lower, at

x_{0} / 2 = 0.707 x_{0}

. The positive and negative areas are equal, so the average over a full cycle is zero.

The heating effect of a current does not depend on its direction, because the heating power $P = I^{2} R$ contains $I^{2}$ . So the fair way to compare an alternating current with a direct current is by the heat they produce. This comparison defines the r.m.s. value (also called the effective value).

Derivation r.m.s. value of a sinusoidal current

Aim. Show that $I_{rms} = I_{0} / 2$ for a current $I = I_{0} sin ω t$ .

Assumptions.

The current is sinusoidal, $I = I_{0} sin ω t$ .
The resistance $R$ stays constant.

Why the mean of

sin^{2} ω t

is one half

\frac{π}{2}

π

\frac{3 π}{2}

2 π

0.5

ω t

y

mean = \frac{1}{2}

sin^{2} ω t

cos^{2} ω t

sin^{2} ω t

and

cos^{2} ω t

are the same curve shifted by a quarter cycle, so they have the same mean. They add up to 1 at every instant, so each mean is

\frac{1}{2}

Heat is produced in a resistor at the rate $P = I^{2} R$ . Square the alternating current.
$P = I^{2} R = I_{0}^{2} R sin^{2} ω t$
Over a whole cycle the mean heating power is $R$ times the mean value of $I^{2}$ ( $R$ is constant).
$⟨ P ⟩ = ⟨ I^{2} ⟩ R$
By definition, the steady current $I_{rms}$ gives the same heating. Set the DC power equal to the mean AC power.
$I_{rms}^{2} R = ⟨ I^{2} ⟩ R ⟹ I_{rms} = ⟨ I^{2} ⟩$
This explains the name: it is the root of the mean of the square of the current.
Find the mean of $sin^{2} ω t$ . Over a cycle, $sin^{2} ω t$ and $cos^{2} ω t$ are the same curve shifted by a quarter cycle (see the graph), so they have the same mean. They add up to 1 at every instant, so each mean is one half.
$⟨ sin^{2} ω t ⟩ + ⟨ cos^{2} ω t ⟩ = 1 ⟹ ⟨ sin^{2} ω t ⟩ = \frac{1}{2}$
So the mean square current is half the square of the peak current.
$⟨ I^{2} ⟩ = I_{0}^{2} ⟨ sin^{2} ω t ⟩ = \frac{I _{0}^{2}}{2}$
Take the square root.
$I_{rms} = \frac{I _{0}^{2}}{2} = \frac{I _{0}}{2} = 0.707 I_{0}$

Result

I_{rms} = \frac{I _{0}}{2} (20.2 a), V_{rms} = \frac{V _{0}}{2} (20.2 b)

An alternating current of peak value $I_{0}$ heats a resistor exactly as much as a steady current of $0.707 I_{0}$ . The same steps with $P = V^{2} / R$ give the result for voltage. These relations are true only for sinusoidal AC.

x_{rms} = \frac{x _{0}}{2} = 0.707 x_{0} ⟺ x_{0} = 2 x_{rms} = 1.414 x_{rms}

For example, if the effective (r.m.s.) value of an alternating current is $7 A$ , it produces the same heating effect as a direct current of $7 A$ . Its peak value is $72 = 9.9 A$ .

Peak value compared with r.m.s. value
	Peak value	r.m.s. value
Meaning	the maximum value reached in a cycle	the steady (DC) value that gives the same heating in a resistor
Symbols	$I_{0}, V_{0}$	$I_{rms}, V_{rms}$
When	reached only for an instant, twice per cycle ( $+ x_{0}$ and $- x_{0}$ )	an effective value for the whole cycle
Relation	$x_{0} = 2 x_{rms}$	$x_{rms} = x_{0} / 2 = 0.707 x_{0}$
Used for	insulation and breakdown ratings; drawing the wave	power and heating; readings of AC meters; ratings such as 220 V

Assignment 20.1

The peak voltage of an ac supply is

320 V

. What is the rms value of this voltage?

Use the relation between r.m.s. and peak values for a sinusoidal voltage.
$V_{rms} = \frac{V _{0}}{2}$
Substitute $V_{0} = 320 V$ .
$V_{rms} = \frac{320}{1.414} = 226.3 V$

V_{rms} \approx 226 V

20.1Mean power and maximum powerOften examined

When an alternating current I=I_0\sin\omega t flows through a resistance R, the power delivered to it changes from instant to instant. The power at any instant is

P=I^2R=(I_0\sin\omega t)^2R=I_0^2R\sin^2\omega t

As the current is squared, the power is always positive: the resistor is heated during both half-cycles. The power is largest, P_{\max}=I_0^2R, when \sin^2\omega t=1 (at the peaks of the current), and zero when the current is zero.

Derivation Mean power is half the maximum power

Aim. Find the mean power delivered by a sinusoidal current to a resistor and compare it with the maximum power.

Write the instantaneous power.
P=I_0^2R\sin^2\omega t
\sin^2\omega t varies between 0 and 1. Its mean over a complete cycle is \tfrac12. The identity \sin^2\omega t=\tfrac12(1-\cos2\omega t) shows this quickly, because the mean of \cos2\omega t over a cycle is zero.
\langle\sin^2\omega t\rangle=\tfrac12-\tfrac12\langle\cos2\omega t\rangle=\tfrac12
Take the mean of the power. I_0 and R are constants.
\langle P\rangle=I_0^2R\,\langle\sin^2\omega t\rangle=\tfrac12I_0^2R\qquad(20.3)
The maximum power is P_{\max}=I_0^2R.
\langle P\rangle=\tfrac12P_{\max}
Write it with r.m.s. values, using I_0=\sqrt2\,I_{\mathrm{rms}} and V_{\mathrm{rms}}=I_{\mathrm{rms}}R.
\langle P\rangle=\tfrac12\left(\sqrt2\,I_{\mathrm{rms}}\right)^2R=I_{\mathrm{rms}}^2R=V_{\mathrm{rms}}I_{\mathrm{rms}}=\dfrac{V_{\mathrm{rms}}^2}{R}

Result\langle P\rangle=\tfrac12I_0^2R=\tfrac12P_{\max}=I_{\mathrm{rms}}^2R

The mean power in a resistive load is half the maximum power for a sinusoidal alternating current. With r.m.s. values, the familiar DC power formulas work unchanged for AC in a resistor.

Power in a resistor

The current (dashed) changes sign, but the power P=I^2R never goes negative. Power pulses twice in every cycle and its mean is exactly half the maximum.

Example 20.1

An A.C. circuit consists of a pure resistance of 20\,\Omega and is connected across an A.C. supply of 220\,\mathrm{V}, 50\,\mathrm{Hz}. Calculate (a) the peak value of voltage (b) the peak value of current (c) the equations for voltage and current.

Given

R = 20\,\Omega
V_{\mathrm{rms}} = 220\,\mathrm{V} the supply voltage is an r.m.s. value
f = 50\,\mathrm{Hz}

To find

V_0
I_0
equations for V and I

(a) Peak voltage from the r.m.s. value.
V_0=\sqrt2\,V_{\mathrm{rms}}=1.414\times220=311.1\,\mathrm{V}
(b) Peak current from Ohm's law, which holds at every instant for a resistor.
I_0=\dfrac{V_0}{R}=\dfrac{311.1}{20}=15.56\,\mathrm{A}
(c) Angular frequency.
\omega=2\pi f=2\times3.14\times50=314\,\mathrm{rad\,s^{-1}}
For a pure resistance the voltage and current are in phase, so both have the form x_0\sin\omega t.
V=311.1\sin(314t)\,\mathrm{V},\qquad I=15.56\sin(314t)\,\mathrm{A}
Extra check on power: the mean power is half the maximum power.
\langle P\rangle=\dfrac{V_{\mathrm{rms}}^2}{R}=\dfrac{220^2}{20}=2420\,\mathrm{W}=\tfrac12\left(V_0I_0\right)=\tfrac12(4840\,\mathrm{W})

V_0=311\,\mathrm{V},\quad I_0=15.6\,\mathrm{A},\quad V=311\sin(314t)\,\mathrm{V},\quad I=15.6\sin(314t)\,\mathrm{A}

20.2Rectification

Every electronic circuit (a mobile phone, a computer, a television, the charger of a laptop) needs a steady DC voltage, usually only a few volts. The mains supply is 220 V AC. So the AC supply voltage is first reduced (stepped down) by a step-down transformer and then converted into DC by a circuit called a rectifier.

Rectification: The process of converting AC voltage into DC voltage.
Rectifier: A circuit that converts AC into DC. It uses one or more diodes.

The stages of a DC power supply: the transformer lowers the voltage, the rectifier makes the current flow one way, and the filter capacitor smooths it.

A rectifier works because a diode (a p-n junction) conducts in one direction only. When its p-side is at a higher potential than its n-side, it is forward biased: its resistance is very low and current flows. When the polarity is reversed, it is reverse biased: its resistance is very high and practically no current flows. So a diode acts like a one-way valve, or an automatic switch, for current.

There are two types of rectification: half-wave rectification and full-wave rectification.

20.2.1Half-wave rectificationOften examined

In half-wave rectification an AC signal is converted into DC by passing one half-cycle of the waveform and blocking the other half. A half-wave rectifier is easily built with one diode D connected in series with the load resistance R (Fig 20.3).

Fig 20.3: half-wave rectifier. The diode D (p-side P, n-side N) is in series with the load R. The input is AC; the output has only the positive half-cycles.

Working

Positive half-cycle. The diode D is forward biased. It offers a very low resistance and current flows through the resistor R. A forward-biased diode acts as a closed switch (Fig 20.4 a), so the output voltage follows the input.
Negative half-cycle. The diode D is reverse biased. It offers a very high resistance and no current flows through R, so the output voltage is zero. A reverse-biased diode acts as an open switch (Fig 20.4 b).
The cycle repeats. The current through R always flows in the same direction, but only during alternate half-cycles.

Fig 20.4: effective circuit of a half-wave rectifier: (a) forward-biased diode as a closed switch, current flows; (b) reverse-biased diode as an open switch, no current.

Half-wave rectification

Fig 20.5 redrawn. Only the positive half-cycles reach the load. The negative half-cycles are blocked, so the output is zero for half of every cycle. One pulse per cycle.

The output of a half-wave rectifier is pulsating DC: it flows in one direction only, but its value rises and falls and is zero for half of the time. There is one pulse in each input cycle, so for a 50\,\mathrm{Hz} supply the output pulses 50 times per second.

Simple and cheap: only one diode is needed.
Wasteful: the negative half of every cycle, and so half the input power, is thrown away.
The gaps between the pulses are long, so the output is hard to smooth.

Filters are used with rectifiers to transform the pulsating DC waveform into a nearly constant DC waveform. A capacitor connected across the load can be used as a filter (see smoothing below).

20.2.2Full-wave rectification: the bridge rectifierOften examined

In full-wave rectification the complete cycle of the AC signal is converted into pulsating DC: the positive half-cycle is passed as it is and the negative half-cycle is turned the right way up. A full-wave rectifier can be built in two ways:

Centre-tapped full-wave rectifier: a transformer whose secondary coil has a connection at its midpoint (a centre tap) and two diodes. Each diode conducts during one half-cycle.
Bridge rectifier: four diodes arranged as a bridge. No centre tap is needed. Your book discusses only this circuit.

Construction

The circuit consists of four diodes D_1, D_2, D_3 and D_4 connected in a closed loop to form a bridge (Fig 20.6 a). The AC input (the secondary of the transformer) is connected across two opposite corners A and B. The load resistor R is connected across the other two corners. At one of these corners the bars (n-sides) of two diodes meet: this is the positive output terminal. At the other the triangles (p-sides) meet: this is the negative output terminal.

Fig 20.6 (a): full-wave bridge rectifier. AC input at A and B; load R across the other two corners. The small hump shows two wires crossing without joining.

Working during the positive half-cycle

During the positive half-cycle of the secondary voltage, end A becomes positive and end B negative. This makes diodes D_1 and D_2 forward biased and diodes D_3 and D_4 reverse biased. So D_1 and D_2 conduct while D_3 and D_4 do not. The current flows from A through D_1, through the load resistor R (from top to bottom), through D_2 and back to B, the negative terminal of the input.

Fig 20.6 (b): positive half-cycle. D_1 and D_2 conduct. Path: A, D_1, R (top to bottom), D_2, B.

Working during the negative half-cycle

During the negative half-cycle, end A becomes negative with respect to end B. Now diodes D_3 and D_4 are forward biased and D_1 and D_2 are reverse biased. So D_3 and D_4 conduct while D_1 and D_2 do not. The current flows from B through D_3, through the load resistor R (again from top to bottom), through D_4 and back to A, which is now the negative terminal of the input.

Fig 20.6 (c): negative half-cycle. D_3 and D_4 conduct. Path: B, D_3, R (top to bottom), D_4, A. The current in R has the same direction as before.

So one pair of diodes (D_1 and D_2) allows current during the positive half-cycle and the other pair (D_3 and D_4) during the negative half-cycle. The current through the load resistor is in the same direction (top to bottom) during both half-cycles. Hence a rectified output voltage is obtained across R for the whole cycle (Fig 20.7).

Full-wave rectification

Fig 20.7 redrawn. Both half-cycles reach the load and both drive current the same way through it, so the negative halves appear flipped upward. Two pulses per cycle.

Each half-cycle gives one output pulse, so there are two pulses per input cycle: for 50\,\mathrm{Hz} mains the output pulses 100 times per second.

Half-wave and full-wave rectification compared
Feature	Half-wave rectifier	Full-wave bridge rectifier
Diodes used	1	4
Part of each input cycle used	one half only	both halves
Output pulses per input cycle	1 (frequency f, e.g. 50 Hz)	2 (frequency 2f, e.g. 100 Hz)
Output during the negative half-cycle	zero	the half-cycle turned upward
Efficiency	low (half the input is wasted)	high
Smoothing	harder (long gaps)	easier (short gaps)

20.2Smoothing with a capacitor (filtering)Often examined

The output of a rectifier is pulsating DC, but electronic circuits need a nearly constant voltage. A capacitor connected in parallel with the load smooths out the pulsating DC into a more stable, nearly constant output. This process is called filtering or smoothing (Fig 20.8 a).

Fig 20.8 (a): full-wave bridge rectifier with a smoothing capacitor C connected in parallel with the load R.

