Introduction to Alternating Currents and Circuits

Q: Why does current lag voltage in an inductor?

An inductor opposes any change in current by a back emf L di/dt. The applied voltage is largest when the current is changing fastest, which is when the current passes through zero, so the current reaches its peak a quarter cycle later and lags by 90 degrees.

Q: Why does a capacitor block DC but allow AC?

Capacitive reactance is XC = 1/(2πfC). For steady DC the frequency is zero, the reactance is infinite and the capacitor simply charges and then stops current. For AC it charges and discharges continuously, so current flows, and more easily at higher frequency.

PhysicsAlternating CurrentsFor JEE aspirants

An alternating current (AC) changes its magnitude continuously and reverses its direction periodically, usually as $i = i_{0} sin (ω t + ϕ)$ . This page covers alternating current from the ground up: period and frequency, phasors, average and rms values, the ac generator, how a pure resistor, inductor and capacitor behave in an AC circuit, reactance, and the transformer. These basics carry sure-shot questions in NEET and JEE Main every year.

On this page1What is AC2Phasors3Average and rms4AC generator5Pure R6Pure L7Pure C8Reactance and power9Transformer

Key Formulas - Quick Reference

★ Must learn $i = i_{0} sin (ω t + ϕ)$ , $v = V_{0} sin (ω t + ϕ)$ ; $ω = 2 π f = \frac{2 π}{T}$
Average over a full cycle $= 0$ ; over a half cycle $i_{a v g} = \frac{2 i _{0}}{π} \approx 0.637 i_{0}$
★ Must learn $i_{r m s} = \frac{i _{0}}{2} \approx 0.707 i_{0}$ , $V_{r m s} = \frac{V _{0}}{2}$ ; in general $i_{r m s} = \overline{i^{2}}$
Generator emf $e = N B A ω sin ω t$ , peak $e_{0} = N B A ω$
★ Must learnInductive reactance $X_{L} = ω L = 2 π f L$ ; capacitive reactance $X_{C} = \frac{1}{ω C} = \frac{1}{2 π f C}$ (both in ohm)
★ Must learnPhase: R, $v$ and $i$ in phase; L, $i$ lags $v$ by $\frac{π}{2}$ ; C, $i$ leads $v$ by $\frac{π}{2}$
Average power: $P = V_{r m s} I_{r m s} = I_{r m s}^{2} R$ for R; $P = 0$ for pure L or pure C
★ Must learnTransformer: $\frac{V _{s}}{V _{p}} = \frac{N _{2}}{N _{1}} = \frac{I _{p}}{I _{s}}$ (ideal); efficiency $η = \frac{P _{o u t}}{P _{in}} \times 100%$

1. What Is Alternating Current?

So far we have met only direct current (DC), which flows in one direction. Its usual source is a battery: current leaves the positive terminal, passes through the external circuit and returns to the negative terminal. A DC may be steady or may vary in size, but it never reverses.

Most electric power produced and used in the world is alternating current (AC): its magnitude changes continuously with time and its direction reverses periodically. The commonest form is sinusoidal:

i = i_{0} sin (ω t + ϕ)

Figure 1: DC may vary in size but never changes direction; AC reverses direction periodically. The sinusoidal form

i = i_{0} sin (ω t + ϕ)

is the one produced by generators and used in all formulas here.

Symbol	Name	Meaning
$i$	Instantaneous current	Value of the current at time $t$
$i_{0}$	Peak current (current amplitude)	Largest value of the current in a cycle
$T$	Time period	Time after which the current repeats, $T = \frac{2 π}{ω}$
$f$	Frequency	Number of cycles per second, $f = \frac{1}{T}$ ; unit hertz (Hz)
$ω$	Angular frequency	$ω = 2 π f$ , unit $rad s^{- 1}$
$ω t + ϕ$	Phase	$ϕ$ is the initial phase (phase at $t = 0$ )

The current is positive for half the period and negative for the other half, so its direction reverses every half period. An alternating voltage has the same form, $v = V_{0} sin (ω t + ϕ)$ , where $V_{0}$ is the peak voltage. It is produced by an ac generator (ac dynamo, Section 4). The frequency of the mains supply in India is $50 Hz$ , so the current reverses $100$ times every second.

An AC circuit is a resistor, inductor, capacitor or any combination of them connected to an ac source. The source is drawn as a circle containing the symbol $\sim$ .

Why is AC preferred for power supply? AC voltages can be stepped up or down easily and efficiently by a transformer (Section 6). Power is sent over long lines at a very high voltage and small current, which keeps the $I^{2} R$ heating loss in the wires small, and is stepped down again for homes. AC generators are also simpler to build than DC ones.

