Alternating Current: Class 12 Physics · Chapter 7

Physics Part 1 · Companion notes, simulations and practice · Official NCERT chapter PDF ↗

Published 28 Sep 2026 · Updated 28 Sep 2026

120 min
1 simulation
16 practice questions
34 MCQs

What you will learn in Alternating Current

Describe an alternating voltage by its amplitude, angular frequency and period, and explain why ac is used for power supply
Derive the average power in a resistor and define and use rms values of current and voltage
Represent sinusoidal voltages and currents by rotating phasors and read phase relations from phasor diagrams
Derive the current, reactance and phase for a pure inductor and a pure capacitor, and show that neither takes average power
Use a phasor diagram to find the impedance and phase angle of a series LCR circuit
Explain resonance in a series LCR circuit, calculate the resonant frequency and relate the sharpness of resonance to R
Calculate average power and power factor, and explain wattless current and power-factor correction
Explain how a transformer works, apply the turns ratio, and describe its losses and its role in power transmission

Original companion text aligned to NCERT Physics Class 12, Chapter 7, pp. 177–200. Not affiliated with or endorsed by NCERT.

7.1 Introduction

The battery in your phone pushes charge round its circuits in one direction only, at an almost constant voltage. The socket on the wall behaves quite differently. The potential difference between its terminals rises to a maximum, falls through zero, becomes negative, and returns, fifty times every second in India, and the current it drives reverses each time the voltage does. A steady one-way current, like the battery's, is a direct current (dc). The socket, by contrast, provides an alternating voltage (प्रत्यावर्ती वोल्टता), one that rises and falls sinusoidally with time, and the current it pushes to and fro is an alternating current, or ac (प्रत्यावर्ती धारा).Here “ac” only describes how a quantity changes with time. So “ac voltage” is an alternating potential difference and “ac current” an alternating current, even though the letters strictly stand for “alternating current”.

3/4Every half cycle the voltage changes sign, and the current it drives reverses.

Fig. 7.1A battery gives a steady voltage; the mains voltage alternates.

Two graphs of voltage against time, one above the other. Step 1: the top graph shows a battery's 12 volt output as a flat horizontal line. Step 2: the lower graph adds the mains voltage, a smooth sine wave that swings between plus 325 volts and minus 325 volts. Step 3: the parts of the wave above the axis are shaded blue with an arrow and the words one way, and the parts below are shaded orange with a reversed arrow, showing that the current changes direction every half cycle. Step 4: a bracket over the first full wave marks one cycle, which lasts 20 milliseconds.

Why do power companies supply ac rather than dc? Mostly because of the transformer (Section 7.8). With an alternating voltage, a device with no moving parts can raise or lower the voltage with very little waste. Power stations use this to send energy hundreds of kilometres at a very high voltage and a small current, which keeps heating in the lines low, and then bring the voltage down again before it enters homes. Ac can also do things that dc cannot: a radio picks one station out of the dozens reaching its aerial with a circuit that answers strongly to a single frequency and barely to the rest (Section 7.6.2).

We write a sinusoidal voltage as $v = v_{m} \sin ω t$ . Here $v$ is the instantaneous value, $v_{m}$ is the peak value or amplitude, and $ω$ is the angular frequency in radians per second. One cycle lasts a period $T = 2 π / ω$ , and the frequency is $ν = 1 / T = ω / 2 π$ . For the Indian mains $ν = 50 H z$ , so $T = 20 m s$ and $ω = 100 π \approx 314 r a d / s$ .

Where does such a voltage come from? In Chapter 6 you met the ac generator: a coil turning steadily in a magnetic field has an induced emf $ε = N B A ω \sin ω t$ . The simulation below lets you turn one yourself.

SimulationA turning coil makes ac

A rectangular coil turns between two magnet poles, driving current through a meter whose needle swings left and right. A graph traces the emf and the flux through the coil against time over a fixed four-second window. Readouts show the speed in rpm, the frequency, the peak emf and the emf at this instant; when paused, the coil angle is shown and the coil can be dragged to any angle. Sliders change the speed, the number of turns and the field.

Fig. 7.2The emf of a coil turning at a steady rate is a sine wave: it reverses every half turn.

Try this

Before you press Play, drag the coil until the EMF now reading is as large as it gets. Note the coil angle, then drag the coil half a turn further. What has happened to the emf?
Now press Play and set Speed to 60 rpm. How do the Frequency and Peak EMF readings, and the number of cycles in the graph window, compare with those at 30 rpm?
Watch the emf trace through one full turn. For what fraction of the turn is it positive? What must its average over a whole turn be?

Timeline

How ac won the “war of the currents”

1882
Dc power for a city
Edison's Pearl Street station in New York supplies dc to nearby buildings; it can only reach customers a short distance away.
1888
The induction motor
Tesla patents motors that run on alternating current; Westinghouse buys the patents.
1893
A fair lit by ac
Westinghouse's company lights the World's Columbian Exposition in Chicago with alternating current.
1896
Power from Niagara
Generators at Niagara Falls send ac to the city of Buffalo, many kilometres away, using transformers.

Quick check

1 of 3

Some countries run their mains at

60 H z

. What is the period of that supply, in milliseconds?

Show answer

Answer 16.667 ms

T = 1 / ν = 1 / 60 s \approx 0.0167 s = 16.7 m s

, a little shorter than the

20 m s

of a

50 H z

supply.

7.2 AC voltage applied to a resistor

Connect a resistor $R$ across an ac source. How large is the current, and how is it timed relative to the voltage?

DerivationCurrent through a resistor

Let the voltage of the source be

v = v_{m} \sin ω t

(7.1)

Kirchhoff's loop rule holds at every instant, just as it does for dc: the source voltage equals the potential drop $i R$ across the resistor. So $v_{m} \sin ω t = i R$ , and

i = i_{m} \sin ω t

(7.2)

with the current amplitude

i_{m} = \frac{v_{m}}{R}

(7.3)

Equation (7.3) is just $V = I R$ written for peak values, so a resistor obeys Ohm's law on ac as it does on dc. Nothing in (7.2) delays the current either. Whenever $v$ passes through zero so does $i$ , and both reach their crests together: the current and the voltage are in phase.

3/3Peaks, troughs and zero crossings line up: current and voltage are in phase.

Fig. 7.3In a resistor the current rises and falls in step with the voltage.NCERT Fig. 7.2 · p. 178

On the left, a circuit: an ac source, drawn as a circle containing a small wave, connected to a 5 ohm resistor, with an arrow for the current i and the label 10 volts peak. On the right, a graph against time. Step 1: the source voltage v, a sine wave reaching plus and minus 10 volts. Step 2: the current i is added as a dashed sine wave of peak 2 amperes, which crosses zero and reaches its peaks at exactly the same moments as the voltage. Step 3: dotted vertical lines through the peaks and troughs and dots at the zero crossings show that both curves peak and cross zero together.

An ammeter that simply averaged the current over a cycle would read zero, since the current flows one way for half the time and the other way for the rest. Yet an electric iron on the mains certainly gets hot. The way out is that heating goes as $i^{2}$ , which is positive whichever way the charge moves.

DerivationAverage power in a resistor

The rate at which electrical energy turns into heat at any instant is

p = i^{2} R = i_{m}^{2} R \sin^{2} ω t

(7.4)

The average of a quantity $F (t)$ over one period $T$ is defined as

⟨ F ⟩ = \frac{1}{T} \int_{0}^{T} F (t) d t

Writing $\overset{ˉ}{p}$ for the average power,

\overset{ˉ}{p} = ⟨ i^{2} R ⟩ = ⟨ i_{m}^{2} R \sin^{2} ω t ⟩

(7.5a)

Neither $i_{m}$ nor $R$ changes with time, so both come outside the average:

\overset{ˉ}{p} = i_{m}^{2} R ⟨ \sin^{2} ω t ⟩

(7.5b)

To average $\sin^{2} ω t$ , use $\sin^{2} ω t = \frac{1}{2} (1 - \cos 2 ω t)$ . The term $\cos 2 ω t$ runs through exactly two whole cycles in the time $T = 2 π / ω$ , so its positive and negative parts cancel:

⟨ \cos 2 ω t ⟩ = \frac{1}{T} {[\frac{\sin 2 ω t}{2 ω}]}_{0}^{T} = \frac{\sin 4 π}{2 ω T} = 0

Hence $⟨ \sin^{2} ω t ⟩ = \frac{1}{2}$ , and

\overset{ˉ}{p} = \frac{1}{2} i_{m}^{2} R

(7.5c)

The average power is exactly half of the peak power $i_{m}^{2} R$ .