How the capacitor smooths the output

Charging. As the rectified voltage rises to the peak of the input waveform, the diodes conduct and the capacitor charges up to the peak voltage V_0.
Discharging. When the input voltage starts to drop, it falls below the capacitor's voltage. The diodes become reverse biased and stop conducting. Now the capacitor discharges slowly through the load, supplying charge (current) to it.
Recharging. Before the capacitor has lost much charge, the next pulse arrives, rises above the capacitor voltage and recharges it to V_0.
So the capacitor's discharge fills in the gaps between the peaks of the rectified waveform. The voltage fluctuations are reduced and the output becomes smoother. The small remaining rise and fall is called the ripple.

Smoothing by a capacitor

Fig 20.8 (b) redrawn. After each peak the diodes switch off and the capacitor discharges slowly through the load. With a large time constant RC the voltage hardly falls before the next pulse recharges it (small ripple). With a small RC it falls a long way (large ripple).

Effect of the capacitance and the load resistance

The capacitor discharges through the load exactly like the RC circuits of Unit 19: its voltage falls as V=V_0e^{-t/RC}. How quickly it falls is set by the time constant

\tau=RC

R: load resistance in ohms; C: capacitance in farads; \tau in seconds. In one time constant the voltage falls to 37\% of its starting value.

**Larger capacitance C**: the capacitor holds more charge at the same voltage, so its voltage falls more slowly between pulses. The ripple is smaller and the smoothing is better. But a larger capacitor also draws a larger charging current through the diodes each time it recharges.
**Larger load resistance R** (a smaller load current): the capacitor discharges more slowly, so the ripple is smaller.
**Smaller load resistance R** (a larger load current): the capacitor discharges faster, which reduces its ability to smooth the output. The ripple is larger.
Good smoothing needs a time constant much longer than the gap between the pulses: RC\gg\dfrac{1}{2f}=0.01\,\mathrm{s} for full-wave rectification of 50\,\mathrm{Hz} mains, and RC\gg0.02\,\mathrm{s} for half-wave.

20.3Self-induction and self-inductanceOften examined

From Grade 11 you know Faraday's law: a changing magnetic flux through a coil induces an emf in it. A coil carrying a current produces its own magnetic field, and this field passes through the coil itself. So when the current in a coil changes, its own flux changes, and an emf is induced in the same coil.

Consider a coil connected to a battery through a switch and a rheostat (Fig 20.9 a). If we move the rheostat quickly, the current through the coil changes with time. This changes the magnetic field and hence the flux through the coil, which finally induces an emf in the coil itself.

Fig 20.9 (a): self-induction. Changing the current with the rheostat changes the coil's own magnetic flux, and an emf \varepsilon_L is induced in the same coil.

Self-induction: The phenomenon in which a changing current induces an emf in the coil itself.

The self-induced emf in a coil is proportional to the time rate of change of current through the coil:

\varepsilon_L\propto\dfrac{\Delta I}{\Delta t}\qquad\Longrightarrow\qquad\varepsilon_L=-L\dfrac{\Delta I}{\Delta t}

(i)

The constant L is the self-inductance of the coil. Equation (i) can also be written as

L=\dfrac{-\varepsilon_L}{\Delta I/\Delta t}

(20.4)

Self-inductance L: The ratio of the emf induced in a coil to the time rate of change of current through it.

The negative sign: back emf

The negative sign in equation (20.4) is in accordance with Lenz's law. It shows that the self-induced emf opposes the change which produces it. When the current is increasing, the induced emf acts against the current and slows its growth. When the current is decreasing, the induced emf acts in the direction of the current and tries to keep it flowing. Because it opposes the applied emf while the current is being built up, the self-induced emf is also called the back emf.

Unit of self-inductance: the henry

From equation (20.4), the unit of self-inductance is volt per (ampere per second). This unit is called the henry (H).

1\,\mathrm{H}=1\,\mathrm{V\,s\,A^{-1}}=1\,\mathrm{Wb\,A^{-1}}=1\,\Omega\,\mathrm{s}

One henry (self-inductance): The self-inductance of a coil is one henry if an emf of one volt is induced in the coil when the current through it changes at the rate of one ampere per second.

Smaller units are common: 1\,\mathrm{mH}=10^{-3}\,\mathrm{H} and 1\,\mu\mathrm{H}=10^{-6}\,\mathrm{H}.

Factors on which self-inductance depends

The self-inductance depends only on how the coil is made, not on the current in it. For a long coil (solenoid) it can be calculated. This derivation is not in your book, but it gives the factors that FBISE papers ask for.

Derivation Self-inductance of a long solenoid

Aim. Find L for a solenoid of N turns, length l and cross-sectional area A.

Assumptions.

The solenoid is long, so the field inside is uniform: B=\mu_0nI with n=N/l turns per unit length.
The core is air (or vacuum).

The flux through each turn is the field times the area.
\Phi=BA=\mu_0\dfrac{N}{l}IA
All N turns carry this flux, so the total flux linkage is N\Phi.
N\Phi=\dfrac{\mu_0N^2A}{l}\,I
Faraday's law: the induced emf is minus the rate of change of flux linkage. Only the current changes.
\varepsilon_L=-\dfrac{\Delta(N\Phi)}{\Delta t}=-\dfrac{\mu_0N^2A}{l}\,\dfrac{\Delta I}{\Delta t}
Compare with \varepsilon_L=-L\,\Delta I/\Delta t.
L=\dfrac{\mu_0N^2A}{l}
With a core of relative permeability \mu_r (for example soft iron), the field and so L become \mu_r times larger.
L=\dfrac{\mu_r\mu_0N^2A}{l}

ResultL=\dfrac{\mu_r\mu_0N^2A}{l}\qquad\text{and in general}\qquad N\Phi=LI

L grows as the square of the number of turns and in proportion to the area. It falls if the same turns are spread over a longer length, and a soft-iron core makes it hundreds of times larger. It does not depend on the current.

**Number of turns N**: L\propto N^2.
**Area of cross-section A** of the coil: L\propto A.
**Length l** of the coil for the same number of turns: L\propto1/l (a closely wound coil has more inductance).
Nature of the core, that is, its relative permeability \mu_r: a soft-iron core increases L greatly.

Inductors and chokes

Because of their self-inductance, coils are also known as inductors or chokes. Inductors are widely used in electrical technology. In AC circuits they oppose the current like resistors (their opposition is called inductive reactance, section 20.4.2), but they waste almost no energy as heat.

20.3Mutual induction and mutual inductanceOften examined

Consider two neighbouring coils (Fig 20.9 b). Coil 1 is connected to a source of changing current: an AC supply, or a DC source whose current is varied. Coil 2 is connected to a galvanometer. The coils are placed so that the magnetic flux of coil 1 also passes through (links) coil 2.

Fig 20.9 (b): mutual induction between two coils. The changing flux of coil 1 passes through coil 2 and induces an emf \varepsilon_2 in it, shown by the galvanometer G.

When AC flows through coil 1, its magnitude and direction keep changing, so the flux of coil 1 changes. This flux also links coil 2, so the changing flux through coil 2 induces an emf in it, and the galvanometer needle deflects. This is mutual induction.

Mutual induction: The phenomenon in which a changing current in one coil induces an emf in a neighbouring coil.

The emf induced in coil 2 is proportional to the time rate of change of current through coil 1:

\varepsilon_2=-M\dfrac{\Delta I_1}{\Delta t}

(ii)

M=\dfrac{-\varepsilon_2}{\Delta I_1/\Delta t}

(20.5)

Mutual inductance M: The ratio of the emf induced in one coil (the secondary) to the time rate of change of current in the neighbouring coil (the primary).

The negative sign shows Lenz's law: the current induced in coil 2 is in such a direction that it opposes the change of current in coil 1.

Unit of mutual inductance

The unit of mutual inductance is also the henry (H), because M, like L, is an emf divided by a rate of change of current: 1\,\mathrm{H}=1\,\mathrm{V\,s\,A^{-1}}.

One henry (mutual inductance): The mutual inductance between two coils is one henry if a current changing at the rate of 1\,\mathrm{A\,s^{-1}} in one coil induces an emf of 1\,\mathrm{V} in the other coil.

Factors on which mutual inductance depends

Number of turns of both coils: M\propto N_1N_2.
Area of cross-section of the coils.
Distance between the coils: the closer they are, the more flux they share and the larger M is.
Relative orientation: M is largest when the coils are coaxial (one wound over the other) and nearly zero when their axes are at right angles.
Nature of the core: a common soft-iron core increases M greatly.

20.4Phase of AC: lead and lagOften examined

In a circuit with only resistance, the alternating current and voltage rise and fall together. Through capacitors or inductors they do not peak at the same time. The voltage may be passing through its zero point while the current has already passed through its zero point, or is yet to pass through it, in the same direction.

Phase: The phase of an alternating quantity at an instant is the angle \omega t in x=x_0\sin\omega t. It tells how far the cycle has advanced, so it fixes both the value of the quantity and the direction in which it is changing.
Phase difference \Phi: The angle between the zero points (or between the peaks) of two alternating quantities of the same frequency. In the book's words: the fraction of a period difference between the peaks, expressed in degrees. It is measured in degrees or radians.
Lead and lag: The quantity which passes through its zero point (or its peak) earlier is said to be leading, while the other is said to be lagging.

\Phi=\dfrac{\Delta t}{T}\times360^\circ=\dfrac{\Delta t}{T}\times2\pi\ \mathrm{rad}

\Delta t: time gap between the two peaks; T: time period.

Phase difference

Both waves have the same frequency. The voltage reaches its peak (and its zero) earlier, so the voltage leads and the current lags by the phase difference \Phi (here 60^\circ). The gap stays the same cycle after cycle.

If V=V_0\sin\omega t and I=I_0\sin(\omega t-\Phi), the current lags the voltage by \Phi (the voltage leads the current by \Phi). If I=I_0\sin(\omega t+\Phi), the current leads. Since both alternating quantities have the same frequency, the phase difference between them remains the same all the time.

It is customary to state the angle by which the voltage leads the current. In circuits of resistors, inductors and capacitors this angle is never more than 90^\circ.

Phasors

A phasor is a rotating arrow that represents an alternating quantity. Its length equals the peak value x_0, and it rotates anticlockwise about its tail with angular velocity \omega. At time t it makes the angle \theta=\omega t with the horizontal, and its vertical projection is x_0\sin\omega t, which is exactly the instantaneous value. As the phasor turns, its projection traces out the sine wave.

A phasor of length x_0 rotating anticlockwise at angular velocity \omega. Its vertical projection x=x_0\sin\omega t traces the sine wave.

Phasors of quantities with the same frequency rotate together, so the angle between them (the phase difference) stays fixed. A phasor diagram is a snapshot of them at one instant.
The phasor that is ahead in the anticlockwise direction leads.
Phasors add like vectors. This is how voltages that are not in phase are added (section 20.5).
In a series circuit the current is the same in every component, so the current phasor is drawn along the horizontal as the reference.

Phasor diagrams for a pure resistor, a pure inductor and a pure capacitor, with the current I as the reference.

Phase relations for single elements
Circuit element	Phase relation	Equations
Resistor R	V and I in phase (\Phi=0)	V=V_0\sin\omega t, I=I_0\sin\omega t
Inductor L	V leads I by 90^\circ (\pi/2)	I=I_0\sin\omega t, V=V_0\sin(\omega t+\pi/2)
Capacitor C	V lags I by 90^\circ (I leads V)	V=V_0\sin\omega t, I=I_0\sin(\omega t+\pi/2)

20.4.1AC through a resistorOften examined

Consider a circuit containing a resistor R connected across an alternating voltage source (Fig 20.10 a).

Fig 20.10 (a): AC through a resistor.

Derivation Current in a pure resistor

Aim. Find the current through R and its phase relative to the applied voltage.

Let the applied alternating voltage be
V=V_0\sin\omega t\qquad(\text{i})
Ohm's law holds at every instant.
I=\dfrac{V}{R}=\dfrac{V_0\sin\omega t}{R}
The constant V_0/R is the peak current I_0.
I_0=\dfrac{V_0}{R}\quad\Longrightarrow\quad I=I_0\sin\omega t\qquad(\text{ii})
Compare (i) and (ii): the voltage and the current have the same phase angle \omega t.
\Phi=\omega t-\omega t=0

ResultI=I_0\sin\omega t,\qquad I_0=\dfrac{V_0}{R}

In a resistor, the applied voltage and the circuit current are in phase with each other. They pass through their zero values at the same instant and reach their positive and negative peaks at the same instant.

Resistor: voltage and current in phase

Fig 20.10 (b) redrawn. V and I pass through zero together and reach their peaks together: phase difference zero.

Fig 20.10 (c): phasor diagram for a resistor. V and I point in the same direction.

The resistance does not depend on the frequency: a resistor opposes DC and AC of any frequency equally. The electrical energy is changed into heat at the mean rate \langle P\rangle=V_{\mathrm{rms}}I_{\mathrm{rms}}=I_{\mathrm{rms}}^2R.

20.4.2AC through an inductorOften examined

An inductor is an electrical component which opposes changes in the electric current passing through it. It consists of a conductor, such as a wire, usually wound into a coil. An ideal (pure) inductor has inductance L and zero resistance.

Fig 20.11 (a): AC through an inductor of inductance L.

Let an alternating current I=I_0\sin\omega t flow through the inductor. As the current changes, a back emf of magnitude L\,\Delta I/\Delta t is induced in the coil, and at every instant it opposes the change in current. There is no potential drop across the inductor due to resistance (an ideal inductor has zero resistance), so the applied voltage has only to overcome the back emf: applied alternating voltage = back emf. The energy used to build up the current is returned during the decay of the current.

Derivation Voltage across a pure inductor

Aim. For a current I=I_0\sin\omega t, find the applied voltage V and its phase relative to I.

Assumptions.

The inductor is ideal (zero resistance).
\Delta I/\Delta t is taken over a very small time interval, so it is the instantaneous rate of change.

Let the current through the inductor be
I=I_0\sin\omega t\qquad(\text{i})
The applied voltage equals the magnitude of the back emf.
V=L\dfrac{\Delta I}{\Delta t}=L\dfrac{\Delta(I_0\sin\omega t)}{\Delta t}=LI_0\dfrac{\Delta(\sin\omega t)}{\Delta t}\qquad(\text{ii})
The rate of change of \sin\omega t is \omega\cos\omega t. (The slope of a sine curve is largest at its zeros and zero at its peaks, which is exactly how a cosine behaves.)
\dfrac{\Delta(\sin\omega t)}{\Delta t}=\omega\cos\omega t
Substitute into (ii).
V=\omega LI_0\cos\omega t
The constant \omega LI_0 is the peak voltage V_0. Also \cos\omega t=\sin(\omega t+\pi/2).
V=V_0\cos\omega t=V_0\sin\left(\omega t+\dfrac{\pi}{2}\right)\qquad(20.6)
Compare the phases of (20.6) and (i).
\Phi=\left(\omega t+\dfrac{\pi}{2}\right)-\omega t=\dfrac{\pi}{2}=90^\circ

ResultV=V_0\sin\left(\omega t+\dfrac{\pi}{2}\right),\qquad V_0=\omega LI_0

In an inductor, the voltage leads the current by \pi/2 radians (90^\circ). Equivalently, the current lags the voltage by 90^\circ.