Key idea

AC reverses direction every half cycle; one cycle takes

T = \frac{2 π}{ω} = \frac{1}{f}

, and in India

f = 50 Hz

2. Phasors: Picturing an Alternating Quantity

A phasor is a vector that rotates anticlockwise about the origin with angular speed $ω$ . Its length equals the peak value ( $V_{0}$ or $i_{0}$ ) and its projection on the vertical axis gives the instantaneous value. At time $t$ a phasor of length $i_{0}$ makes angle $ω t$ with the horizontal axis, so its projection is $i_{0} sin ω t$ .

Figure 2: A phasor is an arrow of length equal to the peak value, rotating anticlockwise at

ω

. Its projection on the vertical axis is the instantaneous value

i = i_{0} sin ω t

(drawn at

ω t = 5 0^{\circ}

The angle between two phasors is the phase difference between the two quantities. It stays the same as they rotate, because both rotate at the same $ω$ .
A phasor that is ahead (anticlockwise) of another leads it; one that is behind (clockwise) lags.
Phasors of quantities in series (for example voltages across parts carrying the same current) are added like vectors. This is how the impedance of a series LCR circuit is found (next concept).

3. Average and RMS Values of Alternating Current

3.1 Average (mean) value

The mean value of a current over a time $T$ is $\overset{ˉ}{i} = \frac{1}{T} \int_{0}^{T} i d t$ . For $i = i_{0} sin (ω t + ϕ)$ over one complete cycle:

\overset{ˉ}{i} = \frac{i _{0}}{T} \int_{0}^{T} sin (ω t + ϕ) d t = - \frac{i _{0}}{ω T} [cos (ω t + ϕ)]_{0}^{T} = - \frac{i _{0}}{2 π} [cos (2 π + ϕ) - cos ϕ] = 0

The positive half cancels the negative half, so the average over any whole number of cycles is zero. That is why a DC instrument (moving coil meter) shows no deflection for AC. The average is therefore defined over a half cycle:

i_{a v g} = \frac{\int _{0}^{T /2} i _{0} sin ω t d t}{T /2} = \frac{2 i _{0}}{π} \approx 0.637 i_{0}, V_{a v g} = \frac{2 V _{0}}{π} \approx 0.637 V_{0}

3.2 Root mean square (rms) value

The rms value is the square root of the mean of the square of the current: $i_{r m s} = \overline{i^{2}}$ .

Mean of the square over one period: $\overline{i^{2}} = \frac{1}{T} \int_{0}^{T} i_{0}^{2} sin^{2} (ω t + ϕ) d t$ .
Use $sin^{2} θ = \frac{1 - cos 2 θ}{2}$ : the $cos 2 θ$ term averages to zero over a period, leaving $\overline{i^{2}} = \frac{i _{0}^{2}}{2}$ .
Take the square root:
$i_{r m s} = \frac{i _{0}}{2} \approx 0.707 i_{0}, V_{r m s} = \frac{V _{0}}{2} \approx 0.707 V_{0}$

Figure 3:

i^{2} = i_{0}^{2} sin^{2} ω t

never goes negative and is symmetric about

\frac{i _{0}^{2}}{2}

, so

\overline{i^{2}} = \frac{i _{0}^{2}}{2}

and

i_{r m s} = \overline{i^{2}} = \frac{i _{0}}{2}

. It repeats twice per cycle of

i

Figure 4: Over a full cycle the positive and negative halves cancel, so

\overset{ˉ}{i} = 0

. Over a half cycle

\overset{ˉ}{i} = \frac{2 i _{0}}{π} \approx 0.637 i_{0}

, and the rms value is

\frac{i _{0}}{2} \approx 0.707 i_{0}

(always larger than the half-cycle average).

★ Must learn

Meaning of rms value. The rms value of an alternating current is that steady (DC) current which produces the same heat in a given resistance in the same time. Heat in one period in a resistor $R$ :

U = \int_{0}^{T} i^{2} R d t = R i_{0}^{2} \int_{0}^{T} sin^{2} (ω t + ϕ) d t = i_{r m s}^{2} R T

So AC currents and voltages are quoted and measured as rms values. The rms value is also called the effective or virtual value.

When we say the household supply is $220 V$ AC, we mean $V_{r m s} = 220 V$ . Its peak value is $2202 \approx 311 V$ , and the voltage swings between $+ 311 V$ and $- 311 V$ (peak-to-peak $622 V$ ).