It is convenient to describe an alternating current by the steady current that would heat the same resistor at the same average rate. Call it $I$ . Setting $I^{2} R$ equal to $\frac{1}{2} i_{m}^{2} R$ gives $I^{2} = \frac{1}{2} i_{m}^{2} = ⟨ i^{2} ⟩$ . So $I$ is found by squaring the current, taking the mean of the square over a cycle, and then taking the square root. That recipe gives it its name: the root-mean-square value, or rms value, also called the effective value (Fig. 7.4).

4/4The square root of the mean square is $I = i_{m} / \sqrt{2} \approx 0.707 i_{m}$ .

Fig. 7.4Square, take the mean, take the root: the rms current is

0.707

of the peak current.NCERT Fig. 7.3 · p. 179

A graph against time over two cycles. Step 1: a dashed sine curve for the current i, swinging between plus i m and minus i m. Step 2: the current fades and a solid curve for i squared appears; it is never negative, rising from zero to i m squared twice as often as the current. Step 3: a dashed horizontal line marks the mean of i squared at half of i m squared, and the parts of the curve above the line are shaded to show that they exactly fill the dips below it. Step 4: the current curve returns with a solid horizontal line at I equals i m over root 2, about 0.707 of the peak, labelled the root of the mean square.

I = \sqrt{⟨ i^{2} ⟩} = \sqrt{\frac{1}{2} i_{m}^{2}} = \frac{i_{m}}{\sqrt{2}} \approx 0.707 i_{m}

(7.6)

In terms of $I$ , the average power takes the familiar dc form:

P = \overset{ˉ}{p} = \frac{1}{2} i_{m}^{2} R = I^{2} R

(7.7)

Definition

rms (effective) value

The value of the steady current that heats a given resistor at the same average rate as the alternating current: $I = \sqrt{⟨ i^{2} ⟩}$ , which is $i_{m} / \sqrt{2}$ for a sine wave. Likewise $V = v_{m} / \sqrt{2}$ for a voltage.

ExampleA current of peak value $10 A$ has an rms value of $7.07 A$ .

Voltages are treated just like currents, giving the rms voltage:

V = \frac{v_{m}}{\sqrt{2}} \approx 0.707 v_{m}

(7.8)

Dividing both sides of $v_{m} = i_{m} R$ by $\sqrt{2}$ turns Ohm's law into a relation between rms values:

V = I R

(7.9)

When you read “ $230 V$ ” on a socket, or any ac reading on a meter, the number is an rms value; the mains voltage actually peaks at $\sqrt{2} \times 230 V \approx 325 V$ . Quoting rms values pays off because, with them, a resistor on ac obeys the familiar dc formulas exactly: $V = I R$ and $P = I^{2} R = V^{2} / R = V I$ . The power printed on an appliance is, similarly, its average power.

Example 7.1NCERT Example 7.1 · p. 180

A room heater on the mains

A room heater is marked $1000 W$ , $230 V$ , $50 H z$ . Treat its element as a pure resistor. (a) What is the resistance of the hot element? (b) What rms current does it draw, and how large does the current get at its peak? (c) What is the largest rate at which it takes energy at any instant? (d) If it runs for 3 hours every day of a 30-day month, what does it cost at ₹8 per unit (kWh)?

Try it yourself first. Then reveal the solution one step at a time.

Example

Reading an ac current from its equation

The current in an $8.0 Ω$ resistor is $i = 5.0 \sin (100 π t)$ , with $i$ in amperes and $t$ in seconds. Find (a) the frequency, (b) the rms current, (c) the average power and (d) the largest instantaneous power.

Try it yourself first. Then reveal the solution one step at a time.

Try this

Read the rating plates

You’ll need

three or four appliances (or their boxes) with rating labels
a calculator

Predict first

All the appliances are rated for

230 V

. Which will draw the largest rms current?

They all draw the same current, because the voltage is the sameThe one with the largest power ratingThe one with the smallest power ratingThe one with the longest cable

Collect three or four appliances, or their boxes, that carry a rating label, for example an electric iron, a kettle, an LED bulb and a phone charger. Read the labels on unplugged appliances only.
Note the rated voltage, frequency and power of each.
Work out the rms current each draws, $I = P / V$ . For the heating appliances (iron, kettle, heater) also work out the hot resistance, $R = V^{2} / P$ .

Your appliances

Appliance	Rated power P (W)	Rated voltage V (V)	I = P/V (A)

Your entries are saved automatically.

The currents are proportional to the power ratings: roughly $4$ to $8 A$ for irons, kettles and heaters, and only a few hundredths of an ampere for LED bulbs and chargers. The formula $R = V^{2} / P$ makes sense only for appliances that are simply heating elements, which behave like pure resistors. Chargers, fans and LED lamps contain other components, so their current is not in phase with the voltage (you will see what that means in Section 7.7).

Quick check

1 of 4

An ac voltmeter connected across a lamp reads

120 V

. What is the peak voltage across the lamp, in volts?

Show answer

Answer 169.7 V

Meters read rms values, so

v_{m} = \sqrt{2} \times 120 V \approx 170 V

7.3 Representing ac by rotating vectors: phasors

A resistor keeps current and voltage in step, but the next two elements do not. In an inductor the current reaches its peak after the voltage; in a capacitor it gets there first. Combining sine waves that are out of step by drawing them point by point is slow and error-prone. A simple geometric picture does the job far more easily.

Imagine an arrow of length $v_{m}$ , pivoted at the origin and turning anticlockwise at a steady angular speed $ω$ . At time $t$ it makes an angle $ω t$ with the horizontal axis, so its projection on the vertical axis is $v_{m} \sin ω t$ , which is exactly the instantaneous voltage. As the arrow turns, its projection traces out the sine curve (Fig. 7.5). A rotating arrow used in this way is called a phasor (फेज़र). We print phasors in bold: $𝐕$ for a voltage, $𝐈$ for a current.

1/5At $ω t = 30^{\circ}$ the projection is $v_{m} \sin 30^{\circ} = v_{m} / 2$ .

Fig. 7.5The vertical projection of a rotating phasor is the instantaneous value

v_{m} \sin ω t

.NCERT Fig. 7.4 · p. 181

On the left, a circle with horizontal and vertical axes and a blue arrow of length v m pivoted at the centre, with a curved arrow showing that it turns anticlockwise at angular speed omega. On the right, a graph of v against the angle omega t from 0 to 360 degrees, with a faint sine curve as a guide. Step 1: the arrow is at 30 degrees; a purple bar on the vertical axis shows its vertical projection, half of v m, and a dashed line carries this height across to a dot on the graph. Step 2: at 90 degrees the arrow points straight up and the dot is at the top of the curve. Step 3: at 180 degrees the arrow points left, the projection is zero and the dot is on the axis. Step 4: at 270 degrees the arrow points down and the dot is at the bottom of the curve. Step 5: after a full turn the arrow is back along the horizontal axis and the traced curve is one complete sine wave.

Definition

phasor

A rotating arrow that represents a sinusoidally varying quantity: its length is the amplitude, it turns anticlockwise at the angular frequency $ω$ , and its projection on the vertical axis gives the instantaneous value.

ExampleThe voltage $v = 325 \sin 314 t$ volts is represented by a phasor of length $325 V$ turning at $314 r a d / s$ .

A current $i = i_{m} \sin (ω t + ϕ)$ is represented by a second phasor, of length $i_{m}$ , that turns at the same $ω$ but keeps an angle $ϕ$ ahead of the voltage phasor. Because both phasors rotate together, the angle between them never changes. So we usually draw a phasor diagram at one instant and leave the rotation understood. Two things can be read straight off it. The lengths give the amplitudes, and the angle between two phasors gives their phase difference. A phasor turned anticlockwise from another is ahead of it, or leads it; one turned clockwise lags behind.

For a resistor, $𝐕$ and $𝐈$ point in the same direction at every instant, so the phase difference between them is zero. We will need the other two cases, where the phasors are a right angle apart, in the next two sections.

7.4 AC voltage applied to an inductor

Now replace the resistor by a coil. We take an ideal one, pure inductance $L$ and no resistance at all (a coil of thick copper wire comes close), connected across the source $v = v_{m} \sin ω t$ . What current flows?