Inductor: voltage leads current by 90°

Fig 20.11 (b) redrawn. The voltage reaches its peak a quarter cycle (90^\circ) before the current. The voltage is largest when the current is zero, because that is when the current changes fastest.

Fig 20.11 (c): phasor diagram for an inductor. V is 90^\circ ahead of I (anticlockwise).

Inductive reactance

Inductive reactance X_L: The opposition offered by an inductor to the flow of AC. SI unit: ohm (\Omega).

Derivation Inductive reactance X_L=\omega L

Aim. Express the opposition of an inductor in terms of L and the frequency f.

In analogy with Ohm's law, the reactance is the peak voltage divided by the peak current (the same ratio holds for r.m.s. values).
V_0=I_0X_L\quad\Longrightarrow\quad X_L=\dfrac{V_0}{I_0}
Use V_0=\omega LI_0 from the previous derivation.
X_L=\dfrac{\omega LI_0}{I_0}=\omega L
Use \omega=2\pi f.
X_L=2\pi fL\qquad(20.7)

ResultX_L=\omega L=2\pi fL

The reactance of a coil is proportional to the frequency of the AC and to its inductance. For DC, f=0 and so X_L=0: an ideal inductor offers no opposition to a steady current.

X_L grows with frequency because at a higher frequency the current changes faster, so the back emf is larger and opposes the current more. A larger inductance also gives a larger back emf. Inductive reactance has the same dimensions as resistance, so it is measured in ohms.

Power in a pure inductor

The instantaneous power is P=VI=V_0I_0\sin\omega t\cos\omega t=\tfrac12V_0I_0\sin2\omega t. It is positive for a quarter cycle (energy flows into the magnetic field) and negative for the next quarter (the same energy flows back to the source). So the mean power over a cycle is zero: a pure inductor consumes no energy.

Power in a pure inductor

The power is positive for a quarter cycle (the inductor stores energy in its magnetic field) and negative for the next quarter (it gives the same energy back). The areas cancel, so the mean power is zero.

20.4.3AC through a capacitorOften examined

Consider an alternating voltage applied across a capacitor of capacitance C (Fig 20.12 a). No charge crosses the insulator between the plates. But the capacitor is charged in one direction and then in the other as the voltage reverses. Electrons move to and fro in the connecting wires, charging and discharging the plates, and this to-and-fro motion is an alternating current in the circuit.

Fig 20.12 (a): AC through a capacitor of capacitance C.

Derivation Current through a capacitor

Aim. For an applied voltage V=V_0\sin\omega t, find the current and its phase relative to V.

Let the applied alternating voltage be
V=V_0\sin\omega t\qquad(\text{i})
The basic relation q=CV holds at every instant, so the charge on the capacitor is
q=CV=CV_0\sin\omega t
Current is the rate of flow of charge.
I=\dfrac{\Delta q}{\Delta t}=\dfrac{\Delta(CV_0\sin\omega t)}{\Delta t}=CV_0\dfrac{\Delta(\sin\omega t)}{\Delta t}
Use \Delta(\sin\omega t)/\Delta t=\omega\cos\omega t and \cos\omega t=\sin(\omega t+\pi/2).
I=\omega CV_0\cos\omega t=\omega CV_0\sin\left(\omega t+\dfrac{\pi}{2}\right)
The constant \omega CV_0 is the peak current I_0.
I=I_0\sin\left(\omega t+\dfrac{\pi}{2}\right)\qquad(20.8),\qquad I_0=\omega CV_0

ResultI=I_0\sin\left(\omega t+\dfrac{\pi}{2}\right),\qquad I_0=\omega CV_0

In a capacitor, the voltage lags the current by \pi/2 radians (90^\circ); in other words, the current leads the voltage by 90^\circ.

Capacitor: current leads voltage by 90°

Fig 20.12 (b) redrawn. The current reaches its peak a quarter cycle before the voltage. The current is largest when V=0 (voltage changing fastest) and zero when V is at its peak (capacitor momentarily fully charged).

Fig 20.12 (c): phasor diagram for a capacitor. V is 90^\circ behind I.

Capacitive reactance

Capacitive reactance X_C: The opposition offered by a capacitor to the flow of AC. SI unit: ohm (\Omega).

Derivation Capacitive reactance X_C=1/(\omega C)

Aim. Express the opposition of a capacitor in terms of C and the frequency f.

In analogy with Ohm's law:
V_0=I_0X_C\quad\Longrightarrow\quad X_C=\dfrac{V_0}{I_0}
For a capacitor I_0=\omega CV_0.
X_C=\dfrac{V_0}{\omega CV_0}=\dfrac{1}{\omega C}
Use \omega=2\pi f.
X_C=\dfrac{1}{2\pi fC}\qquad(20.9)

ResultX_C=\dfrac{1}{\omega C}=\dfrac{1}{2\pi fC}

Capacitive reactance is inversely proportional to the frequency and to the capacitance. For DC (f=0) it is infinite, so a capacitor blocks DC. At high frequencies it is small, so AC passes easily.

X_C falls as the frequency rises because at a higher frequency the capacitor is charged and discharged more times each second, so more charge flows per second: a larger current for the same voltage. A larger capacitance holds more charge at the same voltage, so again more current flows. Capacitive reactance has the same dimensions as resistance, so it is measured in ohms.

As in an inductor, the mean power in a pure capacitor is zero: energy stored in the electric field during one quarter cycle is returned to the source in the next.

Example 20.2

A.C. voltage across a 0.5\,\mu\mathrm{F} capacitor is 16\sin(2\times10^3t)\,\mathrm{V}. Find (a) the capacitive reactance (b) the peak value of current through the capacitor.

Given

C = 0.5\,\mu\mathrm{F}=0.5\times10^{-6}\,\mathrm{F}
V = 16\sin(2\times10^{3}t)\,\mathrm{V}

To find

Compare the given voltage with V=V_0\sin\omega t.
V_0=16\,\mathrm{V},\qquad\omega=2\times10^{3}\,\mathrm{rad\,s^{-1}}
(a) Capacitive reactance.
X_C=\dfrac{1}{\omega C}=\dfrac{1}{(2\times10^{3})(0.5\times10^{-6})}=\dfrac{1}{10^{-3}}=1000\,\Omega
(b) Ohm's law with the reactance.
I_0=\dfrac{V_0}{X_C}=\dfrac{16}{1000}=16\times10^{-3}\,\mathrm{A}=16\,\mathrm{mA}

X_C=1000\,\Omega,\qquad I_0=16\,\mathrm{mA}

Assignment 20.2

The voltage across a 0.01\,\mu\mathrm{F} capacitor is 240\sin(1.25\times10^4t-30^\circ)\,\mathrm{V}. Write the mathematical expression for the current through it.

Read off the peak voltage and the angular frequency.
V_0=240\,\mathrm{V},\qquad\omega=1.25\times10^{4}\,\mathrm{rad\,s^{-1}}
Capacitive reactance with C=0.01\times10^{-6}\,\mathrm{F}.
X_C=\dfrac{1}{\omega C}=\dfrac{1}{(1.25\times10^{4})(1\times10^{-8})}=8000\,\Omega
Peak current.
I_0=\dfrac{V_0}{X_C}=\dfrac{240}{8000}=0.030\,\mathrm{A}=30\,\mathrm{mA}
In a capacitor the current leads the voltage by 90^\circ, so add 90^\circ to the phase of the voltage: -30^\circ+90^\circ=+60^\circ.
I=0.030\sin(1.25\times10^{4}t+60^\circ)\,\mathrm{A}

I=0.030\sin(1.25\times10^{4}t+60^\circ)\,\mathrm{A}

20.4Reactance, resistance and frequencyOften examined

Resistors, inductors and capacitors all oppose AC, and all these oppositions are measured in ohms. But they respond to frequency in completely different ways.

How R, X_L and X_C change with frequency

X_L is a straight line through the origin: zero for DC, growing with frequency. X_C is a hyperbola: very large at low frequency (infinite for DC), small at high frequency. R does not depend on frequency.

The three kinds of opposition compared
	Resistor	Inductor	Capacitor
Opposition	resistance R	inductive reactance X_L	capacitive reactance X_C
Formula	R=V/I	X_L=2\pi fL	X_C=\dfrac{1}{2\pi fC}
Effect of frequency	none	X_L\propto f	X_C\propto1/f
For DC (f=0)	R	zero: passes DC	infinite: blocks DC
Phase of V relative to I	in phase	leads by 90^\circ	lags by 90^\circ
Mean power	I_{\mathrm{rms}}^2R (heat)	zero	zero

Effect of doubling the frequency

Let the frequency change from f to 2f, with L and C unchanged.

X_L'=2\pi(2f)L=2\,(2\pi fL)=2X_L

X_C'=\dfrac{1}{2\pi(2f)C}=\dfrac12\cdot\dfrac{1}{2\pi fC}=\dfrac{X_C}{2}

Doubling the frequency doubles the inductive reactance and halves the capacitive reactance. For the same applied voltage, the current through a pure inductor is halved and the current through a pure capacitor is doubled. The resistance is not affected.

20.5ImpedanceOften examined

Impedance is the combined effect of the resistance and the reactance present in an AC circuit. It is the total opposition to the flow of alternating current, so it is the AC equivalent of resistance. Impedance can be broken down into resistance and reactance; the reactance is due to the presence of inductors and capacitors.

Z=\dfrac{V}{I}=\dfrac{V_{\mathrm{rms}}}{I_{\mathrm{rms}}}=\dfrac{V_0}{I_0}

Impedance Z is measured in ohms (\Omega).

Why impedance is a vector (phasor) sum

In a series circuit the same current flows through every component, but the voltages across them are not in phase: V_R is in phase with I, V_L leads I by 90^\circ and V_C lags I by 90^\circ. Voltages that are 90^\circ apart cannot be added like ordinary numbers. They must be added as phasors (vectors), using Pythagoras' theorem.

Dividing every voltage by the common current I turns the voltage phasor diagram into an impedance triangle. The resistance R lies along the current direction, the reactance (X_L upward, X_C downward) is at right angles to it, and the impedance Z is the resultant, the hypotenuse. So impedance is the vector summation of resistance and reactance:

Z=\sqrt{R^2+X^2},\qquad\tan\varphi=\dfrac{X}{R}

X is the net reactance (X_L, X_C or X_L-X_C) and \varphi is the phase angle between the supply voltage and the current.

20.5RL series AC circuitOften examined

Consider an AC circuit in which a resistor R is connected in series with a coil of pure inductance L (Fig 20.13 a). The same current I flows through both. The supply voltage V is the phasor sum of the two component voltages V_R and V_L.

Fig 20.13 (a): RL series circuit. The same current I flows through R and L.

The voltage drop V_R=IR is in phase with the current. It is represented by the phasor OA, along the current.
The voltage drop V_L=IX_L leads the current by 90^\circ. It is represented by the phasor AB, at right angles to OA.

Derivation Impedance of an RL series circuit

Aim. Find the impedance Z and the phase angle \varphi for R and L in series.

Assumptions.

The coil is a pure inductor (its resistance is included in R).
The current phasor is the reference, drawn horizontally.

Fig 20.13 (b): phasor diagram of the RL series circuit. V leads I by \varphi.

Taking the current as the reference phasor, draw V_R along OA and V_L along AB at 90^\circ to it. The applied voltage V is their phasor sum, OB.
\vec V=\vec V_R+\vec V_L
OAB is a right-angled triangle, so use Pythagoras' theorem.
V^2=V_R^2+V_L^2\quad\Longrightarrow\quad V=\sqrt{V_R^2+V_L^2}
Substitute V_R=IR and V_L=IX_L.
V=\sqrt{(IR)^2+(IX_L)^2}=I\sqrt{R^2+X_L^2}
V/I is the impedance Z.
Z=\dfrac{V}{I}=\sqrt{R^2+X_L^2}\qquad(20.10)
The phase angle \varphi between V (OB) and I (along OA) comes from the same triangle.
\tan\varphi=\dfrac{V_L}{V_R}=\dfrac{IX_L}{IR}=\dfrac{X_L}{R}\quad\Longrightarrow\quad\varphi=\tan^{-1}\dfrac{X_L}{R}

ResultZ=\sqrt{R^2+X_L^2}=\sqrt{R^2+(2\pi fL)^2},\qquad\tan\varphi=\dfrac{X_L}{R}

The current still lags the applied voltage, but by an angle \varphi less than 90^\circ that depends on V_R and V_L. The impedance depends on the resistance R, the inductance L and the supply frequency f.

Fig 20.13 (c): RL series impedance triangle with sides R, X_L and Z. (The book labels it 'Fig 12.13 (c)' by mistake.)

If X_L\gg R, \varphi approaches 90^\circ: the circuit behaves almost like a pure inductor.
If R\gg X_L, \varphi approaches 0: the circuit behaves almost like a pure resistor.

Voltages in an RL circuit

A resistor of 30\,\Omega and an inductor of reactance 40\,\Omega are connected in series across a 100\,\mathrm{V} (r.m.s.) supply. Find the impedance, the current, the voltages V_R and V_L, and the phase angle.

Impedance.
Z=\sqrt{R^2+X_L^2}=\sqrt{30^2+40^2}=50\,\Omega
Current.
I=\dfrac{V}{Z}=\dfrac{100}{50}=2\,\mathrm{A}
Voltage across each component.
V_R=IR=60\,\mathrm{V},\qquad V_L=IX_L=80\,\mathrm{V}
Check by phasor addition.
\sqrt{60^2+80^2}=100\,\mathrm{V}
Phase angle; the current lags the voltage.
\varphi=\tan^{-1}\dfrac{40}{30}=53.1^\circ

Z=50\,\Omega,\quad I=2\,\mathrm{A},\quad V_R=60\,\mathrm{V},\quad V_L=80\,\mathrm{V},\quad\varphi=53.1^\circ\ (\text{current lags})

20.5RC series AC circuitOften examined

Consider an AC circuit in which a capacitor C and a resistor R are connected in series (Fig 20.14 a). The same current flows through both. The voltage V across the combination is the phasor sum of the two component voltages V_R=IR and V_C=IX_C.

Fig 20.14 (a): RC series circuit.