Waveform (peak $i_{0}$ )	rms value	Average value
Sinusoidal $i_{0} sin ω t$	$\frac{i _{0}}{2}$	$0$ (full cycle), $\frac{2 i _{0}}{π}$ (half cycle)
Full-wave rectified sine	$\frac{i _{0}}{2}$	$\frac{2 i _{0}}{π}$
Half-wave rectified sine	$\frac{i _{0}}{2}$	$\frac{i _{0}}{π}$
Square wave $\pm i_{0}$	$i_{0}$	$0$ (full cycle), $i_{0}$ (half cycle)
Triangular wave $\pm i_{0}$	$\frac{i _{0}}{3}$	$0$ (full cycle), $\frac{i _{0}}{2}$ (half cycle)
Sawtooth $i = i_{0} \frac{t}{T}$	$\frac{i _{0}}{3}$	$\frac{i _{0}}{2}$

Average value

Plain mean of $i$ . Zero over a full cycle of symmetric AC, $\frac{2 i _{0}}{π}$ over a half cycle. Read by a moving coil (DC) meter, which is why such a meter shows zero on AC.

RMS value

Square root of the mean of $i^{2}$ . Never zero; $\frac{i _{0}}{2}$ for a sine. Decides heating and power, and is what AC meters (hot-wire, moving iron) read.

Exam Trick

Peak : rms : half-cycle average $= 1 : 0.707 : 0.637$ . For any sum of sinusoids of different frequencies (or DC plus AC), add the squares of the rms values: $i_{r m s} = i_{1, r m s}^{2} + i_{2, r m s}^{2} + \dots$ . For $a + b sin ω t$ this gives $a^{2} + \frac{b ^{2}}{2}$ at once. Two sinusoids of the same frequency must first be combined into one: $3 sin ω t + 4 cos ω t = 5 sin (ω t + 5 3^{\circ})$ .

Quick Recall: tap to check

What does a moving coil ammeter read when connected in an AC circuit?

Zero: it responds to the average current, which is zero over a full cycle.

The mains is

220 V

. What is the peak voltage?

2202 \approx 311 V

Which is larger for a sine wave: the rms value or the half-cycle average?

The rms value (

0.707 i_{0}

against

0.637 i_{0}

4. The AC Generator

An ac generator (dynamo) converts mechanical energy into electrical energy. It works on electromagnetic induction: when a coil rotates in a magnetic field, the flux through it changes and an emf is induced in it.

Figure 5: AC generator. The coil turns in the field

B

, the flux through it changes, and an emf is induced (Faraday's law). Slip rings and brushes connect the rotating coil to the fixed load without twisting the wires, so the output reverses every half turn.

4.1 Construction

Field magnet: produces the magnetic field. A low-power generator uses a permanent magnet; a large one uses an electromagnet.
Armature: a coil of many turns of insulated wire wound on a soft-iron drum, free to rotate about an axle between the poles. The iron supports the coil and strengthens the magnetic field through it.
Slip rings ( $R_{1}$ , $R_{2}$ ): two metal rings fixed to the axle, each joined to one end of the coil; they rotate with it.
Brushes ( $B_{1}$ , $B_{2}$ ): fixed carbon rods or metal strips pressing on the rings. The output current reaches the external load $R_{L}$ through them.

4.2 Working and emf

As the coil rotates anticlockwise, side $ab$ moves up and side $c d$ moves down; Fleming's right-hand rule gives the direction of the induced current, which flows through the load along $B_{1} R_{L} B_{2}$ . After half a turn $ab$ moves down and $c d$ up, so the current reverses and flows along $B_{2} R_{L} B_{1}$ . The direction of the emf therefore changes every half revolution.

Let the coil have $N$ turns of area $A$ and rotate at angular speed $ω = 2 π f$ in a field $B$ . If its normal is along $B$ at $t = 0$ , the angle at time $t$ is $θ = ω t$ .
Flux linkage: $N Φ = N B A cos ω t$ .
Faraday's law:
$e = - \frac{d}{d t} (N B A cos ω t) = N B A ω sin ω t = e_{0} sin ω t$
Peak emf $e_{0} = N B A ω = 2 π f N B A$ , reached when the plane of the coil is parallel to $B$ (flux zero but changing fastest).

Figure 6:

Φ = N B A cos ω t

and

e = - \frac{d Φ}{d t} = N B A ω sin ω t

. The emf is largest when the flux is zero but changing fastest (coil plane parallel to

B

), and zero when the flux is maximum.