DerivationCurrent through a pure inductor

A changing current in the coil induces a back emf $- L d i / d t$ (Chapter 6). With no resistance in the circuit, the loop rule gives

v - L \frac{d i}{d t} = 0

(7.10)

The minus sign of the back emf is Lenz's law: the coil opposes any change in its own current. Rearranging,

\frac{d i}{d t} = \frac{v}{L} = \frac{v_{m}}{L} \sin ω t

(7.11)

The source voltage fixes the rate of change of the current, not the current itself. To find $i$ , integrate with respect to time:

i = \int \frac{v_{m}}{L} \sin ω t d t = - \frac{v_{m}}{ω L} \cos ω t + constant

What about the constant? It would be a steady, dc-like current riding on top of the oscillation, and nothing in this circuit could keep one going: over each cycle the source pushes one way for exactly as long, and as hard, as it pushes the other. So we set the constant to zero. Writing $- \cos ω t = \sin (ω t - π / 2)$ ,

i = i_{m} \sin (ω t - \frac{π}{2}), i_{m} = \frac{v_{m}}{ω L}

(7.12)

Set this beside $i_{m} = v_{m} / R$ for a resistor: here $ω L$ does the job of a resistance. We call it the inductive reactance (प्रेरकीय प्रतिघात),

X_{L} = ω L = 2 π ν L

(7.13)

so that the current amplitude is

i_{m} = \frac{v_{m}}{X_{L}}

(7.14)

Reactance has the unit of resistance, the ohm: $1 H \times 1 s^{- 1} = 1 V s A^{- 1} \times 1 s^{- 1} = 1 Ω$ . A larger coil or a higher frequency means a larger reactance and a smaller current, whereas a resistor opposes every frequency equally.

Compare (7.12) with (7.1). The current reaches each peak a quarter of a period, $T / 4$ , after the voltage does: the current in a pure inductor lags $π / 2$ behind the voltage. Fig. 7.6 shows why, using nothing but (7.11). Where $v$ is largest, $d i / d t$ is largest, so the current is climbing most steeply. Where $v$ passes through zero, $d i / d t = 0$ , so the current is momentarily flat, at a crest or a trough. In the phasor diagram, $𝐈$ trails $𝐕$ by $90^{\circ}$ .

4/4The current peaks $T / 4$ after the voltage; in the phasor diagram $𝐈$ trails $𝐕$ by $90^{\circ}$ .

Fig. 7.6In a pure inductor the voltage sets the slope of the current, so the current peaks a quarter cycle later.NCERT Fig. 7.6 · p. 183

A graph of v and i against omega t over one cycle. Step 1: the voltage v as a blue sine wave. Step 2: short orange line segments show the slope the current must have: steepest and rising where the voltage is at its peak, flat where the voltage crosses zero, and steepest and falling where the voltage is most negative. Dotted lines join the voltage peak and zero crossing to these slopes. Step 3: the dashed orange current curve is drawn through the segments; it is a sine wave shifted to the right. Step 4: a bracket marked T over 4 joins the voltage peak to the later current peak, and a small phasor diagram on the right shows the current phasor I at right angles behind the voltage phasor V, with the words I lags V.

Definition

inductive reactance

The opposition of an inductor to alternating current, $X_{L} = ω L$ , measured in ohms; it grows in proportion to the frequency.

ExampleA $0.10 H$ coil at $50 H z$ has $X_{L} = 2 π \times 50 \times 0.10 = 31.4 Ω$ .

A resistor turns the energy it receives into heat. What does an ideal inductor do with it?

DerivationPower supplied to a pure inductor

The power delivered to the inductor at each instant is

p_{L} = i v = i_{m} \sin (ω t - \frac{π}{2}) v_{m} \sin ω t = - i_{m} v_{m} \cos ω t \sin ω t

p_{L} = - \frac{i_{m} v_{m}}{2} \sin 2 ω t

Over a complete cycle $\sin 2 ω t$ is positive exactly as long as it is negative, so its average is zero:

P_{L} = ⟨ p_{L} ⟩ = - \frac{i_{m} v_{m}}{2} ⟨ \sin 2 ω t ⟩ = 0

An ideal inductor takes no average power.

What happens to the energy is shown in Fig. 7.7. During a quarter cycle in which the size of the current grows, the source does work and the coil stores energy $\frac{1}{2} L i^{2}$ in its magnetic field. During the next quarter the current shrinks and the coil gives the same energy back to the source. Energy surges to and fro twice every cycle, but none is lost.

3/3Energy stored in the magnetic field (+) is handed back to the source (−) a quarter cycle later.

Fig. 7.7Power into an ideal inductor is positive for a quarter cycle and negative for the next: the average is zero.

A graph against omega t over one cycle. Step 1: the voltage v as a solid blue sine wave and the current i as a dashed orange wave a quarter cycle behind it. Step 2: the voltage and current fade and a thick green curve shows the power p equals v times i; it oscillates at twice the frequency, symmetric about zero. Step 3: the loops of the power curve above the axis are shaded green with plus signs and labelled stored; the loops below are shaded orange with minus signs and labelled returned. A note says the average is zero.

Example 7.2NCERT Example 7.2 · p. 183

Is this coil safe on the mains?

A coil in a spare-parts box is labelled “ $120 m H$ , maximum current $5.5 A$ ”; its resistance is negligible. Would it be safe to connect it on its own across the $230 V$ , $50 H z$ mains? Would it be safe on a $230 V$ , $60 H z$ supply?

Try it yourself first. Then reveal the solution one step at a time.

7.5 AC voltage applied to a capacitor

A capacitor has an insulator between its plates, so at first sight no current should get through it at all. The mains supply shows otherwise: a capacitor across a $230 V$ socket carries a steady alternating current for as long as it is connected. The trick is that charge never needs to cross the gap. Charge flowing onto one plate pushes an equal charge off the other, so the wires on both sides carry the same current while the capacitor is charging. With a battery this happens only once, a brief surge that ends when the plates are full. An alternating voltage keeps reversing, so the plates are filled, emptied, filled the other way round and emptied again, fifty times a second, and the current in the wires never stops.

DerivationCurrent in a capacitor circuit

Suppose the capacitor holds charge $q$ at time $t$ . The voltage across its plates is then

v = \frac{q}{C}

(7.15)

The loop rule sets this equal to the source voltage, so $q = C v_{m} \sin ω t$ . The current is the rate of change of this charge:

i = \frac{d q}{d t} = \frac{d}{d t} (C v_{m} \sin ω t) = ω C v_{m} \cos ω t

Using $\cos ω t = \sin (ω t + π / 2)$ ,

i = i_{m} \sin (ω t + \frac{π}{2}), i_{m} = ω C v_{m} = \frac{v_{m}}{1 / ω C}

(7.16)

For the capacitor, the quantity standing where $R$ stood for the resistor is $1 / ω C$ , the capacitive reactance (धारितीय प्रतिघात),

X_{C} = \frac{1}{ω C} = \frac{1}{2 π ν C}

(7.17)

and the current amplitude is

i_{m} = \frac{v_{m}}{X_{C}}

(7.18)

$X_{C}$ is also measured in ohms, but it behaves the opposite way to $X_{L}$ : raising either the frequency or the capacitance makes it smaller, so doubling $ν$ or doubling $C$ halves $X_{C}$ . A large capacitor at a high frequency lets ac through easily; at zero frequency, which is dc, $X_{C}$ is infinite and no steady current flows.

This time the current reaches each peak a quarter period before the voltage: the current in a pure capacitor runs $π / 2$ ahead of the voltage. Fig. 7.8 follows the plates through one cycle. The current is largest when the plates are uncharged and the voltage is changing fastest; it is zero at the moments the plates hold their greatest charge. In the phasor diagram, $𝐈$ is $90^{\circ}$ ahead of $𝐕$ .

5/5The current peaks $T / 4$ before the voltage: $𝐈$ is $90^{\circ}$ ahead of $𝐕$ .

Fig. 7.8In a capacitor the current is largest when the plates are empty: it runs a quarter cycle ahead of the voltage.NCERT Fig. 7.8 · p. 185

On the left, a parallel-plate capacitor with wires to two terminals; on the right, a graph of the voltage v (solid blue) and the current i (dashed orange) against omega t over one cycle, with dots marking the current instant. Step 1, at 0 degrees: the plates carry no charge, the voltage is zero, and a large arrow shows current flowing onto the top plate; the current is at its peak. Step 2, at 90 degrees: the top plate is covered in plus signs and the bottom plate in minus signs, the voltage is at its peak and the current is zero. Step 3, at 180 degrees: the plates are uncharged again and the arrow points the other way; the current is at its negative peak. Step 4, at 270 degrees: the charges are reversed, minus on top, the voltage is most negative and the current is zero. Step 5: the capacitor is replaced by a phasor diagram with the current phasor I at right angles ahead of the voltage phasor V, and a bracket marked T over 4 on the graph shows that the current peak comes before the voltage peak.

Definition

capacitive reactance

The opposition of a capacitor to alternating current, $X_{C} = 1 / ω C$ , measured in ohms; it falls as the frequency rises.

ExampleA $10 μ F$ capacitor at $50 H z$ has $X_{C} = 1 / (2 π \times 50 \times 10^{- 5}) = 318 Ω$ .