The voltage drop V_R=IR is in phase with the current: phasor OA.
The voltage drop V_C=IX_C lags behind the current by 90^\circ: phasor AB, drawn downward.

Derivation Impedance of an RC series circuit

Aim. Find the impedance Z and the phase angle \varphi for R and C in series.

Fig 20.14 (b): phasor diagram of the RC series circuit. V lags I by \varphi.

The applied voltage is the phasor sum OB of V_R and V_C.
\vec V=\vec V_R+\vec V_C
Pythagoras' theorem in the right-angled triangle OAB.
V^2=V_R^2+V_C^2
Substitute V_R=IR and V_C=IX_C. The book writes (-IX_C)^2 to show that V_C points downward; the sign disappears on squaring.
V=\sqrt{(IR)^2+(-IX_C)^2}=I\sqrt{R^2+X_C^2}
Divide by I to get the impedance.
Z=\dfrac{V}{I}=\sqrt{R^2+X_C^2}\qquad(20.11)
Phase angle from the same triangle.
\tan\varphi=\dfrac{V_C}{V_R}=\dfrac{X_C}{R}

ResultZ=\sqrt{R^2+X_C^2}=\sqrt{R^2+\left(\dfrac{1}{2\pi fC}\right)^2},\qquad\tan\varphi=\dfrac{X_C}{R}

In an RC series circuit the current leads the supply voltage by an angle \varphi between 0 and 90^\circ. The impedance is measured in ohms and decreases when the frequency increases.

Fig 20.14 (c): RC series impedance triangle.

RL and RC series circuits compared
	RL series	RC series
Reactive voltage	V_L=IX_L leads I (drawn up)	V_C=IX_C lags I (drawn down)
Impedance	Z=\sqrt{R^2+X_L^2}	Z=\sqrt{R^2+X_C^2}
Phase angle	\tan\varphi=X_L/R	\tan\varphi=X_C/R
Current compared with V	lags by \varphi	leads by \varphi
When f increases	Z increases	Z decreases

20.5RLC series circuit

A circuit in which a resistor R, an inductor L and a capacitor C are connected in series is called an RLC series circuit. All the elements share the same current I. (The book gives this circuit in a 'For your information' box.)

RLC series circuit.

V_R=IR is in phase with I.
V_L=IX_L leads I by 90^\circ (drawn up).
V_C=IX_C lags I by 90^\circ (drawn down).

V_L and V_C point in opposite directions, so they partly cancel. Their resultant is V_L-V_C (taking V_L>V_C). The supply voltage is the phasor sum of V_R and V_L-V_C.

Derivation Impedance of an RLC series circuit

Aim. Find the impedance of R, L and C in series.

Phasor diagram of the RLC series circuit (book's 'For your information' box). V_L and V_C are opposite; the supply voltage is the sum of V_R and V_L-V_C.

Combine the two opposite reactive voltages first.
V_L-V_C=I(X_L-X_C)
This resultant is at 90^\circ to V_R, so use Pythagoras' theorem.
V^2=V_R^2+(V_L-V_C)^2
Substitute and take out the common factor I.
V=\sqrt{(IR)^2+I^2(X_L-X_C)^2}=I\sqrt{R^2+(X_L-X_C)^2}
Divide by I.
Z=\dfrac{V}{I}=\sqrt{R^2+(X_L-X_C)^2}
Phase angle between the supply voltage V_s and the current.
\tan\theta=\dfrac{V_L-V_C}{V_R}=\dfrac{X_L-X_C}{R}

ResultZ=\sqrt{R^2+(X_L-X_C)^2},\qquad\tan\theta=\dfrac{X_L-X_C}{R}

The net reactance is X_L-X_C. If X_L>X_C the circuit behaves inductively and the current lags the voltage; if X_C>X_L it behaves capacitively and the current leads. The RL and RC formulas are the special cases X_C=0 and X_L=0.

Formula sheet

Alternating current and voltage

I=I_0\sin\omega t,\qquad V=V_0\sin\omega t

I_0, V_0: peak values; \omega: angular frequency; t: time

Units: A, V

Angular frequency, frequency and period

\omega=2\pi f=\dfrac{2\pi}{T},\qquad f=\dfrac1T

f: frequency; T: time period

Units: \omega in \mathrm{rad\,s^{-1}}, f in Hz, T in s

Average value from a graph

\text{average}=\dfrac{\text{net area under the curve in time }T}{T}

areas below the time axis count as negative; zero over a full cycle of a sine wave

same unit as the quantity

r.m.s. values (sinusoidal AC)

I_{\mathrm{rms}}=\dfrac{I_0}{\sqrt2}=0.707\,I_0,\qquad V_{\mathrm{rms}}=\dfrac{V_0}{\sqrt2}

I_0, V_0: peak values

Units: A, V

Instantaneous power in a resistor

P=I^2R=I_0^2R\sin^2\omega t

R: resistance

Unit: W

Mean power in a resistor

\langle P\rangle=\tfrac12I_0^2R=\tfrac12P_{\max}=I_{\mathrm{rms}}^2R=V_{\mathrm{rms}}I_{\mathrm{rms}}

P_{\max}=I_0^2R

Unit: W

Time constant of smoothing capacitor

\tau=RC

R: load resistance; C: smoothing capacitance

Unit: s

Self-induced emf

\varepsilon_L=-L\dfrac{\Delta I}{\Delta t}

L: self-inductance; \Delta I/\Delta t: rate of change of current

Unit: V

Self-inductance

L=\dfrac{-\varepsilon_L}{\Delta I/\Delta t},\qquad N\Phi=LI

N\Phi: flux linkage of the coil

Units: henry, 1\,\mathrm{H}=1\,\mathrm{V\,s\,A^{-1}}

Self-inductance of a solenoid

L=\dfrac{\mu_r\mu_0N^2A}{l}

N: turns; A: area; l: length; \mu_r: relative permeability of the core

Unit: H

Mutually induced emf

\varepsilon_2=-M\dfrac{\Delta I_1}{\Delta t}

M: mutual inductance; I_1: current in coil 1

Unit: V

Mutual inductance

M=\dfrac{-\varepsilon_2}{\Delta I_1/\Delta t}

Unit: H

Phase difference

\Phi=\dfrac{\Delta t}{T}\times360^\circ

\Delta t: time between the peaks of the two quantities

Units: degree or radian

Resistor

I=I_0\sin\omega t,\qquad I_0=\dfrac{V_0}{R}

V and I in phase

Unit: A

Inductor

V=V_0\sin\left(\omega t+\dfrac{\pi}{2}\right),\qquad V_0=\omega LI_0

V leads I by 90^\circ

Unit: V

Inductive reactance

X_L=\dfrac{V_0}{I_0}=\omega L=2\pi fL

L: inductance; f: frequency

Unit: \Omega

Capacitor

I=I_0\sin\left(\omega t+\dfrac{\pi}{2}\right),\qquad I_0=\omega CV_0

I leads V by 90^\circ

Unit: A

Capacitive reactance

X_C=\dfrac{V_0}{I_0}=\dfrac{1}{\omega C}=\dfrac{1}{2\pi fC}

C: capacitance; f: frequency

Unit: \Omega

Impedance

Z=\dfrac{V}{I}=\dfrac{V_{\mathrm{rms}}}{I_{\mathrm{rms}}}

total opposition of an AC circuit

Unit: \Omega

RL series circuit

Z=\sqrt{R^2+X_L^2},\qquad\tan\varphi=\dfrac{X_L}{R}

current lags the voltage by \varphi

Unit: \Omega

RC series circuit

Z=\sqrt{R^2+X_C^2},\qquad\tan\varphi=\dfrac{X_C}{R}

current leads the voltage by \varphi

Unit: \Omega

RLC series circuit

Z=\sqrt{R^2+(X_L-X_C)^2},\qquad\tan\theta=\dfrac{X_L-X_C}{R}

X_L-X_C: net reactance

Unit: \Omega

Key terms

Direct current (DC): Current that flows in one direction only.
Alternating current (AC): Current that changes its direction (polarity) at regular intervals of time; for mains AC it varies as I=I_0\sin\omega t.
Instantaneous value: The value of an alternating quantity at a particular instant.
Cycle: One complete set of positive and negative values of an alternating quantity.
Time period: The time taken to complete one cycle of an alternating quantity.
Frequency: The number of cycles completed in one second; unit hertz (Hz). In Pakistan AC has f=50\,\mathrm{Hz}.
Angular frequency: \omega=2\pi f, the rate at which the phase angle increases, in \mathrm{rad\,s^{-1}}.
Peak value: The maximum value of an alternating current or voltage (I_0 or V_0).
Average value: The average of all values of a waveform over a period of time; zero over a full cycle of a sine wave.
Root-mean-square (r.m.s.) value: The steady current which produces the same heat in a resistor as the alternating current in the same time; I_{\mathrm{rms}}=I_0/\sqrt2.
Mean power: The average rate of energy transfer over a cycle; in a resistor it is half the maximum power.
Rectification: The process of converting AC voltage into DC voltage.
Rectifier: A circuit, using diodes, that converts AC into DC.
Half-wave rectification: Rectification in which one half-cycle of the AC passes and the other half is blocked; uses one diode.
Full-wave rectification: Rectification in which the complete cycle of AC is converted into pulsating DC.
Bridge rectifier: A full-wave rectifier made of four diodes arranged as a bridge.
Pulsating DC: Current that flows in one direction only but whose value rises and falls.
Filtering (smoothing): Use of a capacitor in parallel with the load to turn pulsating DC into a nearly constant DC.
Ripple: The small rise and fall that remains in a smoothed DC output.
Self-induction: The phenomenon in which a changing current induces an emf in the coil itself.
Self-inductance: The ratio of the emf induced in a coil to the time rate of change of current through it, L=-\varepsilon_L/(\Delta I/\Delta t).
Back emf: The self-induced emf, which opposes the change in current that produces it (Lenz's law).
Mutual induction: The phenomenon in which a changing current in one coil induces an emf in a neighbouring coil.
Mutual inductance: The ratio of the emf induced in one coil to the time rate of change of current in the other, M=-\varepsilon_2/(\Delta I_1/\Delta t).
Henry: The SI unit of self and mutual inductance: 1\,\mathrm{H}=1\,\mathrm{V\,s\,A^{-1}}.
Inductor: A component, usually a coil of wire, that opposes changes in the current through it.
Choke: An inductor used to offer a high reactance to AC above a certain frequency without appreciably limiting DC or low-frequency current, and with very little power loss.
Phase: The angle \omega t that shows the stage an alternating quantity has reached in its cycle.
Phase difference: The fraction of a period between the peaks of two alternating quantities of the same frequency, expressed as an angle.
Lead and lag: The quantity that reaches its zero or peak earlier leads; the other lags.
Phasor: A rotating arrow whose length is the peak value and whose vertical projection gives the instantaneous value of an alternating quantity.
Reactance: The opposition of an inductor or a capacitor to AC; it depends on frequency and consumes no mean power.
Inductive reactance: The opposition offered by an inductor to AC, X_L=2\pi fL.
Capacitive reactance: The opposition offered by a capacitor to AC, X_C=1/(2\pi fC).
Impedance: The combined effect of resistance and reactance in an AC circuit, Z=V/I; the vector sum of resistance and reactance.
Impedance triangle: A right-angled triangle with sides R and X and hypotenuse Z.
RLC series circuit: A circuit in which a resistor, an inductor and a capacitor are connected in series and carry the same current.

Multiple choice questionsp. 121–122

A full wave bridge rectifier consists of

Ano diode
Bone diode
CTwo diodes
Dfour diodesAnswer

Why

A bridge rectifier uses four diodes (D_1 to D_4) connected as a bridge. Option C is tempting: two diodes are used in the centre-tapped full-wave rectifier, not in the bridge. A half-wave rectifier uses one diode.

Unit of reactance is

Ahenry
BohmsAnswer
Cvolt
Dampere

Why

Reactance is the ratio of voltage to current, X=V_0/I_0, exactly like resistance, so it is measured in ohms. The henry is the unit of inductance L, not of the reactance X_L=2\pi fL.

The unit of self-inductance, henry (H), is equal to:

A\mathrm{A\,s^{-1}\,V^{-1}}
B\mathrm{V\,A\,s^{-1}}
C\mathrm{V\,s^{-1}\,A^{-1}}
D\mathrm{V\,s\,A^{-1}}Answer

Why

From L=\dfrac{\varepsilon_L}{\Delta I/\Delta t}, the unit is \dfrac{\mathrm{V}}{\mathrm{A\,s^{-1}}}=\mathrm{V\,s\,A^{-1}}. Option C has the second in the wrong place.

A capacitor is perfectly insulator for

Adirect currentAnswer
Balternating current
Cdirect as well as alternating current
Dneither for direct current nor for alternating current

Why

For DC the frequency is zero, so X_C=\dfrac{1}{2\pi fC} is infinite: once the capacitor is charged no steady current flows. With AC the capacitor charges and discharges continuously, so an alternating current does flow in the circuit.

The peak value of alternating current is 5\sqrt2\,\mathrm{A}. The mean square value of current will be

A5\,\mathrm{A}
B2.5\,\mathrm{A}
C5\sqrt2\,\mathrm{A}
D5^2\,\mathrm{A}Answer

Why

The mean square value is the mean of I^2 over a cycle: \langle I^2\rangle=\dfrac{I_0^2}{2}=\dfrac{(5\sqrt2)^2}{2}=\dfrac{50}{2}=25=5^2. Its unit is \mathrm{A^2} (printed as A in option D). The root mean square value is the square root of this, I_{\mathrm{rms}}=I_0/\sqrt2=5\,\mathrm{A}, which is option A.

Note. The question literally asks for the mean square value, which is 25\,\mathrm{A^2}=5^2 (option D). Many keys treat it as a misprint for 'root mean square value', whose answer is option A (5\,\mathrm{A}). In an exam, read the wording: 'mean square' gives D, 'root mean square' gives A.

In AC system we generate sine wave form because

AIt can be easily draw
BIt produces least disturbance in electrical circuitsAnswer
CIt is nature standard
DOther waves cannot be produced easily.

Why

A sine wave changes smoothly, and it is the only waveform whose shape is not changed by resistors, inductors and capacitors (its rate of change is again a sine wave). So it produces the least disturbance (distortion, noise and extra heating) in electrical circuits and machines. A rotating generator does give a sine wave naturally, but 'nature standard' is not a physical reason.