Can a moving coil galvanometer measure the output of an ac generator? No. Its deflection follows the average current, and the average of AC over a full cycle is zero; at $50 Hz$ the coil cannot follow each reversal either. AC is measured with hot-wire or moving-iron meters, which respond to $i^{2}$ and are calibrated in rms values.

Key idea

Generator: rotating coil, changing flux,

e = N B A ω sin ω t

. The emf is maximum when the flux is zero and zero when the flux is maximum.

5. AC Circuits with a Single Element

Kirchhoff's loop rule holds at every instant in an AC circuit. In each case below the source is $v = V_{0} sin ω t$ .

5.1 Pure resistor

Loop rule: $V_{0} sin ω t - i R = 0$ , so

i = \frac{V _{0}}{R} sin ω t = i_{0} sin ω t, i_{0} = \frac{V _{0}}{R}

The voltage across the resistor, $v_{R} = i_{0} R sin ω t$ , and the current reach their maxima together: they are in phase. In the phasor diagram the phasors $V_{0}$ and $i_{0}$ lie along the same line.

Figure 7: Pure resistor.

v

and

i

are in phase:

i = \frac{V _{0}}{R} sin ω t

. The phasors

V_{0}

and

i_{0}

lie along the same line.

Average power: $\overset{ˉ}{P} = \overline{i^{2}} R = \frac{i _{0}^{2} R}{2} = I_{r m s}^{2} R = V_{r m s} I_{r m s}$ .

5.2 Pure inductor

The self-induced emf across an inductor is $L \frac{d i}{d t}$ . Loop rule:

V_{0} sin ω t - L \frac{d i}{d t} = 0 \Rightarrow \frac{d i}{d t} = \frac{V _{0}}{L} sin ω t

Integrating, $i = - \frac{V _{0}}{ω L} cos ω t + C^{'}$ . The average current over a cycle must be zero (there is no DC source), so $C^{'} = 0$ . With $- cos ω t = sin (ω t - \frac{π}{2})$ :

i = \frac{V _{0}}{ω L} sin (ω t - \frac{π}{2}) = i_{0} sin (ω t - \frac{π}{2}), i_{0} = \frac{V _{0}}{X _{L}}

Inductive reactance $X_{L} = ω L = 2 π f L$ is the opposition of an inductor to AC. Its unit is the ohm, and $i_{0} = \frac{V _{0}}{X _{L}}$ (also $I_{r m s} = \frac{V _{r m s}}{X _{L}}$ ) is Ohm's law for an inductor. $X_{L} \propto f$ , so an inductor passes DC ( $f = 0$ , $X_{L} = 0$ ) freely and opposes high frequencies strongly.

The current reaches its maximum a quarter period after the voltage: in a pure inductor the current lags the voltage by $9 0^{\circ}$ .

Figure 8: Pure inductor. The current lags the voltage by

\frac{π}{2}

i = \frac{V _{0}}{ω L} sin (ω t - \frac{π}{2})

;

i_{0}

9 0^{\circ}

clockwise from

V_{0}

5.3 Pure capacitor

Loop rule: $V_{0} sin ω t - \frac{q}{C} = 0$ , so $q = C V_{0} sin ω t$ . Differentiating,

i = \frac{d q}{d t} = ω C V_{0} cos ω t = ω C V_{0} sin (ω t + \frac{π}{2}) = i_{0} sin (ω t + \frac{π}{2}), i_{0} = \frac{V _{0}}{X _{C}}

Capacitive reactance $X_{C} = \frac{1}{ω C} = \frac{1}{2 π f C}$ , in ohm. $X_{C} \propto \frac{1}{f}$ : a capacitor blocks DC, since $X_{C} \to \infty$ when $f = 0$ , and passes high frequencies easily.

The current reaches its maximum a quarter cycle before the voltage: in a pure capacitor the current leads the voltage by $9 0^{\circ}$ . Physically, the current is largest when the capacitor is uncharged and the voltage across it is zero.

Figure 9: Pure capacitor. The current leads the voltage by

\frac{π}{2}

i = ω C V_{0} sin (ω t + \frac{π}{2})

;

i_{0}

9 0^{\circ}

anticlockwise from

V_{0}

5.4 Reactance, frequency and power

Figure 10: For

R = 20 Ω

L = 0.1 H

C = 100 μ F

R

does not depend on

f

;

X_{L} = 2 π f L

grows linearly (zero for DC);