DerivationPower supplied to a pure capacitor

The instantaneous power is

p_{C} = i v = i_{m} \cos ω t v_{m} \sin ω t = \frac{i_{m} v_{m}}{2} \sin 2 ω t

(7.19)

Once again $\sin 2 ω t$ averages to zero over a cycle, so $P_{C} = 0$ . The capacitor stores energy $q^{2} / 2 C$ in its electric field for a quarter cycle and returns it during the next. The power curve has the same shape as the inductor's in Fig. 7.7, with the storing and returning quarters swapped.

Compare

Resistor, inductor and capacitor on ac

Resistor

Opposition $R$ , the same at every frequency
Current in phase with voltage
Average power $I^{2} R$ , turned into heat

Inductor

Reactance $X_{L} = ω L$ , rises with frequency
Current lags voltage by $π / 2$
Average power zero

Capacitor

Reactance $X_{C} = 1 / ω C$ , falls with frequency
Current leads voltage by $π / 2$
Average power zero

Fig. 7.9 compares the three for $R = 100 Ω$ , $L = 0.25 H$ and $C = 20 μ F$ .A memory aid: in “ELI the ICE man”, the voltage (E) comes before the current (I) in an inductor (L), and the current comes before the voltage in a capacitor (C). At low frequencies the capacitor's reactance is huge and the inductor's tiny; at high frequencies it is the other way round. At about $71 H z$ the two reactances are equal, a coincidence that turns out to matter a great deal in Section 7.6.2.

4/4At $ν \approx 71 H z$ the two reactances are equal, about $112 Ω$ each.

Fig. 7.9

X_{L}

grows with frequency,

X_{C}

shrinks, and

R

stays the same; at one frequency

X_{L} = X_{C}

A graph of opposition in ohms, from 0 to 300, against frequency from 0 to 200 hertz. Step 1: a dotted horizontal line at 100 ohms, labelled R, shows that the resistance does not depend on frequency. Step 2: a straight orange line through the origin shows the inductive reactance X L equals omega L rising in proportion to frequency. Step 3: a dashed blue curve shows the capacitive reactance X C equals 1 over omega C, very large at low frequency and falling steadily as the frequency rises. Step 4: a dot marks where the line and the curve cross, at about 71 hertz and 112 ohms, labelled X L equals X C.

Example 7.3NCERT Example 7.3 · p. 185

Resistor or capacitor? Two lamp boxes

Two identical filament lamps are each wired in series with a component hidden in a sealed box. Box P hides a resistor and box Q a capacitor, chosen so that both lamps glow equally brightly on the $230 V$ , $50 H z$ mains. Predict what happens (a) when both boxes are moved to a $230 V$ dc supply; (b) when, back on the mains, the hidden part in each box is replaced by two identical copies of itself joined in parallel; (c) which box gets warm in use.

Try it yourself first. Then reveal the solution one step at a time.

Example 7.4NCERT Example 7.4 · p. 185

The capacitor across the mains inside an appliance

Many appliances have a small “X-capacitor” connected permanently across the mains leads, to short-circuit high-frequency interference. A typical one is $0.47 μ F$ . On the $230 V$ , $50 H z$ mains, find (a) its reactance, (b) the rms and peak currents through it and (c) the average power it takes. (d) The same appliance is sold in a country with a $230 V$ , $60 H z$ supply. How does the current change?

Try it yourself first. Then reveal the solution one step at a time.

Example 7.5NCERT Example 7.5 · p. 186

Why a jammed solenoid burns out

Many machines open a valve or a lock with an ac solenoid: a coil that pulls a soft-iron plunger into itself when it is switched on. One such coil has a resistance of $40 Ω$ and runs on the $230 V$ , $50 H z$ mains. Its inductance is $1.2 H$ with the plunger pulled in, but only $0.20 H$ with the plunger out. (a) Why does the plunger change the inductance so much? (b) Find the rms current and the heating power in the wire with the plunger in and with it out. (c) The maker warns that the coil burns out within minutes if the plunger jams and cannot move in. Explain why.

Try it yourself first. Then reveal the solution one step at a time.

Quick check

1 of 4

Find the reactance of a

10 μ F

capacitor at

50 H z

, in ohms.

Show answer

Answer 318.3 Ω

X_{C} = 1 / (2 π ν C) = 1 / (2 π \times 50 \times 10 \times 10^{- 6}) \approx 318 Ω

7.6 AC voltage applied to a series LCR circuit

Most real circuits contain more than one kind of element. Connect a resistor $R$ , an inductor $L$ and a capacitor $C$ in series with the source $v = v_{m} \sin ω t$ . At time $t$ the capacitor holds charge $q$ and the current is $i = d q / d t$ . Going round the loop, the source voltage must equal the sum of the three potential drops:

L \frac{d i}{d t} + i R + \frac{q}{C} = v

(7.20)

We want the current and its phase relative to the source voltage. The equation can be solved with calculus, but a phasor diagram reaches the answer with nothing more than a right-angled triangle.

7.6.1 Phasor-diagram solution

Charge cannot pile up anywhere along a series loop, so the current in $R$ , in $L$ and in $C$ is one and the same function of time. That makes the current the natural reference for the phasor diagram.

DerivationImpedance and phase angle of a series LCR circuit

Write the current as

i = i_{m} \sin (ω t + ϕ)

(7.21)

where $ϕ$ , still unknown, is the phase of the current relative to the source voltage. The results of Sections 7.2, 7.4 and 7.5 tell us how to draw each element's voltage phasor against the current phasor $𝐈$ :

resistor: $𝐕_{R}$ points along $𝐈$ ;
inductor: $𝐕_{L}$ is turned $90^{\circ}$ anticlockwise from $𝐈$ ;
capacitor: $𝐕_{C}$ is turned $90^{\circ}$ clockwise from $𝐈$ .

Their lengths follow from each element's opposition to the current:

v_{R m} = i_{m} R, v_{L m} = i_{m} X_{L}, v_{C m} = i_{m} X_{C}

(7.22)

The loop rule (7.20) says that at every instant the three voltages add up to the source voltage,

v_{R} + v_{L} + v_{C} = v

(7.23)

Each instantaneous voltage is the vertical projection of its phasor, and the projection of a sum is the sum of the projections. So the phasors must add in the same way:

𝐕_{R} + 𝐕_{L} + 𝐕_{C} = 𝐕

(7.24)

Fig. 7.10 does this addition tip to tail. Because $𝐕_{L}$ and $𝐕_{C}$ are antiparallel, their sum is simply their difference, $∣ v_{C m} - v_{L m} ∣$ , drawn perpendicular to $𝐈$ . Placing it at the tip of $𝐕_{R}$ closes a right-angled triangle whose longest side is $𝐕$ , so

v_{m}^{2} = v_{R m}^{2} + (v_{C m} - v_{L m})^{2} = i_{m}^{2} [R^{2} + (X_{C} - X_{L})^{2}]

so the current amplitude is

i_{m} = \frac{v_{m}}{\sqrt{R^{2} + (X_{C} - X_{L})^{2}}}

(7.25a)

This has the shape of Ohm's law,

i_{m} = \frac{v_{m}}{Z}

(7.25b)

with the resistance replaced by the impedance (प्रतिबाधा)

Z = \sqrt{R^{2} + (X_{C} - X_{L})^{2}}

(7.26)

The angle of this triangle at the origin, between $𝐕$ and the current direction, is the phase angle. Opposite side over adjacent side gives

\tan ϕ = \frac{v_{C m} - v_{L m}}{v_{R m}} = \frac{X_{C} - X_{L}}{R}

(7.27)

6/7 $𝐕$ closes the triangle: $v_{m} = \sqrt{240^{2} + 100^{2}} = 260 V$ ; the current lags by $ϕ = {22.6}^{\circ}$ .

Fig. 7.10Adding the voltage phasors of a series LCR circuit gives a right-angled triangle; dividing by

i_{m}

gives the impedance triangle.NCERT Fig. 7.11 · p. 187

On the left, a series circuit: an ac source on the bottom wire with a 120 ohm resistor, an inductor of reactance 90 ohm and a capacitor of reactance 40 ohm, carrying a current of peak 2.0 amperes. On the right, a phasor diagram. Step 1: the current phasor I is drawn along the horizontal axis as the reference. Step 2: the resistor's voltage phasor V R, 240 volts, is added along the same direction. Step 3: the inductor's voltage phasor V L, 180 volts, points straight up, 90 degrees ahead of I. Step 4: the capacitor's voltage phasor V C, 80 volts, points straight down, 90 degrees behind I. Step 5: the three voltage phasors fade and the difference of V L and V C, 100 volts upward, is drawn from the tip of V R. Step 6: the source phasor V, 260 volts, is drawn from the origin to that point, at the angle phi above I. Step 7: every side is divided by the current, giving the impedance triangle with R equals 120 ohm, X L minus X C equals 50 ohm and hypotenuse Z equals 130 ohm.