An alternating voltage is given by 20\sin(157t). The frequency of alternating voltage is

A50\,\mathrm{Hz}
B25\,\mathrm{Hz}Answer
C100\,\mathrm{Hz}
D75\,\mathrm{Hz}

Why

Comparing with V=V_0\sin\omega t gives \omega=157\,\mathrm{rad\,s^{-1}}. Then f=\dfrac{\omega}{2\pi}=\dfrac{157}{6.28}=25\,\mathrm{Hz}. Option A (50 Hz) would need \omega=314\,\mathrm{rad\,s^{-1}}.

Peak value of an alternating quantity (voltage or current) can be found if we know it's

Aphase
Br.m.s valueAnswer
Cfrequency
Dboth (a) & (b)

Why

For a sinusoidal quantity x_0=\sqrt2\,x_{\mathrm{rms}}, so the r.m.s. value alone fixes the peak value. The phase and the frequency say nothing about the size of the wave, so options A, C and D are wrong.

Short questionsp. 122

13 marks

Define (a) mutual-induction (b) self-induction.

Solution

(a) Mutual induction is the phenomenon in which a changing current in one coil induces an emf in a neighbouring coil. The changing magnetic flux of the first coil (primary) also links the second coil (secondary), so an emf \varepsilon_2=-M\,\dfrac{\Delta I_1}{\Delta t} is induced in it, where M is the mutual inductance. A transformer works on this principle.

(b) Self-induction is the phenomenon in which a changing current in a coil induces an emf in the coil itself, because the coil's own magnetic flux changes. The induced (back) emf is \varepsilon_L=-L\,\dfrac{\Delta I}{\Delta t}, where L is the self-inductance. The minus sign (Lenz's law) shows that this emf opposes the change in current.

23 marks

Does the SI unit used for mutual-inductance and self-inductance are same? Explain briefly.

Solution

Yes. Both have the same SI unit, the henry (H), because both are defined in the same way: an induced emf divided by a rate of change of current.

L=\dfrac{\varepsilon_L}{\Delta I/\Delta t},\qquad M=\dfrac{\varepsilon_2}{\Delta I_1/\Delta t}\qquad\Longrightarrow\qquad1\,\mathrm{H}=\dfrac{1\,\mathrm{V}}{1\,\mathrm{A\,s^{-1}}}=1\,\mathrm{V\,s\,A^{-1}}

A coil has a self-inductance of 1 H if an emf of 1 V is induced in it when its own current changes at 1\,\mathrm{A\,s^{-1}}. Two coils have a mutual inductance of 1 H if 1 V is induced in one when the current in the other changes at 1\,\mathrm{A\,s^{-1}}. The only difference is where the emf appears: in the same coil, or in the neighbouring coil.

33 marks

Draw the circuit for a half wave rectifier and full wave rectifier.

Solution

Half-wave rectifier

Half-wave rectifier: one diode D in series with the load R.

One diode D is connected in series with the load R. It conducts only during the positive half-cycles, so only they appear across R.

Full-wave (bridge) rectifier

Full-wave bridge rectifier: four diodes; AC input at A and B; load R across the other two corners.

Four diodes form a bridge. The AC input is connected to corners A and B and the load R to the other two corners. D_1 and D_2 conduct in the positive half-cycle and D_3 and D_4 in the negative half-cycle; in both, the current through R flows from top to bottom.

43 marks

What is the use of a single diode for the half-wave rectification of an alternating current?

Solution

A diode conducts only when it is forward biased, so it acts as a one-way valve. In a half-wave rectifier a single diode is connected in series with the load resistor R. During the positive half-cycle the diode is forward biased, acts as a closed switch, and current flows through R. During the negative half-cycle it is reverse biased, acts as an open switch, and no current flows. So the diode passes one half of each cycle and blocks the other half: the current through the load flows in one direction only (pulsating DC).

Half-wave rectification

Fig 20.5 redrawn. Only the positive half-cycles reach the load. The negative half-cycles are blocked, so the output is zero for half of every cycle. One pulse per cycle.

52026 Annual I, Set B, Q.2 (xiii) OR3 marks

Distinguish graphically between half-wave and full-wave rectification.

Solution

Half-wave rectification: only the positive half-cycles appear in the output. The output is zero during each negative half-cycle, so there is one pulse per input cycle.

Half-wave rectification

Fig 20.5 redrawn. Only the positive half-cycles reach the load. The negative half-cycles are blocked, so the output is zero for half of every cycle. One pulse per cycle.

Full-wave rectification: both half-cycles appear in the output, the negative ones turned upward, so the current through the load is in one direction all the time. There are two pulses per input cycle.

Full-wave rectification

Fig 20.7 redrawn. Both half-cycles reach the load and both drive current the same way through it, so the negative halves appear flipped upward. Two pulses per cycle.

So a half-wave output has gaps as long as its pulses, while a full-wave output has no gaps, twice the pulse frequency (2f instead of f) and twice the average value (2V_0/\pi instead of V_0/\pi).

63 marks

Is the frequency content of the output of a half wave rectifier and full wave rectifier the same? Explain briefly.

Solution

No. In a half-wave rectifier only one half of each input cycle reaches the load, so the output has one pulse per cycle and repeats at the input frequency f. In a full-wave rectifier both halves reach the load, so there are two pulses per cycle and the output repeats at 2f.

\begin{aligned}f_{\text{half-wave output}}&=f=50\,\mathrm{Hz}\\ f_{\text{full-wave output}}&=2f=100\,\mathrm{Hz}\end{aligned}\qquad(\text{for }50\,\mathrm{Hz}\text{ mains})

Both outputs also contain a steady DC part (average values V_0/\pi and 2V_0/\pi). Because the full-wave ripple has twice the frequency and shorter gaps, it is easier to smooth with a capacitor.

73 marks

In half wave rectifier, the half of the signal is blocked after the rectification. Why?

Solution

A half-wave rectifier has only one diode, in series with the load. During the negative half-cycle of the input, the p-side of the diode is at a lower potential than its n-side, so the diode is reverse biased. It then offers a very high resistance and acts as an open switch: no current flows through the load and the output is zero. The diode conducts only when it is forward biased, in the positive half-cycle. Hence one half of every cycle (the negative half) is blocked. To use both halves, four diodes are arranged as a bridge (full-wave rectifier).

82026 Annual I, Set D, Q.2 (xiii) OR3 marks

Distinguish between root-mean-square (r.m.s.) and peak values for a sinusoidal alternating current.

Solution

Peak value I_0: the maximum value of the alternating current. It is reached twice in each cycle, as +I_0 and -I_0, and only for an instant.

r.m.s. value I_{\mathrm{rms}}: the steady (DC) current which produces the same heat in a resistor as the alternating current does in the same time. It is the square root of the mean of I^2 over a cycle. For a sinusoidal current:

I_{\mathrm{rms}}=\dfrac{I_0}{\sqrt2}=0.707\,I_0\qquad\text{or}\qquad I_0=\sqrt2\,I_{\mathrm{rms}}

AC meters read r.m.s. values, and power is calculated with them (\langle P\rangle=I_{\mathrm{rms}}^2R). For example, a current of I_{\mathrm{rms}}=7\,\mathrm{A} has a peak value of 9.9\,\mathrm{A} and heats a resistor like 7\,\mathrm{A} DC.

93 marks

What is impedance? How is it related to resistance, reactance, and frequency? Also find its SI unit?

Solution

Impedance Z is the combined (total) opposition of the resistance and the reactance in an AC circuit: Z=V/I.

Resistance and reactance: the voltage across a reactance is 90^\circ out of phase with the voltage across the resistance, so R and X add as vectors (at right angles):

Z=\sqrt{R^2+X^2},\qquad X=X_L-X_C

Frequency: the reactances depend on frequency, X_L=2\pi fL and X_C=\dfrac{1}{2\pi fC}. So for an RL circuit Z=\sqrt{R^2+(2\pi fL)^2} increases with f, and for an RC circuit Z=\sqrt{R^2+(1/2\pi fC)^2} decreases with f.

SI unit: Z=V/I, so its unit is \mathrm{V\,A^{-1}}, the ohm (\Omega).

103 marks

What is the difference between reactance and impedance?

Solution

Reactance X is the opposition to AC offered by an inductor (X_L=2\pi fL) or a capacitor (X_C=1/2\pi fC). Impedance Z is the total opposition of a whole AC circuit: the vector sum of its resistance and reactance, Z=\sqrt{R^2+X^2}.

Reactance	Impedance
due to inductors and capacitors only	due to resistance and reactance together
V and I differ in phase by exactly 90^\circ	phase difference \varphi between 0 and 90^\circ, \tan\varphi=X/R
no mean power is used	mean power I_{\mathrm{rms}}^2R is used in the resistance
always depends on frequency	depends on frequency through its reactance part

Both are measured in ohms. For a pure inductor or a pure capacitor the impedance equals the reactance.

113 marks

Why does a capacitor block DC but allow AC to pass through?

Solution

The plates of a capacitor are separated by an insulator, so no charge can cross the gap.

DC: when a steady voltage is applied, charge flows only until the capacitor is charged to the supply voltage; then the current stops. So a capacitor blocks DC. In terms of reactance, f=0 gives X_C=\dfrac{1}{2\pi fC}=\infty.

AC: the voltage keeps reversing, so the capacitor is charged, discharged and charged the other way again and again. Electrons keep moving to and fro in the connecting wires, which is an alternating current in the circuit. The reactance X_C=1/(2\pi fC) is finite, and smaller at higher frequencies, so AC passes, more easily at high frequency.

122026 Annual I, Set B, Q.2 (v) OR2026 Annual I, Old syllabus (Curriculum 2006), Q.2 (xiii) OR, old syllabus3 marks

How does the reactance of a capacitor change with an increase in frequency?

Solution

The capacitive reactance decreases when the frequency increases, because it is inversely proportional to the frequency:

X_C=\dfrac{1}{2\pi fC}\qquad\Longrightarrow\qquad X_C\propto\dfrac1f

Reason: at a higher frequency the capacitor is charged and discharged more times per second, so more charge flows per second for the same voltage. A larger current for the same voltage means a smaller opposition. For example, doubling the frequency halves the reactance: X_C'=\dfrac{1}{2\pi(2f)C}=\dfrac{X_C}{2}. For DC (f=0) X_C is infinite, and at very high frequencies it approaches zero.

Capacitive reactance against frequency

X_C is inversely proportional to f: when the frequency is doubled, the reactance is halved.

133 marks

What is the phase difference between voltage and current in a (a) capacitor (b) Inductor?

Solution

(a) Capacitor: the voltage lags the current by 90^\circ (\pi/2 rad), that is, the current leads. If V=V_0\sin\omega t, then I=I_0\sin(\omega t+\pi/2). The current is largest when the voltage is zero, because the voltage is then changing fastest.

(b) Inductor: the voltage leads the current by 90^\circ (\pi/2 rad). If I=I_0\sin\omega t, then V=V_0\sin(\omega t+\pi/2). The back emf of the coil delays the current.

Phasor diagrams with the current I as the reference: inductor (V ahead of I) and capacitor (V behind I).

Memory aid: ELI the ICE man (in L, E before I; in C, I before E).

143 marks

How does the capacitance of a capacitor affect the current through it in an AC circuit?

Solution

For a given supply voltage and frequency, the current through a capacitor is directly proportional to its capacitance:

I=\dfrac{V}{X_C}=\dfrac{V}{1/(2\pi fC)}=2\pi fC\,V\qquad\Longrightarrow\qquad I\propto C

A larger capacitance has a smaller reactance, X_C=1/(2\pi fC). Physically, a larger capacitor stores more charge at the same voltage, so more charge flows to and fro in every cycle, and the current is larger. Doubling C halves X_C and doubles the current; a very small capacitance allows only a very small current.

Comprehensive questionsp. 122–123

What is half wave rectification? Explain.

Solution

Definition

Rectification is the process of converting AC voltage into DC voltage. In half-wave rectification an AC signal is converted into DC by passing one half-cycle of the waveform and blocking the other half.

Circuit

Half-wave rectifier: a single diode D in series with the load resistance R.

A half-wave rectifier is made with one diode D connected in series with the load resistance R across the AC supply (usually the secondary of a step-down transformer).

Working

Positive half-cycle: the p-side of the diode is at the higher potential, so D is forward biased. It offers a very low resistance and acts as a closed switch. Current flows through R and the output voltage follows the input.
Negative half-cycle: the polarity reverses and D is reverse biased. It offers a very high resistance and acts as an open switch. No current flows through R, so the output voltage is zero.
The process repeats every cycle. The current through R always flows in the same direction, but only in alternate half-cycles.

Effective circuit: (a) forward-biased diode as a closed switch; (b) reverse-biased diode as an open switch.

Input and output waveforms

Half-wave rectification

Fig 20.5 redrawn. Only the positive half-cycles reach the load. The negative half-cycles are blocked, so the output is zero for half of every cycle. One pulse per cycle.

Nature of the output

The output is pulsating DC: it flows in one direction but its value rises and falls, and it is zero for half the time. There is one pulse per input cycle, so the output repeats at the supply frequency (50 Hz for mains). Its average value is only V_0/\pi\approx0.32\,V_0.

Advantage: simple and cheap, only one diode.
Disadvantage: half of every cycle is wasted (low efficiency), and the long gaps make the output hard to smooth.

To obtain a nearly constant DC, a filter such as a capacitor is connected in parallel with the load. It charges to the peak voltage and discharges slowly through the load during the gaps.

22026 Annual I, Set D, Q.5

With the help of a diagram, explain the operation of a bridge rectifier.

Solution

Bridge rectifier

A bridge rectifier is a full-wave rectifier made of four diodes D_1, D_2, D_3 and D_4 connected in a closed loop (a bridge). The AC input is applied across two opposite corners A and B, and the load resistor R is connected across the other two corners.

Circuit diagram

Full-wave bridge rectifier. The small hump shows two wires crossing without joining.

Positive half-cycle

End A becomes positive and end B negative. Diodes D_1 and D_2 are forward biased and conduct; D_3 and D_4 are reverse biased and do not conduct. Current flows from A through D_1, through the load R from top to bottom, through D_2, and back to B.

Positive half-cycle: current path A, D_1, R, D_2, B.

Negative half-cycle

End A becomes negative with respect to B. Now D_3 and D_4 are forward biased and conduct, while D_1 and D_2 are reverse biased. Current flows from B through D_3, through the load R again from top to bottom, through D_4, and back to A.

Negative half-cycle: current path B, D_3, R, D_4, A.

Output

One pair of diodes conducts in each half-cycle, and in both half-cycles the current through the load flows in the same direction. So the output across R is full-wave rectified (pulsating) DC: the negative half-cycles appear turned upward.