X_{C} = \frac{1}{2 π f C}

falls as a hyperbola (infinite for DC). They cross at

f = 50.3 Hz

, where

X_{L} = X_{C} = 31.6 Ω

Element	Opposition	Phase of $i$ relative to $v$	Average power	Behaviour with DC
Resistor $R$	$R$ (independent of $f$ )	In phase	$V_{r m s} I_{r m s}$	Same as AC
Inductor $L$	$X_{L} = ω L$	Lags by $\frac{π}{2}$	Zero	Short circuit ( $X_{L} = 0$ )
Capacitor $C$	$X_{C} = \frac{1}{ω C}$	Leads by $\frac{π}{2}$	Zero	Open circuit ( $X_{C} \to \infty$ )

The instantaneous power delivered to any element is $p = v i$ . In a resistor it is never negative. In a pure inductor or capacitor energy is stored for a quarter cycle (in the magnetic field $\frac{1}{2} L i^{2}$ or the electric field $\frac{1}{2} C v^{2}$ ) and fully returned in the next quarter cycle, so the average power is zero.

Figure 11:

p = v i

. In a resistor

p = V_{0} I_{0} sin^{2} ω t \geq 0

, average

\frac{V _{0} I _{0}}{2} = V_{r m s} I_{r m s}

. In a pure

L

p = - \frac{V _{0} I _{0}}{2} sin 2 ω t

(in a pure

C

the sign is

+

). Either way the energy stored in one quarter cycle is returned in the next, so the average power is zero.

Exam Trick

ELI the ICE man. In an inductor (L) the voltage E comes before the current I (ELI: current lags). In a capacitor (C) the current I comes before E (ICE: current leads). For the instantaneous current, write $i = \frac{V _{0}}{X} sin (phase of v - \frac{π}{2})$ for L and $+ \frac{π}{2}$ for C.

JEE Advanced

Complex impedance. Write $v = V_{0} e^{j ω t}$ . Then $Z_{R} = R$ , $Z_{L} = j ω L$ and $Z_{C} = \frac{1}{j ω C} = - \frac{j}{ω C}$ , where $j = - 1$ is a $9 0^{\circ}$ anticlockwise turn. Series impedances add and parallel ones combine like resistors, and $∣ Z ∣$ and $ar g Z$ give the amplitude ratio and phase directly. This turns any AC network, even a parallel LC tank, into algebra. Also note: since $X_{L} \to 0$ and $X_{C} \to \infty$ as $f \to 0$ , the long-time DC state of a circuit is found by shorting inductors and removing capacitors.

Figure 12: One element, three steps: find the opposition (

R

ω L

\frac{1}{ω C}

), divide the peak voltage by it, then shift the phase (

0

- \frac{π}{2}

+ \frac{π}{2}

). Mnemonic: ELI the ICE man.

Key idea

Divide by the right opposition (

R

ω L

\frac{1}{ω C}

) to get the current, then shift its phase by

0

- \frac{π}{2}

+ \frac{π}{2}

Quick Recall: tap to check

What is the reactance of a capacitor connected to a steady DC source?

Infinite (

f = 0

), so no steady current flows: the capacitor acts as an open circuit.

In which element does the current lead the voltage?

In a capacitor, by

9 0^{\circ}

How does

X_{L}

change if the frequency is doubled?

It doubles (

X_{L} \propto f

);

X_{C}

would halve.

What is the average power in a pure inductor?

Zero: stored energy is returned every quarter cycle.

6. The Transformer

A transformer changes an alternating voltage from low to high (step-up) or from high to low (step-down) without changing its frequency.

Principle: mutual induction. A changing current in one coil sets up a changing flux that induces an emf in a second coil linked with it.
Construction: two coils of insulated copper wire, the primary (input, $N_{1}$ turns) and the secondary (output, $N_{2}$ turns), wound on the same laminated soft-iron core. The high permeability of soft iron keeps almost all the flux inside the core, so it passes through both coils.
Core type: primary and secondary on separate limbs of the core. Shell type: one coil wound over the other on the same limb.

Figure 13: Step-up transformer (

N_{2} > N_{1}

). The same changing flux

Φ

links every turn of both coils, so

\frac{V _{s}}{V _{p}} = \frac{N _{2}}{N _{1}}

. Lamination cuts eddy-current loss. A transformer works only with AC (or a changing current).

6.1 Voltage and current ratios

With the secondary open, the same flux $Φ$ links each turn of both coils:

V_{p} = - N_{1} \frac{d Φ}{d t}, V_{s} = - N_{2} \frac{d Φ}{d t} \Rightarrow \frac{V _{s}}{V _{p}} = \frac{N _{2}}{N _{1}}

The ratio $\frac{N _{2}}{N _{1}}$ is the turns ratio (transformation ratio). For an ideal transformer (no losses), input power $=$ output power: $V_{p} I_{p} = V_{s} I_{s}$ , so

\frac{I _{p}}{I _{s}} = \frac{V _{s}}{V _{p}} = \frac{N _{2}}{N _{1}}

Step-up transformer

$N_{2} > N_{1}$ , so $V_{s} > V_{p}$ and $I_{s} < I_{p}$ . Used at power stations before long-distance transmission.