Divide every side of the voltage triangle by $i_{m}$ and you get the impedance triangle shown in the last step of Fig. 7.10: a right-angled triangle with sides $R$ and $∣ X_{C} - X_{L} ∣$ , hypotenuse $Z$ , and the angle $ϕ$ between $R$ and $Z$ . Sketching one is the quickest way to start any series-circuit problem, and it gives a second useful result, $\cos ϕ = R / Z$ .

Definition

impedance

The total opposition of an ac circuit to current, $Z = v_{m} / i_{m} = V / I$ , measured in ohms; for a series LCR circuit $Z = \sqrt{R^{2} + (X_{C} - X_{L})^{2}}$ .

Example $R = 120 Ω$ , $X_{L} = 90 Ω$ and $X_{C} = 40 Ω$ in series give $Z = 130 Ω$ .

In Fig. 7.10, $X_{L} = 90 Ω$ beats $X_{C} = 40 Ω$ , so $\tan ϕ = (40 - 90) / 120 = - 0.417$ and $ϕ = - {22.6}^{\circ}$ : the current trails the source voltage. The sign of $X_{C} - X_{L}$ settles the question for any series circuit. More inductive than capacitive reactance makes $ϕ$ negative; such a circuit is called inductive, and its current lags. More capacitive reactance makes $ϕ$ positive; the circuit is capacitive, and its current leads.

Everything above assumes the circuit has been running long enough to settle into a regular rhythm, with the current repeating itself at the source frequency. This settled behaviour is the steady state. In the first moments after switching on, the current also depends on how things started: whether the capacitor held some charge, or where in its cycle the source happened to be. That extra, irregular part is the transient, and it fades within a few cycles as the resistance turns its energy into heat.On an oscilloscope the transient shows up as a short, lopsided stretch of the trace just after the switch is closed. Phasors describe only the steady state, which is why we never needed to say at which instant a diagram was drawn.

InteractivePhasor lab

An interactive phasor diagram beside graphs of voltage and current. Tabs choose a resistor, an inductor, a capacitor or a series LCR circuit driven by a 10 volt peak source. The phasors turn slowly, or can be paused and stepped by 30 degrees. Sliders set the frequency from 10 to 250 hertz and the resistance, inductance and capacitance. Readouts give the reactances, impedance, peak current, the phase of the current, the power factor and the average power; in the LCR tab they also give the peak voltages across each element and the resonant frequency, and a status line says whether the circuit is capacitive, inductive or at resonance.

Fig. 7.11Choose an element, change the frequency and the component values, and watch the phasors and the waveforms of voltage and current.

Try this

Explore the phasor lab

Choose L only. Which phasor is ahead, $𝐕$ or $𝐈$ ? Raise the frequency from $50 H z$ to $100 H z$ and watch $X_{L}$ and the peak current.
Choose C only and do the same. Which way does the current change this time?
Choose Series LCR. The status line says whether $X_{C}$ or $X_{L}$ is larger. Slide the frequency until the current reads “in phase”. Compare that frequency with the resonance readout $f_{0}$ , and compare the $V_{L}$ and $V_{C}$ readings with the $10 V$ source.

L only: $𝐈$ trails $𝐕$ by $90^{\circ}$ . Doubling the frequency doubles $X_{L}$ (from $31.4 Ω$ to $62.8 Ω$ ) and halves the peak current (from $318 m A$ to $159 m A$ ). C only: $𝐈$ is $90^{\circ}$ ahead, and doubling the frequency halves $X_{C}$ (from $159 Ω$ to $79.6 Ω$ ) and doubles the current (from about $63 m A$ to $126 m A$ ). Series LCR: the current comes into phase at about $79.6 H z$ , which is $1 / (2 π \sqrt{L C})$ for $L = 200 m H$ and $C = 20 μ F$ . There $V_{L}$ and $V_{C}$ are equal, about $16.7 V$ each, larger than the source voltage, and they cancel.

Example 7.6NCERT Example 7.6 · p. 190

Two voltmeter readings that add up to too much

A lamp of resistance $100 Ω$ (treat it as a pure resistor) is connected in series with a choke of inductance $0.50 H$ and negligible resistance across the $230 V$ , $50 H z$ mains. (a) Find the rms current. (b) A student measures the rms voltage across the lamp and across the choke with an ac voltmeter, adds the two readings and gets far more than $230 V$ . Is the meter faulty?

Try it yourself first. Then reveal the solution one step at a time.

7.6.2 Resonance

Ride in an old bus and you may notice a window pane that rattles loudly at one particular engine speed and stays quiet at others. The pane has a natural frequency of its own. When the engine shakes it at that frequency, each small shake adds to the last and the vibration grows large. A system driven at or near its natural frequency responds strongly; this is resonance (अनुनाद). A series LCR circuit has a natural frequency too.

DerivationThe resonant frequency

Write out the impedance in (7.25a) with both reactances in full:

i_{m} = \frac{v_{m}}{Z} = \frac{v_{m}}{\sqrt{R^{2} + {(\frac{1}{ω C} - ω L)}^{2}}}

The squared bracket can never be negative, so $Z$ can never be smaller than $R$ . At low frequencies $1 / ω C$ is the larger term and at high frequencies $ω L$ is (Fig. 7.9), so exactly one angular frequency $ω_{0}$ balances them and empties the bracket:

ω_{0} L = \frac{1}{ω_{0} C}

ω_{0} = \frac{1}{\sqrt{L C}}, ν_{0} = \frac{1}{2 π \sqrt{L C}}

(7.28)

At $ω_{0}$ the source sees only the resistance. The current amplitude reaches its ceiling, $v_{m} / R$ , and since $\tan ϕ = 0$ the current rises and falls exactly in step with the source voltage. $ω_{0}$ is the resonant frequency of the circuit.

What does the resistance change? Fig. 7.12 answers this for a circuit with $L = 0.50 m H$ and $C = 8.0 n F$ driven by a source of peak voltage $10 V$ . Since $R$ appears nowhere in (7.28), the curves for $R = 40 Ω$ and $R = 100 Ω$ peak at the same place, $ω_{0} = 1 / \sqrt{L C} = 5.0 \times 10^{5} r a d / s$ . What $R$ controls is the height of the peak, $v_{m} / R$ ( $0.25 A$ against $0.10 A$ ), and its width: the smaller resistance gives the taller, narrower peak. To the left of $ω_{0}$ the capacitor's reactance wins and the current leads; to the right the inductor's wins and the current lags.

3/4With $R = 100 Ω$ the peak is at the same $ω_{0}$ but lower ( $0.10 A$ ) and broader.

Fig. 7.12The current peaks at

ω_{0} = 1 / \sqrt{L C}

; a smaller resistance gives a taller, sharper peak.NCERT Fig. 7.14 · p. 189

A graph of peak current in amperes, from 0 to 0.3, against angular frequency from 0 to 10 times 10 to the 5 radians per second, for a circuit with L equals 0.50 millihenry, C equals 8.0 nanofarad and a 10 volt source. Step 1: a blue curve for R equals 40 ohm rises slowly, peaks sharply at 0.25 ampere and falls again. Step 2: a dotted line marks the peak at omega 0 equals 5 times 10 to the 5 radians per second, with the label i m equals v m over R. Step 3: a dashed orange curve for R equals 100 ohm peaks at the same frequency but only at 0.10 ampere, and is broader. Step 4: the region left of the peak is shaded and labelled capacitive, i leads v, and the region to the right is labelled inductive, i lags v.

The sharpness of the peak is measured by the quality factor of the circuit, a pure number:

Q = \frac{ω_{0} L}{R} = \frac{1}{ω_{0} C R}

For the curves above, $ω_{0} L = 250 Ω$ , so $Q = 6.25$ when $R = 40 Ω$ and $Q = 2.5$ when $R = 100 Ω$ . At resonance the voltage across the inductor is $i_{m} X_{L} = (v_{m} / R) ω_{0} L = Q v_{m}$ , and the same across the capacitor. A high- $Q$ circuit builds up voltages across $L$ and $C$ many times the source voltage, even though the two cancel each other in the loop.

Dozens of stations reach a radio's aerial at once, each on its own carrier frequency. The set picks out one with its tuning circuit, a series LCR circuit in which every station's signal acts as a separate small source. Take $L = 0.20 m H$ and $R = 20 Ω$ , and turn the tuning capacitor to $156 p F$ : this puts $ν_{0}$ at $900 k H z$ . A station on $900 k H z$ now drives about 37 times as much current as an equally strong one on $1250 k H z$ , and about 45 times as much as one on $610 k H z$ , so it is the only one you hear. Turning the knob changes $C$ and slides $ν_{0}$ along the dial; a higher $Q$ makes the choice sharper.