Full-wave rectification

Fig 20.7 redrawn. Both half-cycles reach the load and both drive current the same way through it, so the negative halves appear flipped upward. Two pulses per cycle.

Advantages

The whole input cycle is used, so the efficiency is high and the average output (2V_0/\pi) is twice that of a half-wave rectifier.
No centre-tapped transformer is needed.
The output pulses at twice the supply frequency (100 Hz for 50 Hz mains), so it is easier to smooth with a capacitor.

32026 Annual I, Set B, Q.2 (iii) OR

Analyze the effect of a single capacitor in smoothing current flow.

Solution

Need for smoothing

The output of a rectifier is pulsating DC: it flows in one direction, but its value keeps rising and falling. Electronic circuits need a nearly constant DC. A single capacitor connected in parallel with the load smooths the output; this is called filtering.

Circuit

Bridge rectifier with a smoothing capacitor C in parallel with the load R.

Action of the capacitor

As the rectified voltage rises to its peak, the diodes conduct and the capacitor charges up to the peak voltage V_0.
When the rectified voltage starts to fall, it drops below the capacitor voltage. The diodes become reverse biased, and the capacitor discharges slowly through the load, supplying the current.
Before the capacitor has discharged much, the next pulse arrives and recharges it to V_0.
So the capacitor fills the gaps between the peaks. The output stays close to V_0 with only a small variation called the ripple.

Smoothing by a capacitor

Effect of capacitance

The capacitor discharges through the load as V=V_0e^{-t/RC}, so the fall between pulses depends on the time constant \tau=RC. A larger capacitance stores more charge, discharges more slowly and gives a smaller ripple: better smoothing. However, a larger capacitance also draws larger charging current pulses through the diodes.

Effect of load resistance

A larger load resistance (smaller load current) makes the capacitor discharge more slowly, so the ripple is smaller. A smaller load resistance (larger load current) makes it discharge faster, which reduces its ability to smooth the output: the ripple is larger.

\tau=RC\gg\text{time between pulses},\qquad\Delta V\approx\dfrac{V_0}{2fRC}\ \ (\text{full-wave})

Good smoothing needs RC\gg0.01\,\mathrm{s} for full-wave rectification of 50 Hz mains. The ripple \Delta V falls as C, R or f increases.

Conclusion

A single capacitor converts pulsating DC into nearly steady DC. The smoothing is better for a larger capacitance, a larger load resistance and a full-wave (rather than half-wave) rectified input, whose gaps are half as long.

42026 Annual I, Set B, Q.2 (iv)2026 Annual I, Old syllabus (Curriculum 2006), Q.6 OR, old syllabus

Explain the phenomenon of self-induction.

Solution

Self-induction

Self-induction is the phenomenon in which a changing current in a coil induces an emf in the coil itself.

Explanation

A coil connected to a battery through a switch and a rheostat.

A coil is connected to a battery through a switch and a rheostat. The current produces a magnetic field, and its flux passes through the coil itself. If the rheostat is moved quickly, the current changes, so the magnetic field and the flux through the coil change. By Faraday's law, the changing flux induces an emf in the same coil.

Self-induced emf and self-inductance

The induced emf is proportional to the rate of change of current:

\varepsilon_L=-L\dfrac{\Delta I}{\Delta t}\qquad\Longrightarrow\qquad L=\dfrac{-\varepsilon_L}{\Delta I/\Delta t}

The constant L is the self-inductance of the coil: the ratio of the emf induced in the coil to the time rate of change of current through it.

Back emf

The negative sign is due to Lenz's law: the self-induced emf opposes the change that produces it. When the current is switched on or increased, the induced emf opposes the growth of current; when the current is switched off or decreased, it tries to keep the current flowing. For this reason it is called the back emf. Because the coil opposes changes of current, it is called an inductor or a choke.

Unit

1\,\mathrm{H}=1\,\mathrm{V\,s\,A^{-1}}

The self-inductance of a coil is one henry if an emf of one volt is induced in it when the current through it changes at the rate of one ampere per second.

Factors on which L depends

L=\dfrac{\mu_r\mu_0N^2A}{l}\qquad(\text{long solenoid})

number of turns N (L\propto N^2);
area of cross-section A of the coil (L\propto A);
length l of the coil for the same turns (L\propto1/l);
nature of the core, its relative permeability \mu_r (a soft-iron core increases L greatly).

Self-inductance does not depend on the current itself. Inductors are used as chokes in AC circuits, where they limit alternating current with very little loss of energy.

A sinusoidal alternating voltage of angular frequency \omega is connected across an inductor. Find mathematical expression for instantaneous voltage and instantaneous current.

Solution

Circuit

An alternating voltage source connected across a pure inductor L.

Let an ideal inductor of inductance L (zero resistance) be connected to an alternating source of angular frequency \omega. The changing current induces a back emf L\,\Delta I/\Delta t in the coil. Since there is no resistance, the applied voltage at every instant just balances this back emf.

Derivation Instantaneous current and voltage in an inductor

Aim. Find I and V as functions of time.

Let the instantaneous current be
I=I_0\sin\omega t\qquad(\text{i})
Applied voltage = back emf.
V=L\dfrac{\Delta I}{\Delta t}=LI_0\dfrac{\Delta(\sin\omega t)}{\Delta t}
The rate of change of \sin\omega t is \omega\cos\omega t.
V=\omega LI_0\cos\omega t
Write V_0=\omega LI_0 and \cos\omega t=\sin(\omega t+\pi/2).
V=V_0\sin\left(\omega t+\dfrac{\pi}{2}\right)\qquad(\text{ii})
Equivalently, taking the applied voltage as the reference, V=V_0\sin\omega t, the current is a quarter cycle behind it.
I=\dfrac{V_0}{\omega L}\sin\left(\omega t-\dfrac{\pi}{2}\right)

ResultI=I_0\sin\omega t,\qquad V=V_0\sin\left(\omega t+\dfrac{\pi}{2}\right),\qquad V_0=\omega LI_0

The instantaneous voltage leads the instantaneous current by \pi/2 radians (90^\circ): the current lags the voltage by 90^\circ.

Phase relation

Inductor: voltage leads current by 90°

Fig 20.11 (b) redrawn. The voltage reaches its peak a quarter cycle (90^\circ) before the current. The voltage is largest when the current is zero, because that is when the current changes fastest.

Phasor diagram: V leads I by 90^\circ.

The voltage is greatest when the current is zero, because the current is then changing fastest and the back emf is largest. The inductance opposes changes in current, so the current reaches each stage of its cycle a quarter cycle after the voltage.

Inductive reactance

X_L=\dfrac{V_0}{I_0}=\omega L=2\pi fL

The opposition of the inductor grows with frequency. For DC (\omega=0) it is zero. The mean power taken by a pure inductor over a cycle is zero, because the energy stored in its magnetic field is returned to the source.

62026 Annual I, Set B, Q.5

Explain the term impedance of an AC circuit. Find its expression for the RL series circuit.

Solution

Impedance

Impedance is the combined effect of the resistance and the reactance in an AC circuit. It is the total opposition to alternating current: Z=V/I. Its SI unit is the ohm. Because the voltages across resistance and reactance are 90^\circ out of phase, they are added as phasors (vectors), not as ordinary numbers.

RL series circuit

RL series circuit.

A resistor R and a pure inductor L are in series, so the same current I flows through both. V_R=IR is in phase with I, and V_L=IX_L leads I by 90^\circ.

Derivation Impedance of the RL series circuit

Aim. Find Z in terms of R and X_L.

Phasor diagram of the RL series circuit.

Taking I as the reference, draw V_R along OA and V_L along AB, perpendicular to OA. The supply voltage V is the resultant OB.
\vec V=\vec V_R+\vec V_L
Pythagoras' theorem in the right-angled triangle OAB.
V^2=V_R^2+V_L^2
Substitute V_R=IR and V_L=IX_L.
V=\sqrt{(IR)^2+(IX_L)^2}=I\sqrt{R^2+X_L^2}
Divide by I.
Z=\dfrac{V}{I}=\sqrt{R^2+X_L^2}
The angle between V and I.
\tan\varphi=\dfrac{V_L}{V_R}=\dfrac{X_L}{R}

ResultZ=\sqrt{R^2+X_L^2}=\sqrt{R^2+(2\pi fL)^2}

The impedance depends on R, L and the frequency f. The current lags the applied voltage by \varphi=\tan^{-1}(X_L/R), which lies between 0 and 90^\circ.

Impedance triangle: Z is the vector sum of R and X_L.

For example, with R=30\,\Omega and X_L=40\,\Omega, Z=\sqrt{30^2+40^2}=50\,\Omega, not 70\,\Omega.

72026 Annual I, Set B, Q.5

In an RL series circuit will the current lag or lead the applied alternating voltage? Explain the answer with a phasor diagram.

Solution

Answer

In an RL series circuit the current lags behind the applied voltage by an angle \varphi, where 0<\varphi<90^\circ.

Explanation

RL series circuit.

R and L are in series, so the same current I flows through both. The current is therefore taken as the reference phasor (horizontal).
The voltage across the resistor, V_R=IR, is in phase with I: phasor OA along the current.
The voltage across the inductor, V_L=IX_L, leads I by 90^\circ: phasor AB, perpendicular to OA.
The applied voltage V is the phasor sum OB. It lies between V_R and V_L, so it is ahead of the current by the angle \varphi. Equivalently, the current lags the voltage by \varphi.

Phasor diagram

Phasor diagram of the RL series circuit: the supply voltage V leads the current I by \varphi.

Angle of lag

\tan\varphi=\dfrac{V_L}{V_R}=\dfrac{X_L}{R}=\dfrac{2\pi fL}{R},\qquad V=\sqrt{V_R^2+V_L^2},\qquad Z=\sqrt{R^2+X_L^2}

The lag is less than 90^\circ because of the resistance. If X_L\gg R the angle approaches 90^\circ (like a pure inductor); if R\gg X_L it approaches 0 (like a pure resistor). The inductance causes the lag because its back emf opposes every change in current.

Describe the phase of A.C and how phase lags and leads in A.C circuits.

Solution

Phase

The phase of an alternating quantity x=x_0\sin\omega t at an instant is the angle \omega t. It shows how far the quantity has advanced in its cycle, so it fixes both its value and the direction in which it is changing.

Phase difference

When two alternating quantities of the same frequency (such as the voltage and the current in a circuit) do not reach their zero values or peaks at the same time, they have a phase difference \Phi: the fraction of a period between their peaks, expressed in degrees or radians. Because the frequencies are the same, \Phi stays constant.

\Phi=\dfrac{\Delta t}{T}\times360^\circ

Phase difference

Lead and lag

The quantity that passes through its zero (or peak) earlier leads; the other lags. For V=V_0\sin\omega t and I=I_0\sin(\omega t-\Phi) the current lags the voltage by \Phi. It is customary to quote the angle by which the voltage leads the current; in RLC circuits it is never more than 90^\circ.

Phasor representation

A rotating phasor and the sine wave traced by its projection.

An alternating quantity is represented by a phasor: an arrow of length equal to its peak value, rotating anticlockwise at angular velocity \omega. Its vertical projection is the instantaneous value. The angle between two phasors is their phase difference, and the one ahead (anticlockwise) leads.

Lead and lag in AC circuits

Phasor diagrams for a resistor, an inductor and a capacitor, with the current I as reference.

Resistor: V and I are in phase (\Phi=0).
Inductor: V leads I by 90^\circ; the back emf delays the current.
Capacitor: V lags I by 90^\circ; the capacitor must receive charge before its voltage can rise.
RL series: V leads I by \varphi, with \tan\varphi=X_L/R and 0<\varphi<90^\circ.
RC series: V lags I by \varphi, with \tan\varphi=X_C/R.
RLC series: V leads if X_L>X_C, lags if X_C>X_L, and is in phase at resonance.

Memory aid: ELI the ICE man: in an inductor the emf (E) comes before the current (I); in a capacitor the current comes before the emf.

Describe impedance as vector summation of resistances and reactances.

Solution

Impedance

Impedance Z is the total opposition of an AC circuit to the current, Z=V/I, measured in ohms. It is the combined effect of the resistance R and the reactances X_L and X_C.

Why a vector sum

In a series circuit the same current flows through all the components, but their voltages are out of phase: V_R is in phase with I, V_L leads by 90^\circ and V_C lags by 90^\circ. So the voltages must be added as phasors. Dividing each voltage by the common current I gives a triangle of R, X and Z: the impedance triangle. The resistance is drawn along the current, the reactance at right angles (up for X_L, down for X_C), and Z is the resultant.

RL circuit

RL impedance triangle.

Z=\sqrt{R^2+X_L^2},\qquad\tan\varphi=\dfrac{X_L}{R}

RC circuit

RC impedance triangle.

Z=\sqrt{R^2+X_C^2},\qquad\tan\varphi=\dfrac{X_C}{R}

RLC circuit

X_L and X_C point in opposite directions, so the net reactance is X_L-X_C:

Z=\sqrt{R^2+(X_L-X_C)^2},\qquad\tan\theta=\dfrac{X_L-X_C}{R}

Example and conclusion

For R=30\,\Omega and X_L=40\,\Omega in series, Z=\sqrt{30^2+40^2}=50\,\Omega, not 30+40=70\,\Omega. Impedance is therefore the vector (phasor) sum of resistance and reactance, never their simple sum. The angle of the triangle is the phase angle between the supply voltage and the current.

Numerical problemsp. 123

Convert every quantity to SI units first (\mu\mathrm{H}, mH, \mu\mathrm{F}, mA). Reactances and impedances then come out in ohms.

What is the rms value of the voltage of an ac supply having peak voltage 300\,\mathrm{V}?

Solution

Given

V_0 = 300\,\mathrm{V} peak voltage

To find

V_{\mathrm{rms}}

For a sinusoidal voltage the r.m.s. value is the peak value divided by \sqrt2.
V_{\mathrm{rms}}=\dfrac{V_0}{\sqrt2}
Substitute.
V_{\mathrm{rms}}=\dfrac{300}{1.414}=212.1\,\mathrm{V}

V_{\mathrm{rms}}\approx212\,\mathrm{V}

Book answer. 212\,\mathrm{V}

An inductor with an inductance of 100\,\mu\mathrm{H} passes a current of 10\,\mathrm{mA} when its terminal voltage is 6.3\,\mathrm{V}. Calculate the frequency of A.C supply.