Step-down transformer

$N_{2} < N_{1}$ , so $V_{s} < V_{p}$ and $I_{s} > I_{p}$ . Used at substations and in phone chargers and adaptors.

6.2 Efficiency and energy losses

η = \frac{output power}{input power} \times 100%

Real transformers are very efficient ( $90$ to $98%$ , larger ones even higher) but never $100%$ , because of:

Loss	Cause	Remedy
Flux leakage	Some primary flux does not pass through the secondary	Wind one coil over the other; good core design
Copper ( $I^{2} R$ ) loss	Heating of the windings	Thick copper wire for the high-current coil
Eddy current loss	Currents induced in the iron core heat it	Laminated core with insulated sheets
Hysteresis loss	Repeated magnetisation of the core	Soft iron with a narrow hysteresis loop

Exam Trick

A transformer trades voltage for current, never power. If the voltage is stepped up $n$ times, the current falls $n$ times. Sending power $P$ at voltage $V$ means current $\frac{P}{V}$ and line loss $(\frac{P}{V})^{2} R$ : raising $V$ by $10$ times cuts the loss $100$ times. A transformer does not work on steady DC, since the flux does not change.

Key idea

\frac{V _{s}}{V _{p}} = \frac{N _{2}}{N _{1}} = \frac{I _{p}}{I _{s}}

for an ideal transformer; losses come from leakage, copper, eddy currents and hysteresis.

Figure 14: Revision map for alternating current basics: signal, mean values, the three single elements and the two AC devices.

7. Solved Examples

Solved Example 1

A direct current of

a

ampere is superimposed on an alternating current

i = b sin ω t

in the same wire. What is the effective (rms) value of the resulting current?

Solution:

The current at any instant is $i = a + b sin ω t$ . Its mean square over one period is

\overline{i^{2}} = \frac{1}{T} \int_{0}^{T} (a^{2} + 2 ab sin ω t + b^{2} sin^{2} ω t) d t

Over a period $\frac{1}{T} \int_{0}^{T} sin ω t d t = 0$ and $\frac{1}{T} \int_{0}^{T} sin^{2} ω t d t = \frac{1}{2}$ , so $\overline{i^{2}} = a^{2} + \frac{b ^{2}}{2}$ .

Answer: $i_{e f f} = a^{2} + \frac{b ^{2}}{2}$ .

Solved Example 2

The voltage in an AC circuit is

V = 2202 sin (314 t - ϕ)

volt. Find (a) the peak and rms voltage, (b) the average voltage, (c) the frequency.

Solution:

(a) Peak $V_{0} = 2202 = 311.1 V$ ; rms $V_{r m s} = \frac{V _{0}}{2} = 220 V$ .

(b) Over a full cycle the average is zero. Over a half cycle $V_{a v g} = \frac{2 V _{0}}{π} = \frac{2 ( 311.1 )}{π} = 198.1 V$ .

Answer: (a) $311 V$ , $220 V$ ; (b) $0$ over a cycle, $198 V$ over a half cycle; (c) $50 Hz$ .

Solved Example 3

The current in a circuit rises linearly as

i = i_{0} \frac{t}{T}

from

t = 0

t = T

. Find the rms current over this interval.

Solution:

\overline{i^{2}} = \frac{1}{T} \int_{0}^{T} \frac{i _{0}^{2} t ^{2}}{T ^{2}} d t = \frac{i _{0}^{2}}{T ^{3}} \cdot \frac{T ^{3}}{3} = \frac{i _{0}^{2}}{3}

Answer: $i_{r m s} = \frac{i _{0}}{3} \approx 0.577 i_{0}$ (the value for any sawtooth or triangular wave).

Solved Example 4

An inductor of inductance

L = 5 H

is connected to an AC source

v = 10 sin (10 t + \frac{π}{6})

volt. Find (i) the inductive reactance, (ii) the peak and rms voltage, (iii) the peak and rms current, (iv) the instantaneous current.

Solution:

(i) $X_{L} = ω L = 10 \times 5 = 50 Ω$ .

(ii) $V_{0} = 10 V$ ; $V_{r m s} = \frac{10}{2} = 7.07 V$ .