InteractiveResonance lab

Two tabs. Resonance curve: a graph of peak current against angular frequency for a series LCR circuit with a 0.50 millihenry inductor and a 10 volt source, with sliders for the drive frequency, the resistance and the capacitance. A dot marks the current at the drive frequency and a dotted line marks the resonant frequency; a button pins the present curve as a dashed outline for comparison. Readouts give the reactances, impedance, current, phase, the peak voltages across each element and the quality factor. Radio tuner: a dial from 500 to 1500 kilohertz with three stations, a tuning-capacitor slider and a resistance slider; bars show the current each station drives in the circuit and readouts name the loudest station.

Fig. 7.13Sweep the drive frequency across the resonance peak, compare resistances, then tune a radio.

Try this

Find the peak, then tune a radio

In Resonance curve, drag the drive frequency until the dot on the curve sits on top of the peak. Compare the drive $ω$ with the $ω_{0}$ readout, and the $V_{L}$ and $V_{C}$ readings with the $10 V$ source.
Press Pin this curve, then raise $R$ from $40 Ω$ to $100 Ω$ . How do the height and width of the peak change? Does $ω_{0}$ move?
Now make $C$ smaller. Which way does the peak move?
Open Radio tuner and turn the tuning capacitor until Station B ( $900 k H z$ ) is loudest. What capacitance does it need? Then raise $R$ to $300 Ω$ . Can you still hear B on its own?

The dot reaches the top of the curve when the drive frequency equals $ω_{0} = 5.00 \times 10^{5} r a d / s$ . There, with $R = 40 Ω$ , $V_{L} = V_{C} = 62.5 V$ , more than six times the source voltage, and the current is in phase. With $R = 100 Ω$ the peak drops to $0.10 A$ and spreads out, but stays at the same $ω_{0}$ . A smaller $C$ raises $ω_{0} = 1 / \sqrt{L C}$ , so the peak moves to the right. Station B needs about $156 p F$ , from $C = 1 / (4 π^{2} ν^{2} L)$ . With $R = 20 Ω$ its current is about 37 times that of the next station; with $R = 300 Ω$ only about 2.7 times, so the neighbouring stations would be heard as well.

7.7 Power in an ac circuit: the power factor

A kettle, a ceiling-fan motor and the choke of a tube light can all draw the same current from the same $230 V$ supply and yet take very different amounts of power. For a resistor, $P = V I$ . In general, though, the current is shifted in phase from the voltage, and that shift changes the average power.

DerivationAverage power in an ac circuit

Let the source voltage be $v = v_{m} \sin ω t$ and the current $i = i_{m} \sin (ω t + ϕ)$ , as in a series LCR circuit. The power delivered by the source at each instant is

p = v i = v_{m} i_{m} \sin ω t \sin (ω t + ϕ)

Using the identity $2 \sin A \sin B = \cos (A - B) - \cos (A + B)$ ,

p = \frac{v_{m} i_{m}}{2} [\cos ϕ - \cos (2 ω t + ϕ)]

(7.29)

The first term is constant. The second oscillates at twice the source frequency and averages to zero over a cycle. So the average power is

P = \frac{v_{m} i_{m}}{2} \cos ϕ = \frac{v_{m}}{\sqrt{2}} \frac{i_{m}}{\sqrt{2}} \cos ϕ = V I \cos ϕ

(7.30a)

Since $V = I Z$ , this can also be written

P = I^{2} Z \cos ϕ

(7.30b)

So two circuits with the same rms voltage and the same rms current can take quite different powers: the phase angle enters through its cosine.

The factor $\cos ϕ$ is called the power factor (शक्ति गुणांक). The supply must be able to deliver the product $V I$ , but only the fraction $\cos ϕ$ of it is actually used. Fig. 7.14 shows how a growing phase angle eats into the average power: more and more of each cycle is spent handing energy back to the source.

2/3The current lags by $60^{\circ}$ ( $ϕ = - 60^{\circ}$ ). For part of each cycle $p < 0$ and energy flows back to the source: $P = V I \cos 60^{\circ} = V I / 2$ .

Fig. 7.14The larger the phase difference, the more of each cycle the power is negative, and the smaller the average power

V I \cos ϕ

A graph against omega t over one cycle showing the voltage v as a solid blue sine wave, the current i as a dashed orange wave, and the instantaneous power p as a thick green curve, with a dashed black line at its average. Step 1: phi equals 0, cos phi equals 1; the current is in step with the voltage, p is never negative and its average is half its peak. Step 2: the current lags by 60 degrees, so phi equals minus 60 degrees and cos phi equals 0.5; small loops of p dip below the axis, shaded orange, and the average line is lower, at half its previous height. Step 3: the current lags by 90 degrees, phi equals minus 90 degrees and cos phi equals 0; the loops above and below the axis are equal and the average line lies on the axis.

Definition

power factor

The factor $\cos ϕ$ in the average power $P = V I \cos ϕ$ , where $ϕ$ is the phase difference between voltage and current; for a series LCR circuit $\cos ϕ = R / Z$ .

ExampleA motor drawing $10 A$ from $230 V$ at power factor $0.8$ takes $1840 W$ .

Table 7.1 collects the cases you will meet most often.

Table 7.1Power factor and average power for common circuitsNCERT p. 191

Circuit	Phase angle φ	Power factor	Average power	Comment
Pure resistor	$0$	$1$	$V I = I^{2} R$	All of VI is used
Pure inductor or capacitor	$\pm 90^{\circ}$	$0$	$0$	Current is wattless
Series RL, RC or LCR	$\tan^{- 1} \frac{X_{C} - X_{L}}{R}$	$R / Z$	$I^{2} R$	Only the resistor dissipates energy
Series LCR at resonance	$0$	$1$	$I^{2} R$ , with $I = V / R$	Largest power for that circuit

Two entries deserve a comment. A current that is a quarter cycle out of step with the voltage, as in an ideal inductor or capacitor, moves energy out and back without delivering any; it is called a wattless current (वाटहीन धारा). And for any series circuit, the impedance triangle gives $\cos ϕ = R / Z$ , so (7.30b) becomes $P = I^{2} Z (R / Z) = I^{2} R$ : however the reactances are arranged, every watt ends up as heat in the resistance.

For any combination of $R$ , $L$ and $C$ , $∣ ϕ ∣$ never exceeds $90^{\circ}$ , so $\cos ϕ$ is never negative: the average power taken by such a circuit is never negative, even though the instantaneous power often is. You can read $\cos ϕ$ as the share of the product $V I$ that the circuit really turns into heat or work: all of it for a pure resistance, none of it for a pure reactance.

Example 7.7NCERT Example 7.7 · p. 191

Fixing a poor power factor

A water pump in a workshop runs from the $230 V$ , $50 H z$ supply. Its motor takes $6.9 k W$ at a power factor of $0.60$ , with the current lagging the voltage. The cable from the local transformer has a total resistance of $0.20 Ω$ . (a) Find the current and the power wasted in the cable. (b) Explain how a capacitor connected in parallel with the motor can raise the power factor of the supply to 1, and find the capacitance needed. (c) What are the current and the cable loss then?

Try it yourself first. Then reveal the solution one step at a time.

2/4 $I_{p} = 30 A$ along $𝐕$ carries the power; $I_{q} = 40 A$ , at right angles behind $𝐕$ , is wattless.

Fig. 7.15Adding a capacitor current that leads

𝐕

90^{\circ}

cancels the wattless part of a lagging current.NCERT Fig. 7.15 · p. 192

A phasor diagram with the voltage phasor V drawn to the right along the horizontal as the reference; phasors turn anticlockwise, so a lagging current points below V. Step 1: the motor current I, 50 amperes, points down and to the right at an angle phi of 53 degrees below V, and the text reads motor current 50 A, lags V by 53 degrees, power factor 0.60. Step 2: I is split into a green part I p of 30 amperes along V and a dashed purple part I q of 40 amperes pointing straight down from the tip of I p to the tip of I, with a right-angle mark; the text reads power part 30 A, wattless part 40 A. Step 3: a capacitor current I C of 40 amperes is drawn from the origin pointing straight up, 90 degrees ahead of V. Step 4: I C slides across and is placed head to tail at the tip of I, pointing up; it ends exactly on V, so the line current, motor current plus capacitor current, is just I p: 30 amperes in phase with V, power factor 1. The motor current is shown faint.

Example 7.8NCERT Example 7.8 · p. 192

How hot will the rheostat get?

In a laboratory, a $10 Ω$ rheostat, a coil of inductance $95.5 m H$ (negligible resistance) and a $531 μ F$ capacitor are joined in series across the $230 V$ , $50 H z$ supply. The rheostat is rated to dissipate at most $1.0 k W$ . (a) What rms current does the circuit draw? (b) Will the rheostat overheat? (c) What is the power factor, and does the current lead or lag the supply voltage, and by how much?

Try it yourself first. Then reveal the solution one step at a time.