Solution

Given

L = 100\,\mu\mathrm{H}=1.0\times10^{-4}\,\mathrm{H}
I = 10\,\mathrm{mA}=1.0\times10^{-2}\,\mathrm{A}
V = 6.3\,\mathrm{V}

To find

The opposition of the inductor is its reactance, found from Ohm's law for AC.
X_L=\dfrac{V}{I}=\dfrac{6.3}{1.0\times10^{-2}}=630\,\Omega
Use X_L=2\pi fL and solve for f.
f=\dfrac{X_L}{2\pi L}=\dfrac{630}{2\times3.14\times1.0\times10^{-4}}
Evaluate.
f=1.003\times10^{6}\,\mathrm{Hz}\approx1.0\times10^{6}\,\mathrm{Hz}

f\approx1.0\times10^{6}\,\mathrm{Hz}\ (1\,\mathrm{MHz})

Book answer. 10^{6}\,\mathrm{Hz}

A coil of pure inductance 318\,\mathrm{mH} is connected in series with a pure resistance of 75\,\Omega. The voltage across resistor is 150\,\mathrm{V} and the frequency of power supply is 50\,\mathrm{Hz}. Calculate impedance of the circuit.

Solution

Given

L = 318\,\mathrm{mH}=0.318\,\mathrm{H}
R = 75\,\Omega
V_R = 150\,\mathrm{V}
f = 50\,\mathrm{Hz}

To find

Inductive reactance of the coil.
X_L=2\pi fL=2\times3.14\times50\times0.318=99.9\,\Omega
Impedance of the RL series circuit.
Z=\sqrt{R^2+X_L^2}=\sqrt{75^2+99.9^2}=\sqrt{5625+9981}
Evaluate.
Z=\sqrt{15606}=124.9\,\Omega

The voltage across the resistor is not needed for Z. It gives the current, I=V_R/R=150/75=2\,\mathrm{A}, and hence the supply voltage V=IZ=2\times124.9=250\,\mathrm{V} and the phase angle \varphi=\tan^{-1}(99.9/75)=53.1^\circ (current lagging).

Z\approx124.9\,\Omega

Book answer. 124.9\,\Omega

A resistor of resistance 30\,\Omega is connected in series with a capacitor of capacitance 79.5\,\mu\mathrm{F} across a power supply of 50\,\mathrm{Hz} and 100\,\mathrm{V}. Find impedance of the circuit.

Solution

Given

R = 30\,\Omega
C = 79.5\,\mu\mathrm{F}=79.5\times10^{-6}\,\mathrm{F}
f = 50\,\mathrm{Hz}
V = 100\,\mathrm{V}

To find

Capacitive reactance.
X_C=\dfrac{1}{2\pi fC}=\dfrac{1}{2\times3.14\times50\times79.5\times10^{-6}}=\dfrac{1}{0.02497}=40.0\,\Omega
Impedance of the RC series circuit.
Z=\sqrt{R^2+X_C^2}=\sqrt{30^2+40^2}=\sqrt{2500}
Evaluate.
Z=50\,\Omega

Extra: the current is I=V/Z=100/50=2\,\mathrm{A}, and it leads the supply voltage by \varphi=\tan^{-1}(40/30)=53^\circ.

Z\approx50\,\Omega

Book answer. 50\,\Omega

FBISE 2026 questions

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

2026 Annual I, Set B, MCQ 71 mark

A.C voltage across a 0.5\,\mu F capacitor is 16\sin(2\times10^3t)\,V, peak value of voltage through the capacitor is:

A2\times10^3
B10^3
C16Answer
D2

Why

Compare with V=V_0\sin\omega t: the peak value is the number in front of the sine, V_0=16\,\mathrm{V}. The number 2\times10^3 is the angular frequency \omega in \mathrm{rad\,s^{-1}}, and 10^3 is the reactance X_C=1/(\omega C)=1000\,\Omega, not a voltage.

Note. 'Peak value of voltage through the capacitor' means the peak voltage across it.

2026 Annual I, Set B, MCQ 91 mark

A bridge rectifier is used to:

AIncrease voltage
BConvert AC to DCAnswer
CStore charge
DConvert DC to AC

Why

A bridge rectifier uses four diodes to convert AC into (pulsating) DC by full-wave rectification. A transformer changes the voltage, a capacitor stores charge, and an inverter changes DC into AC.

2026 Annual I, Set B, MCQ 101 mark

In a purely resistive AC circuit, the phase difference between voltage and current is:

A0Answer
B\dfrac{\pi}{4}
C\dfrac{\pi}{2}
D\dfrac{3\pi}{4}

Why

In a resistor I=V/R at every instant, so the voltage and current pass through zero and reach their peaks together: they are in phase, \Phi=0. A phase difference of \pi/2 belongs to a pure inductor or a pure capacitor.

2026 Annual I, Set B, Q.2 (iii) OR3 marks

Explain the function of capacitor in bridge rectifier circuit.

Solution

The output of a bridge rectifier is pulsating DC. A capacitor connected in parallel with the load smooths it into a nearly constant DC; it acts as a filter.

Bridge rectifier with a smoothing capacitor C across the load R.

As the rectified voltage rises to its peak, the capacitor charges to the peak voltage V_0.
When the rectified voltage falls, the diodes switch off and the capacitor discharges slowly through the load, supplying the current.
The next pulse recharges it before its voltage has fallen much, so the gaps between the peaks are filled in.

Smoothing by a capacitor

The ripple left in the output is small when the time constant RC is much longer than the time between pulses: a larger capacitance or a larger load resistance gives better smoothing, while a small load resistance makes the capacitor discharge faster.

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

What is self-inductance? Also write its SI unit and the factors upon which it depends.

Solution

Self-inductance (1 mark)

When the current in a coil changes, an emf \varepsilon_L=-L\,\Delta I/\Delta t is induced in the coil itself (self-induction). The constant L is the self-inductance: the ratio of the emf induced in the coil to the time rate of change of current through it.

L=\dfrac{-\varepsilon_L}{\Delta I/\Delta t}

SI unit and factors (2 marks)

SI unit: the henry (H), 1\,\mathrm{H}=1\,\mathrm{V\,s\,A^{-1}}. A coil has a self-inductance of 1 H if an emf of 1 V is induced in it when its current changes at 1\,\mathrm{A\,s^{-1}}.

For a long coil L=\dfrac{\mu_r\mu_0N^2A}{l}, so L depends on:

the number of turns N (L\propto N^2);
the area of cross-section A (L\propto A);
the length l of the coil for the same turns (L\propto1/l);
the core material, its relative permeability \mu_r (an iron core increases L).

2026 Annual I, Set B, Q.2 (v) OR3 marks

How does doubling the frequency affect the capacitive reactance? Show mathematically.

Solution

Capacitive reactance is inversely proportional to frequency:

X_C=\dfrac{1}{2\pi fC}

When the frequency is doubled (f'=2f) with the same capacitor:

X_C'=\dfrac{1}{2\pi f'C}=\dfrac{1}{2\pi(2f)C}=\dfrac12\left(\dfrac{1}{2\pi fC}\right)=\dfrac{X_C}{2}

So doubling the frequency halves the capacitive reactance, and for the same voltage the current through the capacitor doubles. Reason: the capacitor is charged and discharged twice as often, so twice as much charge flows per second.

Capacitive reactance against frequency

X_C is inversely proportional to f: when the frequency is doubled, the reactance is halved.

2026 Annual I, Set B, Q.2 (ix) OR2+1 = 3 marks

Draw labelled wave form and phasor diagram between applied voltage and current for pure inductor A.C.

Solution

Waveform (2 marks)

For a pure inductor, if I=I_0\sin\omega t then V=V_0\sin(\omega t+\pi/2) with V_0=\omega LI_0. The voltage reaches its peak a quarter cycle (90^\circ) before the current.

Inductor: voltage leads current by 90°

Fig 20.11 (b) redrawn. The voltage reaches its peak a quarter cycle (90^\circ) before the current. The voltage is largest when the current is zero, because that is when the current changes fastest.

Phasor diagram (1 mark)

Phasor diagram for a pure inductor: V is 90^\circ ahead of I (anticlockwise).

The current phasor is drawn along the horizontal and the voltage phasor at 90^\circ anticlockwise from it: the voltage leads the current by 90^\circ (the current lags).

2026 Annual I, Set B, Q.2 (xiii) OR1.5+1.5 = 3 marks

Distinguish graphically between half-wave and full-wave rectification.

Solution

Half-wave rectification (1.5 marks)

One diode passes only the positive half-cycles; the output is zero during every negative half-cycle. One pulse per input cycle.

Half-wave rectification

Fig 20.5 redrawn. Only the positive half-cycles reach the load. The negative half-cycles are blocked, so the output is zero for half of every cycle. One pulse per cycle.

Full-wave rectification (1.5 marks)

Four diodes (bridge) pass both half-cycles, and the negative halves are turned upward, so the current through the load flows one way all the time. Two pulses per input cycle.

Full-wave rectification

Fig 20.7 redrawn. Both half-cycles reach the load and both drive current the same way through it, so the negative halves appear flipped upward. Two pulses per cycle.

The full-wave output has no gaps, twice the pulse frequency and twice the average value (2V_0/\pi against V_0/\pi).

2026 Annual I, Set B, Q.52+2+2 = 6 marks

In an RL series circuit, will the current lag or lead the applied alternating voltage? Explain the answer with phasor diagram. Also find an expression for the impedance of RL series circuit.

Solution

The current lags the voltage (2 marks)

In an RL series circuit the current lags behind the applied voltage by an angle \varphi, where 0<\varphi<90^\circ. The same current flows through R and L. The voltage across R is in phase with the current, but the voltage across L leads the current by 90^\circ, because the back emf of the coil opposes every change in current. The applied voltage is the phasor sum of the two, so it is ahead of the current by \varphi; that is, the current lags.

RL series circuit.

Phasor diagram (2 marks)

Phasor diagram of the RL series circuit. V_R (OA) is along I, V_L (AB) leads by 90^\circ, and the supply voltage V (OB) leads I by \varphi.

The current I is the reference phasor (horizontal).
V_R=IR is drawn along I (OA).
V_L=IX_L is drawn at 90^\circ ahead of I (AB).
The resultant V (OB) makes the angle \varphi with I: \tan\varphi=\dfrac{V_L}{V_R}=\dfrac{X_L}{R}.

Impedance of the RL series circuit (2 marks)

Derivation Impedance of the RL series circuit

Aim. Find Z=V/I.

From the right-angled triangle OAB (Pythagoras' theorem):
V^2=V_R^2+V_L^2
Substitute V_R=IR and V_L=IX_L.
V^2=(IR)^2+(IX_L)^2=I^2\left(R^2+X_L^2\right)
Take the square root and divide by I.
Z=\dfrac{V}{I}=\sqrt{R^2+X_L^2}

ResultZ=\sqrt{R^2+X_L^2}=\sqrt{R^2+(2\pi fL)^2}

The impedance depends on the resistance, the inductance and the frequency of the supply. It is measured in ohms.

2026 Annual I, Set D, MCQ 111 mark

In a bridge rectifier, current flows through the load:

ADuring both half cycles in the same directionAnswer
BDuring both half cycles in alternating directions
COnly during positive half cycle
DOnly during negative half cycle

Why

D_1 and D_2 conduct in the positive half-cycle and D_3 and D_4 in the negative half-cycle, and both pairs send the current through the load in the same direction. Options C and D describe a half-wave rectifier.

2026 Annual I, Set D, MCQ 121 mark

In a pure inductive AC circuit, the applied voltage leads the current by:

A0
B\dfrac{\pi}{2}Answer
C\dfrac{3\pi}{2}
D\pi

Why

For a pure inductor, I=I_0\sin\omega t gives V=V_0\sin(\omega t+\pi/2): the voltage leads the current by \pi/2 (90^\circ). Remember ELI: in L the emf comes before I.

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

What is mutual inductance? Also write its SI unit and the factors upon which it depends.

Solution

Mutual inductance (1 mark)

When the current in one coil changes, an emf \varepsilon_2=-M\,\Delta I_1/\Delta t is induced in a neighbouring coil (mutual induction). The constant M is the mutual inductance: the ratio of the emf induced in one coil to the time rate of change of current in the other coil.

M=\dfrac{-\varepsilon_2}{\Delta I_1/\Delta t}=\dfrac{N_2\Phi_2}{I_1}

SI unit and factors (2 marks)

SI unit: the henry (H), 1\,\mathrm{H}=1\,\mathrm{V\,s\,A^{-1}}. The mutual inductance of two coils is 1 H if a current changing at 1\,\mathrm{A\,s^{-1}} in one coil induces an emf of 1 V in the other.

For two coaxial coils M=\dfrac{\mu_r\mu_0N_1N_2A}{l}, so M depends on:

the number of turns of both coils (N_1N_2);
the area of cross-section of the coils;
the distance between the coils and their relative orientation (largest when close and coaxial);
the core material, its relative permeability \mu_r (a common iron core increases M).

2026 Annual I, Set D, Q.2 (v) OR3 marks

An AC voltage across a 0.5\,\mu F capacitor is 16\sin(2\times10^3t)\,V. Find its capacitive reactance.

Solution

Given

C = 0.5\,\mu\mathrm{F}=0.5\times10^{-6}\,\mathrm{F}
V = 16\sin(2\times10^{3}t)\,\mathrm{V}

To find

Compare with V=V_0\sin\omega t to read the angular frequency.
\omega=2\times10^{3}\,\mathrm{rad\,s^{-1}},\qquad V_0=16\,\mathrm{V}
Capacitive reactance.
X_C=\dfrac{1}{\omega C}=\dfrac{1}{(2\times10^{3})(0.5\times10^{-6})}=\dfrac{1}{1\times10^{-3}}
Evaluate.
X_C=1000\,\Omega

This is Example 20.2 of the book, which also finds the peak current I_0=V_0/X_C=16\,\mathrm{mA}.

X_C=1000\,\Omega

2026 Annual I, Set D, Q.2 (ix) OR3 marks

An AC circuit consists of pure resistance of 20\,\Omega and is connected across AC supply of 220\,V, 50\,Hz. Calculate: a. Peak value of voltage. b. Peak value of current.

Solution

Given

R = 20\,\Omega
V_{\mathrm{rms}} = 220\,\mathrm{V} the supply voltage is an r.m.s. value
f = 50\,\mathrm{Hz}

To find

a. Peak value of voltage.

Peak value from the r.m.s. value.
V_0=\sqrt2\,V_{\mathrm{rms}}=1.414\times220=311.1\,\mathrm{V}

V_0\approx311\,\mathrm{V}

b. Peak value of current.

Ohm's law holds at every instant in a resistor.
I_0=\dfrac{V_0}{R}=\dfrac{311.1}{20}=15.56\,\mathrm{A}

The frequency is not needed for these two answers; it gives \omega=2\pi f=314\,\mathrm{rad\,s^{-1}} for the equations V=311\sin(314t) and I=15.6\sin(314t) (Example 20.1).