(iii) $I_{0} = \frac{V _{0}}{X _{L}} = \frac{10}{50} = 0.2 A$ ; $I_{r m s} = \frac{0.2}{2} = 0.141 A$ .

(iv) The current lags by $\frac{π}{2}$ : $I (t) = 0.2 sin (10 t + \frac{π}{6} - \frac{π}{2})$ .

Answer: $I (t) = 0.2 sin (10 t - \frac{π}{3}) A$ .

Solved Example 5

A capacitor of capacitive reactance

5 Ω

is connected to an AC source

v = 3 sin (ω t + \frac{π}{6})

volt. Find the rms and peak voltage, the rms and peak current, and the instantaneous current.

Solution:

$V_{0} = 3 V$ , $V_{r m s} = \frac{3}{2} = 2.12 V$ .

$I_{0} = \frac{V _{0}}{X _{C}} = \frac{3}{5} = 0.6 A$ , $I_{r m s} = \frac{3}{5 2} = 0.424 A$ .

The current leads by $\frac{π}{2}$ : $I (t) = 0.6 sin (ω t + \frac{π}{6} + \frac{π}{2})$ .

Answer: $I (t) = 0.6 sin (ω t + \frac{2 π}{3}) A$ .

Solved Example 6

The rms value of the current

i = 3 sin ω t + 4 cos ω t

ampere is
(A)

5 A

(B)

3.54 A

(C)

7 A

(D)

4.95 A

Solution:

Answer: (B). Both terms have the same frequency, so combine them first: $3 sin ω t + 4 cos ω t = 5 sin (ω t + 5 3^{\circ})$ , a sine of peak $5 A$ . Then $i_{r m s} = \frac{5}{2} = 3.54 A$ . Adding the rms values ( $2.12 + 2.83 = 4.95$ ) is the trap.

Solved Example 7

100 W

220 V

bulb runs on the

220 V

AC mains. Find its resistance, the rms current and the peak current.

Solution:

$R = \frac{V _{r m s}^{2}}{P} = \frac{22 0 ^{2}}{100} = 484 Ω$ .

$I_{r m s} = \frac{P}{V _{r m s}} = \frac{100}{220} = 0.455 A$ ; $I_{0} = 2 I_{r m s} = 0.643 A$ .

Answer: $484 Ω$ , $0.455 A$ , $0.643 A$ . The bulb rating always refers to rms values.

Solved Example 8

Find the reactance of a

10 μ F

capacitor at (a)

50 Hz

and (b)

5 kHz

Solution:

$X_{C} = \frac{1}{2 π f C}$ .

(a) $X_{C} = \frac{1}{2 π ( 50 ) ( 10 \times 1 0 ^{- 6} )} = 318 Ω$ . (b) $X_{C} = \frac{1}{2 π ( 5000 ) ( 10 \times 1 0 ^{- 6} )} = 3.18 Ω$ .

Answer: $318 Ω$ and $3.18 Ω$ . A hundred times the frequency gives one hundredth of the reactance; this is why capacitors pass high frequencies.

Solved Example 9

A coil of

100

turns and area

0.10 m^{2}

rotates at

50

revolutions per second in a uniform field of

0.010 T

perpendicular to its axis of rotation. Find the peak emf.

Solution:

$ω = 2 π (50) = 314 rad s^{- 1}$ . $e_{0} = N B A ω = (100) (0.010) (0.10) (314)$ .

Answer: $e_{0} \approx 31.4 V$ .

Solved Example 10

A step-down transformer has

2200

turns in the primary and

110

turns in the secondary. The primary is connected to

220 V

AC and the secondary supplies

2 A

. Assuming it is ideal, find the output voltage and the primary current.

Solution:

$V_{s} = V_{p} \frac{N _{2}}{N _{1}} = 220 \times \frac{110}{2200} = 11 V$ .

$I_{p} = \frac{V _{s} I _{s}}{V _{p}} = \frac{11 \times 2}{220} = 0.1 A$ .

Answer: $11 V$ ; $0.1 A$ .

Solved Example 11

The rms value of a half-wave rectified current of peak

i_{0}

is
(A)

i_{0} / 2

(B)

i_{0} /2

(C)

i_{0} / π

(D)

2 i_{0} / π

Solution:

Answer: (B). The current is $i_{0} sin ω t$ for half the period and zero for the other half. $\overline{i^{2}} = \frac{1}{T} \int_{0}^{T /2} i_{0}^{2} sin^{2} ω t d t = \frac{i _{0}^{2}}{4}$ , so $i_{r m s} = \frac{i _{0}}{2}$ . (C) is its average value.