Example 7.9NCERT Example 7.9 · p. 193

Tuning the supply to the circuit

The laboratory supply of Example 7.8 also has a frequency control; its voltage stays at $230 V$ . (a) At what frequency will the rheostat take the most power? (b) Find that power. Is the rheostat still safe? (c) What rms voltage then appears across the capacitor?

Try it yourself first. Then reveal the solution one step at a time.

Example 7.10NCERT Example 7.10 · p. 193

How a traffic signal knows a car is waiting

At many traffic signals and toll plazas a rectangular loop of wire is buried just under the road surface. The loop is the inductor of a tuned LC circuit, and the detector watches the circuit's resonant frequency. When a car stops over the loop, the signal controller is told that a vehicle is waiting. Explain the physics that lets the buried loop sense the car.

Try it yourself first. Then reveal the solution one step at a time.

7.8 Transformers

A doorbell needs $12 V$ ; the national grid carries hundreds of kilovolts. The same kind of device serves both: a transformer (ट्रांसफ़ॉर्मर), which works by mutual induction (Chapter 6) and can turn one alternating voltage into a larger or a smaller one.

Look inside a transformer and you find an iron core and two separate windings of insulated copper wire (Fig. 7.16). Nothing joins the windings electrically. What couples them is magnetic flux, which the soft-iron core carries from one winding to the other. The winding fed by the supply is called the primary ( $N_{p}$ turns); the one connected to the load is the secondary ( $N_{s}$ turns). Designers either place the two windings on opposite sides of a closed iron frame, as in the figure, or wind one directly on top of the other so that hardly any flux can slip between them.

4/5The changing flux induces an emf in the secondary coil ( $N_{s}$ turns), which drives current through the load.

Fig. 7.16The flux set up by the primary is guided by the iron core through every turn of the secondary.NCERT Fig. 7.16 · p. 194

A thick grey ring represents the soft-iron core. Step 1: the core alone, labelled soft-iron core. Step 2: a primary winding of many turns, drawn as orange strokes across the left side of the ring, is connected to an ac source, with an arrow for its current; it is labelled primary, N p. Step 3: a dashed blue circle with arrowheads runs round inside the ring, showing the magnetic flux Phi guided by the core. Step 4: a secondary winding with fewer turns appears on the right side of the ring, connected to a glowing lamp, labelled secondary, N s. Step 5: the ratio V s over V p equals N s over N p is written in the middle of the ring.

DerivationThe ideal transformer

Start with an ideal transformer: windings with no resistance, a core that wastes no energy, and a core so good at guiding flux that every line of it passes through every turn of both windings. Let $Φ$ be the flux in the core, and so the flux through each single turn, at time $t$ .

Faraday's law applies to each turn on its own. One loop threaded by a changing flux has an emf $- d Φ / d t$ , and since the core gives every turn the same flux, every turn on the core, primary or secondary, has the same emf. Engineers call this the volts per turn of the transformer. Turns joined in series add their emfs, so each winding's emf is its number of turns times the volts per turn:

ε_{s} = - N_{s} \frac{d Φ}{d t}

(7.31)

ε_{p} = - N_{p} \frac{d Φ}{d t}

(7.32)

Next, link these emfs to the voltages at the terminals. On the secondary side, if the load draws only a small current, almost no voltage is lost inside the winding, so the terminal voltage $v_{s}$ is simply the emf. On the primary side the supply drives current through a winding with no resistance, so the loop rule leaves nothing over: at every instant the back emf of the primary must be exactly as large as the applied voltage $v_{p}$ . The flux in the core adjusts itself until this is true, which means the supply, not the load, sets how fast the flux changes. Taking each terminal voltage in the same sense as the emf of its own winding, both statements read

v_{s} = - N_{s} \frac{d Φ}{d t}

(7.31a)

v_{p} = - N_{p} \frac{d Φ}{d t}

(7.32a)

The volts per turn is the same on both sides, so dividing one equation by the other removes it:

\frac{v_{s}}{v_{p}} = \frac{N_{s}}{N_{p}}

(7.33)

In words, the voltage is shared out in proportion to the number of turns. To find the currents, switch from flux to energy. An ideal transformer has nowhere to lose energy, so whatever power the load takes from the secondary, the primary must draw from the supply at the same instant:

i_{p} v_{p} = i_{s} v_{s}

(7.34)

Combining (7.33) and (7.34),

\frac{i_{p}}{i_{s}} = \frac{v_{s}}{v_{p}} = \frac{N_{s}}{N_{p}}

(7.35)

All these voltages and currents oscillate together at the supply frequency, so the same ratios link their peak values and their rms values:

V_{s} = \frac{N_{s}}{N_{p}} V_{p}, I_{s} = \frac{N_{p}}{N_{s}} I_{p}

(7.36)

Read (7.36) as a trade: whatever factor the voltage is multiplied by, the current is divided by the same factor. With more turns on the secondary ( $N_{s} > N_{p}$ ) the transformer raises the voltage and lowers the current; it is a step-up transformer. With fewer turns on the secondary it is a step-down transformer, lowering the voltage and raising the current. The same transformer can often do either job, depending on which winding is connected to the supply.

Example

A doorbell transformer

A bell transformer steps the $230 V$ mains down to $12 V$ . Its primary has 2300 turns. (a) How many turns does the secondary need? (b) When the bell draws $1.5 A$ , what current flows in the primary if the transformer is ideal? (c) What would the primary current be for a transformer that is $90 %$ efficient?

Try it yourself first. Then reveal the solution one step at a time.

Each simplification in the derivation hides a way in which a real transformer falls short. Resistance in the primary uses up part of the supply voltage, so less is left to balance the emf. Flux that strays out of the core threads the primary but misses the secondary. And a large load current loses some voltage inside the secondary winding, which is why the output of a small adapter sags a little as the load grows. Even so, real transformers come remarkably close to the ideal: large ones pass on more than $95 %$ of the power they receive, and the largest more than $99 %$ . The energy they do lose, nearly all as heat, goes in four main ways (Table 7.2).

Table 7.2Energy losses in a real transformerNCERT p. 195

Loss	What happens	How it is reduced
Flux leakage	Some field lines leak out of the core and never thread the secondary, so its emf falls a little short of what the turns ratio predicts.	Wind one coil directly over the other on a closed core, so that the two share almost all the flux.
Resistance of the windings	Current heats the copper wire at the rate $I^{2} R$ .	Use thicker copper, above all on the winding that carries the larger current (the low-voltage side).
Eddy currents	The changing flux drives loops of current round inside the iron, and the iron warms up.	Build the core from thin, insulated sheets (laminations) that break up those loops.
Hysteresis	The core's magnetisation is reversed twice every cycle, and a little energy turns into heat each time.	Use a soft magnetic material, such as silicon steel, whose magnetisation reverses with little loss.

Transformers are what make long-distance transmission practical. A line of resistance $R_{ℓ}$ carrying a current $I$ loses power $I^{2} R_{ℓ}$ . For a given power $P = V I$ , raising the voltage lowers the current in the same proportion and cuts the loss by the square of that factor. In practice the journey has several stages (Fig. 7.17). A step-up transformer beside the power station feeds the long-distance lines at hundreds of kilovolts. Near a town, substations lower it, in one or two steps, to about $11 k V$ for local distribution, and a small transformer on a pole or in a street kiosk brings it down to the $230 V$ used at home.

4/4A transformer on a pole makes the final step down to $230 V$ .

Fig. 7.17Stepping up the voltage for transmission cuts the current, and with it the heating loss in the lines.NCERT p. 196

A chain of symbols from left to right with the voltage written above each stage and the current below. Step 1: a generator G at the power station produces 11 kilovolts, with a thick arrow showing a large current of 909 amperes. Step 2: a step-up transformer raises the voltage to 220 kilovolts on a transmission line strung between two towers, and a thin arrow shows that the current is now only 45 amperes. Step 3: a substation transformer lowers the voltage to 11 kilovolts. Step 4: a pole transformer lowers it again to 230 volts for a house.

Example

Why the grid runs at high voltage

A power station sends $10 M W$ along a line of total resistance $2.0 Ω$ . Assume a power factor of 1. Compare the power lost in the line if the energy is sent (a) at $11 k V$ , the generator voltage, and (b) at $220 k V$ , after a step-up transformer.

Try it yourself first. Then reveal the solution one step at a time.

Let's explore

What is inside a charger?

You’ll need

an unplugged mains adapter or charger with a rating label
a calculator
a kitchen scale (optional)

Predict first

A label reads INPUT

230 V

~, OUTPUT

12 V

1.0 A

. If the adapter contained an ideal transformer, what current would it draw from the mains at full load?