I_0\approx15.6\,\mathrm{A}

2026 Annual I, Set D, Q.2 (xiii) OR1.5+1.5 = 3 marks

Distinguish between peak value and root mean square (rms) values of alternating voltage.

Solution

Peak value (1.5 marks)

The peak value V_0 is the maximum value of the alternating voltage. It is reached twice in each cycle, as +V_0 and -V_0, and only for an instant. For the 220 V mains, V_0=311\,\mathrm{V}.

r.m.s. value (1.5 marks)

The r.m.s. value V_{\mathrm{rms}} is the steady (DC) voltage that produces the same heat in a resistor as the alternating voltage in the same time. It is the square root of the mean of V^2 over a cycle. AC meters read r.m.s. values, and ratings such as 220 V are r.m.s. values. For a sinusoidal voltage:

V_{\mathrm{rms}}=\dfrac{V_0}{\sqrt2}=0.707\,V_0

Peak and r.m.s. values

Fig 20.2 redrawn. The peak value x_0 is the highest point of the wave. The r.m.s. value sits lower, at x_0/\sqrt2=0.707\,x_0. The positive and negative areas are equal, so the average over a full cycle is zero.

2026 Annual I, Set D, Q.54+2 = 6 marks

Explain the working of a bridge rectifier using four diodes. Also draw the circuit diagram.

Solution

Circuit diagram (2 marks)

Full-wave bridge rectifier with four diodes. AC input at A and B; load R across the other two corners.

Four diodes D_1, D_2, D_3 and D_4 form a bridge. The AC input (secondary of a step-down transformer) is connected across corners A and B, and the load resistor R across the other two corners.

Working (4 marks)

Positive half-cycle: end A is positive and B negative. D_1 and D_2 are forward biased and conduct; D_3 and D_4 are reverse biased. Current flows from A through D_1, through R from top to bottom, through D_2 and back to B.

Positive half-cycle: D_1 and D_2 conduct.

Negative half-cycle: end A is negative and B positive. D_3 and D_4 are forward biased and conduct; D_1 and D_2 are reverse biased. Current flows from B through D_3, through R again from top to bottom, through D_4 and back to A.

Negative half-cycle: D_3 and D_4 conduct.

So the current through the load is in the same direction during both half-cycles, and both halves of the AC appear across R as pulsating DC (full-wave rectification). The output pulses at twice the supply frequency and can be smoothed with a capacitor across R.

Full-wave rectification

Fig 20.7 redrawn. Both half-cycles reach the load and both drive current the same way through it, so the negative halves appear flipped upward. Two pulses per cycle.

2026 Annual I, Old syllabus (Curriculum 2006), MCQ 6, old syllabus1 mark

An alternating voltage is V=20\sin314t. The frequency of alternating voltage is:

A25\,\mathrm{Hz}
B50\,\mathrm{Hz}Answer
C75\,\mathrm{Hz}
D100\,\mathrm{Hz}

Why

Comparing with V=V_0\sin(2\pi ft) gives \omega=2\pi f=314\,\mathrm{rad\,s^{-1}}, so f=\dfrac{314}{2\times3.14}=50\,\mathrm{Hz}. (Book exercise MCQ 7 is the same type with \omega=157, giving 25 Hz.)

2026 Annual I, Old syllabus (Curriculum 2006), MCQ 13, old syllabus1 mark

What is the coefficient of mutual inductance when the magnetic flux changes by 2\times10^{-2}\,Wb and change in current is 0.01\,A.

A0
B\dfrac12\,\mathrm{H}
C2\,\mathrm{H}Answer
D3\,\mathrm{H}

Why

The mutual inductance is the flux linkage of the secondary per unit current in the primary: M=\dfrac{N\Delta\Phi}{\Delta I}=\dfrac{2\times10^{-2}}{0.01}=2\,\mathrm{H} (taking the given flux change as the change in flux linkage, N=1). This matches the paper's formula sheet, M=N\varphi/I.

2026 Annual I, Old syllabus (Curriculum 2006), MCQ 14, old syllabus1 mark

Lenz's law is the consequence of the law of conservation of:

AMass
BEnergyAnswer
CChange
DMomentum

Why

The induced current opposes the change that produces it, so work must be done against it, and this work appears as electrical energy. If it helped the change, energy would be created from nothing. So Lenz's law follows from the law of conservation of energy.

Note. Option C is printed as 'Change'; it was probably meant to be 'Charge'. Either way the answer is B.

2026 Annual I, Old syllabus (Curriculum 2006), MCQ 15, old syllabus1 mark

The average power loss in an inductive circuit is:

A0Answer
BV_mI_m
CV_{rms}I_{rms}
D\dfrac{V_m}{I_m}

Why

In a pure inductor the current lags the voltage by 90^\circ. Energy stored in the magnetic field during one quarter cycle is returned to the source in the next, so the average power is V_{\mathrm{rms}}I_{\mathrm{rms}}\cos90^\circ=0. Option C, V_{\mathrm{rms}}I_{\mathrm{rms}}, is the average power in a pure resistor.

Note. 'Inductive circuit' here means a purely inductive circuit. A real coil with resistance R loses I_{\mathrm{rms}}^2R.

2026 Annual I, Old syllabus (Curriculum 2006), Q.2 (vii), old syllabus1+1+1 = 3 marks

State the factors which affect the inductance of a coil.

Solution

The self-inductance of a long coil (solenoid) is

L=\dfrac{\mu_r\mu_0N^2A}{l}

So it is affected by three factors (one mark each):

Number of turns N: L\propto N^2, so doubling the turns makes L four times larger.
Size of the coil: L is proportional to the area of cross-section A and, for the same number of turns, inversely proportional to the length l (a closely wound coil has more inductance).
Nature of the core: a core of high relative permeability \mu_r, such as soft iron, increases L many times compared with air.

The inductance does not depend on the current flowing in the coil.

2026 Annual I, Old syllabus (Curriculum 2006), Q.2 (x), old syllabus3 marks

Briefly explain impedance in RLC series circuit of AC.

Solution

Impedance Z is the total opposition of an AC circuit, Z=V/I, measured in ohms. In an RLC series circuit the same current I flows through R, L and C, but V_R=IR is in phase with I, V_L=IX_L leads it by 90^\circ and V_C=IX_C lags it by 90^\circ. V_L and V_C are opposite, so the net reactive voltage is V_L-V_C, and the supply voltage is the phasor sum of V_R and V_L-V_C.

Phasor diagram of the RLC series circuit.

\begin{aligned}V&=\sqrt{V_R^2+(V_L-V_C)^2}=I\sqrt{R^2+(X_L-X_C)^2}\\ \Longrightarrow\ Z&=\sqrt{R^2+(X_L-X_C)^2}\end{aligned}

with \tan\theta=(X_L-X_C)/R. If X_L>X_C the current lags; if X_C>X_L it leads. At resonance (X_L=X_C) the impedance is minimum, Z=R, and the current is maximum.

2026 Annual I, Old syllabus (Curriculum 2006), Q.2 (xiii) OR, old syllabus1.5+1.5 = 3 marks

How does doubling the frequency affect the reactance of an inductor and a capacitor?

Solution

Inductor (1.5 marks)

X_L=2\pi fL\quad\Longrightarrow\quad X_L'=2\pi(2f)L=2X_L

Doubling the frequency doubles the inductive reactance, because the current changes twice as fast and the back emf is twice as large.

Capacitor (1.5 marks)

X_C=\dfrac{1}{2\pi fC}\quad\Longrightarrow\quad X_C'=\dfrac{1}{2\pi(2f)C}=\dfrac{X_C}{2}

Doubling the frequency halves the capacitive reactance, because the capacitor charges and discharges twice as often and twice as much charge flows per second.

How R, X_L and X_C change with frequency

2026 Annual I, Old syllabus (Curriculum 2006), Q.4 OR, old syllabus7 marks

A 220\,V, 50\,Hz A.C. supply is applied to series combination of a 2.5\,\Omega resistor, 6\,mH inductor and a 6.5\,\mu F capacitor. Calculate (i) reactance of inductor (ii) reactance of capacitor (iii) impedance of circuit (iv) current in the circuit

Solution

Given

V = 220\,\mathrm{V} r.m.s. supply voltage
f = 50\,\mathrm{Hz}
R = 2.5\,\Omega
L = 6\,\mathrm{mH}=6\times10^{-3}\,\mathrm{H}
C = 6.5\,\mu\mathrm{F}=6.5\times10^{-6}\,\mathrm{F}

To find

(i) Reactance of inductor

Inductive reactance.
X_L=2\pi fL=2\times3.14\times50\times6\times10^{-3}=1.88\,\Omega

X_L\approx1.88\,\Omega

(ii) Reactance of capacitor

Capacitive reactance.
X_C=\dfrac{1}{2\pi fC}=\dfrac{1}{2\times3.14\times50\times6.5\times10^{-6}}=\dfrac{1}{2.042\times10^{-3}}=489.7\,\Omega

X_C\approx490\,\Omega

(iii) Impedance of circuit

Series RLC impedance; the net reactance is X_L-X_C.
Z=\sqrt{R^2+(X_L-X_C)^2}=\sqrt{2.5^2+(1.88-489.7)^2}
Evaluate. The resistance hardly matters because the net reactance is so large.
Z=\sqrt{6.25+2.380\times10^{5}}=487.8\,\Omega

Z\approx488\,\Omega

(iv) Current in the circuit

Ohm's law for AC with r.m.s. values.
I=\dfrac{V}{Z}=\dfrac{220}{487.8}=0.451\,\mathrm{A}

Since X_C\gg X_L, the circuit is strongly capacitive: the current leads the voltage by \theta=\tan^{-1}\dfrac{X_C-X_L}{R}\approx89.7^\circ. The circuit would resonate only at f_0=1/(2\pi\sqrt{LC})\approx806\,\mathrm{Hz}.

I\approx0.45\,\mathrm{A}

Note. The paper does not print the marks for this OR question. Section C carries 26 marks and the first alternative of Q.4 is marked 2+2+3, so this question carries 7 marks.

2026 Annual I, Old syllabus (Curriculum 2006), Q.6 OR, old syllabus6 marks

Explain the phenomes of self-inductance of a coil. What is its unit? On what factors it depends?

Solution

Phenomenon of self-induction

A coil connected to a battery through a switch and a rheostat.

A coil carrying a current produces a magnetic field whose flux passes through the coil itself. If the current is changed, for example by moving the rheostat quickly or by switching on or off, the flux through the coil changes, and by Faraday's law an emf is induced in the same coil. This is self-induction: the phenomenon in which a changing current induces an emf in the coil itself.

\varepsilon_L=-L\dfrac{\Delta I}{\Delta t}\qquad\Longrightarrow\qquad L=\dfrac{-\varepsilon_L}{\Delta I/\Delta t}

The constant L is the self-inductance of the coil: the ratio of the induced emf to the rate of change of current. The minus sign is Lenz's law: the induced emf opposes the change in current, so it is called the back emf. It slows the growth of current when the circuit is switched on and tries to maintain the current when it is switched off.

Unit

1\,\mathrm{H}=1\,\mathrm{V\,s\,A^{-1}}

The unit is the henry. The self-inductance of a coil is one henry if an emf of one volt is induced in it when the current changes at the rate of one ampere per second.

Factors

L=\dfrac{\mu_r\mu_0N^2A}{l}

number of turns N (L\propto N^2);
area of cross-section A of the coil;
length l of the coil for the same number of turns (L\propto1/l);
nature of the core, its relative permeability \mu_r (soft iron increases L greatly).

Note. 'Phenomes' is printed for 'phenomenon'. The paper does not print the marks for this OR question; Section C carries 26 marks and the first alternative of Q.6 is marked 4+2, so this question carries 6 marks.

Exam focus

Explain the working of a bridge rectifier using four diodes, with a labelled circuit diagram and input and output waveforms. (long)
Asked as a full Section C question in 2026 (Set D Q.5, 4+2); bridge-rectifier MCQs also appeared in Sets B and D.
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In an RL series circuit, does the current lag or lead the voltage? Draw the phasor diagram and derive Z=\sqrt{R^2+X_L^2}. (long)
Set B Q.5 (2+2+2) in 2026; the same derivation answers book comprehensive questions 6, 7 and 9.
Go to answer
Define self-inductance and mutual inductance, give the unit henry, and state the factors on which each depends. (short)
Self-inductance came in Set B Q.2 (iv) and old-syllabus Q.2 (vii) and Q.6 OR; mutual inductance in Set D Q.2 (iv). Learn L=\mu_r\mu_0N^2A/l for the factors.
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Distinguish between peak and r.m.s. values and use I_{\mathrm{rms}}=I_0/\sqrt2, V_{\mathrm{rms}}=V_0/\sqrt2. (short)
Set D Q.2 (xiii) OR in 2026, book short question 8 and several numericals (Example 20.1, numerical 1).
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Distinguish graphically between half-wave and full-wave rectification. (short)
Set B Q.2 (xiii) OR in 2026 and an SLO; two labelled sketches earn the marks.
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Explain the function of the smoothing capacitor and the effect of C and the load resistance on the ripple. (short)
Set B Q.2 (iii) OR in 2026 and book comprehensive question 3; the key idea is the time constant RC.
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Show how doubling the frequency changes X_L=2\pi fL and X_C=1/(2\pi fC). (short)
Set B Q.2 (v) OR and old-syllabus Q.2 (xiii) OR in 2026: X_L doubles, X_C halves.
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Derive V=V_0\sin(\omega t+\pi/2) for a pure inductor and draw its waveform and phasor diagram; do the same for a capacitor. (long)
Set B Q.2 (ix) OR asked for the inductor waveform and phasor diagram; Set D MCQ 12 and book comprehensive question 5 test the same result.
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Numericals on reactance and impedance: X_C from V=V_0\sin\omega t, peak values from r.m.s. values, and Z of RL, RC and RLC circuits. (numerical)
Set D Q.2 (v) OR and (ix) OR repeat the book's Examples 20.2 and 20.1; the old syllabus had a full RLC numerical (Q.4 OR).
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Explain impedance in an RLC series circuit and the condition for resonance, f_0=1/(2\pi\sqrt{LC}). (short)
Old-syllabus Q.2 (x) in 2026; it also shows why impedance is a vector sum of resistance and reactance (an SLO).
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Show that the mean power in a resistor is half the maximum power, and why the average power in a pure inductor is zero. (short)
An SLO of the unit; old-syllabus MCQ 15 in 2026 asked for the zero power loss in an inductive circuit.
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