Practice Questions

What is the reactance of a capacitor connected to a constant DC source?Answer: Infinite; the capacitor blocks steady DC
Find the rms voltage and frequency of $v = 100 sin (100 π t)$ volt.Answer: $70.7 V$ ; $50 Hz$
Find the reactance of a $0.2 H$ inductor at $50 Hz$ .Answer: $62.8 Ω$
For $i = 10 sin (100 π t)$ ampere, how long does the current take to rise from zero to its peak?Answer: $\frac{T}{4} = 5 ms$
Find the rms value of a square-wave current that switches between $+ i_{0}$ and $- i_{0}$ .Answer: $i_{0}$
A transformer steps $220 V$ down to $12 V$ . The primary has $1100$ turns. How many turns does the secondary have?Answer: $60$
Why can a moving coil ammeter not measure the current from an ac generator?Answer: Its deflection follows the average current, which is zero over a cycle

Common Mistakes to Avoid

Watch out

Taking the mains $220 V$ as the peak value. It is the rms value; the peak is $311 V$ .
Using $0.707$ for the average value or $0.637$ for the rms value. Average (half cycle) $= \frac{2}{π} i_{0}$ , rms $= \frac{i _{0}}{2}$ .
Adding rms values of two same-frequency sinusoids directly. Combine them into one sine first ( $3 sin ω t + 4 cos ω t$ has peak $5$ ).
Swapping the phase: current lags in an inductor and leads in a capacitor (ELI the ICE man).
Writing $X_{C} = ω C$ or $X_{L} = \frac{1}{ω L}$ . Check with DC: an inductor passes DC ( $X_{L} = 0$ ), a capacitor blocks it ( $X_{C} \to \infty$ ).
Using $f$ where $ω$ is needed: $X_{L} = 2 π f L$ , not $f L$ . In $v = V_{0} sin (314 t)$ the $314$ is $ω$ , not $f$ .
Thinking a transformer increases power. Voltage goes up only as current goes down; $V_{p} I_{p} = V_{s} I_{s}$ at best.
Expecting a transformer to work on a battery. Steady DC gives no changing flux, so no secondary emf.

Frequently Asked Questions

What is alternating current?

Alternating current is current whose magnitude changes continuously with time and whose direction reverses periodically, usually sinusoidally as $i = i_{0} sin (ω t + ϕ)$ . In India the mains frequency is 50 Hz, so the current reverses direction 100 times every second.

What is the difference between the rms value and the average value of AC?

The average of a sinusoidal current over a full cycle is zero; over a half cycle it is $\frac{2 i _{0}}{π} \approx 0.637 i_{0}$ . The rms value is the square root of the mean of $i^{2}$ , $\frac{i _{0}}{2} \approx 0.707 i_{0}$ , and it decides heating and power.

Why is the rms value of AC used instead of the peak value?

The rms value is the steady DC current that would produce the same heat in the same resistor in the same time. So power formulas like $P = V_{r m s} I_{r m s}$ work exactly as in DC, and all AC meters and ratings, such as 220 V mains, are given in rms.

Why does current lag voltage in an inductor?

An inductor opposes any change in current by a back emf $L \frac{d i}{d t}$ . The applied voltage is largest when the current is changing fastest, which is when the current passes through zero, so the current reaches its peak a quarter cycle later and lags by 90 degrees.

Why does a capacitor block DC but allow AC?

Capacitive reactance is $X_{C} = \frac{1}{2 π f C}$ . For steady DC the frequency is zero, the reactance is infinite and the capacitor simply charges and then stops current. For AC it charges and discharges continuously, so current flows, and more easily at higher frequency.

Why is power consumed in a pure inductor or capacitor zero?

The current and voltage are 90 degrees out of phase, so the product $v i$ is positive for one quarter cycle and negative for the next. Energy stored in the field is returned to the source in full, and the average power over a cycle is zero.

Which AC topics are asked in NEET?

NEET regularly asks rms and peak values of mains voltage, reactance of an inductor or capacitor at a given frequency, the phase between current and voltage in L and C circuits, and transformer turns-ratio and efficiency numericals. The ELI the ICE man rule settles most phase questions.

What AC questions come in JEE Main from this topic?

JEE Main uses rms values of combined or non-sinusoidal waveforms, such as DC plus AC or half-wave rectified current, the emf of an ac generator, instantaneous current in a pure inductor or capacitor, and transformer current and power-loss calculations.

Previous year questions on Introduction to Alternating Currents and Circuits

8 questions from past papers, each with a step-by-step solution.

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