About

19 A

About

4.3 m A

1.0 A

About

52 m A

Find an unplugged mains adapter or charger and read its label. Older, heavy adapters (for toys, radios or door bells) often say something like INPUT 230 V~ 50 Hz, OUTPUT 12 V~ 1.0 A.
Work out the turns ratio $N_{s} / N_{p} = V_{s} / V_{p}$ that an ideal transformer inside would need.
Work out the output power $V_{s} I_{s}$ and the primary current an ideal transformer would draw at full load.
Compare the weight of the old adapter with that of a modern phone charger of similar power.

Your adapter

Quantity	Value
Input voltage (V)
Output voltage (V)
Turns ratio Ns/Np
Output power (W)
Ideal primary current (A)

Your entries are saved automatically.

For a $12 V$ output the turns ratio is about $1 : 19$ , and at full load the primary carries only a few tens of milliamperes. Old adapters are heavy because they contain an iron-cored transformer working at $50 H z$ . Modern chargers are light because they first switch the current on and off tens of thousands of times a second and then use a tiny transformer. The emf per turn is proportional to the rate of change of flux, so at a high frequency a much smaller core, carrying much less flux, does the same job.

Table 7.3Quantities met in this chapterNCERT p. 198

Quantity	Symbol	SI unit	Dimensions	Remarks
rms voltage	$V$	volt (V)	$[M L^{2} T^{- 3} A^{- 1}]$	$V = v_{m} / \sqrt{2}$ for a sinusoidal voltage of amplitude $v_{m}$
rms current	$I$	ampere (A)	$[A]$	$I = i_{m} / \sqrt{2}$
inductive reactance	$X_{L}$	ohm (Ω)	$[M L^{2} T^{- 3} A^{- 2}]$	$X_{L} = ω L$
capacitive reactance	$X_{C}$	ohm (Ω)	$[M L^{2} T^{- 3} A^{- 2}]$	$X_{C} = 1 / ω C$
impedance	$Z$	ohm (Ω)	$[M L^{2} T^{- 3} A^{- 2}]$	depends on the elements; $\sqrt{R^{2} + (X_{C} - X_{L})^{2}}$ for series LCR
resonant frequency	$ω_{0}$ (or $ν_{0}$ )	rad/s (or Hz)	$[T^{- 1}]$	$ω_{0} = 1 / \sqrt{L C}$ for series LCR
quality factor	$Q$	none	dimensionless	$Q = ω_{0} L / R = 1 / (ω_{0} C R)$ for series LCR
power factor	$\cos ϕ$	none	dimensionless	$ϕ$ is the phase difference between voltage and current

Recap

Alternating Current: key points and summary

An alternating voltage $v = v_{m} \sin ω t$ reverses direction every half cycle; the Indian mains has $ν = 50 H z$ and $V = 230 V$ rms (Section 7.1).
In a resistor $i = (v_{m} / R) \sin ω t$ : the current is in phase with the voltage (Section 7.2).
The average power in a resistor is $\frac{1}{2} i_{m}^{2} R$ . With rms values $I = i_{m} / \sqrt{2}$ and $V = v_{m} / \sqrt{2}$ , $P = I^{2} R = V^{2} / R = V I$ , as for dc (Section 7.2).
Meter readings, supply voltages and power ratings are rms or average values: $230 V$ rms has a peak of about $325 V$ .
A phasor is a rotating arrow whose vertical projection gives the instantaneous value; lengths give amplitudes and angles give phase differences (Section 7.3).
In a pure inductor $X_{L} = ω L$ and the current is $π / 2$ behind the voltage (Section 7.4).
In a pure capacitor $X_{C} = 1 / ω C$ and the current is $π / 2$ ahead of the voltage. Neither a pure inductor nor a pure capacitor takes any average power (Section 7.5).
Series LCR circuit: $Z = \sqrt{R^{2} + (X_{C} - X_{L})^{2}}$ and $\tan ϕ = (X_{C} - X_{L}) / R$ . Voltages across the elements add as phasors, not as numbers (Section 7.6.1).
Resonance occurs at $ω_{0} = 1 / \sqrt{L C}$ : $X_{L} = X_{C}$ , $Z = R$ and the current is largest and in phase with the voltage. A smaller $R$ gives a sharper peak, measured by $Q = ω_{0} L / R$ (Section 7.6.2).
Average power $P = V I \cos ϕ$ ; $\cos ϕ$ is the power factor. A current with $\cos ϕ = 0$ is wattless (Section 7.7).
A low power factor means a larger current, and larger line losses, for the same power; a capacitor in parallel can correct it (Example 7.7).
An ideal transformer has $V_{s} / V_{p} = N_{s} / N_{p}$ and $I_{s} / I_{p} = N_{p} / N_{s}$ . High-voltage transmission reduces $I^{2} R$ losses; real transformers lose a little energy through flux leakage, winding resistance, eddy currents and hysteresis (Section 7.8).

Flash cards · 8

Formula flashcards

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Our practice questions are written for ncert.ai and follow the NCERT exercises. Find the one that matches the question in your textbook; its answer is with it.

In your NCERT textbook	Practice question here
NCERT Exercises 7.1	Practice Q1
NCERT Exercises 7.2	Practice Q2
NCERT Exercises 7.3	Practice Q3
NCERT Exercises 7.4	Practice Q4
NCERT Exercises 7.5	Practice Q5
NCERT Exercises 7.6	Practice Q6
NCERT Exercises 7.7	Practice Q7
NCERT Exercises 7.8	Practice Q8

Quick check: MCQs on Alternating Current

7.1 Introduction3 questions
7.2 AC voltage applied to a resistor4 questions
7.3 Representing ac by rotating vectors: phasors3 questions
7.4 AC voltage applied to an inductor4 questions
7.5 AC voltage applied to a capacitor4 questions
7.6.1 Phasor-diagram solution4 questions
7.6.2 Resonance4 questions
7.7 Power in an ac circuit: the power factor4 questions
7.8 Transformers4 questions

Glossary · 12

Key terms (English–Hindi)

alternating current (ac)प्रत्यावर्ती धाराpratyavarti dhara: A current (or voltage) that varies sinusoidally with time and reverses direction every half cycle, such as $i = i_{m} \sin ω t$ .First seen in the chapter →

capacitive reactanceधारितीय प्रतिघातdharitiya pratighat: $X_{C} = 1 / ω C$ , the opposition of a capacitor to ac, in ohms; it falls as frequency rises.First seen in the chapter →

impedanceप्रतिबाधाpratibadha: The ratio $Z = V / I$ for an ac circuit; for series LCR, $Z = \sqrt{R^{2} + (X_{C} - X_{L})^{2}}$ .First seen in the chapter →
inductive reactanceप्रेरकीय प्रतिघातprerakiya pratighat: $X_{L} = ω L$ , the opposition of an inductor to ac, in ohms; it rises with frequency.First seen in the chapter →

phasorफेज़रphezar: A rotating arrow representing a sinusoidal quantity: its length is the amplitude and its vertical projection is the instantaneous value.First seen in the chapter →
power factorशक्ति गुणांकshakti gunank: $\cos ϕ$ in $P = V I \cos ϕ$ ; equal to $R / Z$ for a series LCR circuit.First seen in the chapter →

quality factor (Q): $Q = ω_{0} L / R$ , a pure number measuring how sharp the resonance peak is.First seen in the chapter →

resonanceअनुनादanunaad: The large response of a system driven at its natural frequency; in a series LCR circuit, the current peak when $X_{L} = X_{C}$ .First seen in the chapter →
resonant frequencyअनुनादी आवृत्तिanunadi aavritti: The frequency at which a series LCR circuit resonates: $ω_{0} = 1 / \sqrt{L C}$ .First seen in the chapter →
rms valueवर्ग माध्य मूल मानvarg madhya mool maan: The square root of the mean of the square over a cycle; for a sine wave, the peak value divided by $\sqrt{2}$ .

transformerट्रांसफ़ॉर्मरtransformer: A device with two coils on a shared iron core that changes an ac voltage by mutual induction: $V_{s} / V_{p} = N_{s} / N_{p}$ .First seen in the chapter →

wattless currentवाटहीन धाराwatt-heen dhara: A current that is $90^{\circ}$ out of phase with the voltage, so it carries no average power.First seen in the chapter →

Alternating Current: Class 12 Physics · Chapter 7

What you will learn in Alternating Current

7.1 Introduction

7.2 AC voltage applied to a resistor

7.3 Representing ac by rotating vectors: phasors

7.4 AC voltage applied to an inductor

7.5 AC voltage applied to a capacitor

7.6 AC voltage applied to a series LCR circuit

7.6.1 Phasor-diagram solution

7.6.2 Resonance

7.7 Power in an ac circuit: the power factor

7.8 Transformers

Alternating Current: key points and summary

Practice questions with answers (NCERT “Exercises”)

Find your NCERT exercise question

Quick check: MCQs on Alternating Current

Key terms (English–Hindi)

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