Alternating Current

Written and reviewed by Akanksha Tiwari, Physics & Chemistry faculty at UniDrill (M.Sc. Physics, Banaras Hindu University · 8 years teaching Class XI–XII science). Based on NCERT Physics Class XII, Chapter 7.

Why does the mains voltage keep reversing?

Twice every 20 milliseconds, the voltage at your wall socket falls to zero, flips sign and climbs back up. The current does the same. Yet the bulb does not seem to flicker, a heater still gets hot and a phone charger still works. This chapter explains how a quantity that averages to zero can still do useful work, and why the whole world chose this awkward supply over steady dc.

The answer to the second question is the transformer. A transformer can raise or lower an ac voltage with almost no loss, and it cannot do that for dc. Cheap voltage conversion makes it possible to send power hundreds of kilometres at high voltage and then bring it down to a safe level at your home.

Alternating voltage and the rms idea

An alternating voltage varies as a sine wave:

$$v = v_m \sin\omega t, \quad \omega = 2\pi\nu$$

Here vₘ is the peak value (amplitude), ν the frequency and ω the angular frequency. In India the mains frequency is 50 Hz, so the period is 20 ms and $\omega \approx 314$ rad/s.

If you connect this source to a pure resistor R, Kirchhoff's loop rule gives $i = (v_m/R)\sin\omega t = i_m \sin\omega t$. The current rises and falls exactly in step with the voltage. We say they are in phase.

Now the puzzle. Over a full cycle the positive half cancels the negative half, so the average current is zero. But a heater does not care which way the current flows. Heating depends on $i^2$, and $i^2$ is never negative. The average of $\sin^2\omega t$ over a cycle is $1/2$, because sin² and cos² have the same average and they add up to 1. So:

Average power $P = 1/2\,i_m^2 R = I^2 R$, where $I = i_m/\sqrt{2} = 0.707\, i_m$

I is the rms (root mean square) current: square the current, take its mean, then take the square root. It is the steady dc current that would heat the resistor equally. Similarly the rms voltage is $V = v_m/\sqrt{2} = 0.707\, v_m$. The reward for using rms values is that ac power looks exactly like dc power: $P = VI = I^2R = V^2/R$.

When someone says "220 V ac", they mean the rms value. Voltmeters and ammeters for ac are also calibrated to read rms values.

Worked example: a bulb on the mains. A 100 W, 220 V bulb is connected to 220 V rms, 50 Hz mains. Find its resistance, the rms current and the peak current.

Common wrong step. Treating 220 V as the peak value. That would make the rms voltage 156 V and the bulb would glow at only half its rated power. Rated and displayed ac values are always rms.

Pause and predict: a moving-coil dc ammeter is connected in series with a heater on the ac mains. What does it read, and does the heater still get hot?

Check your answer: the meter reads about zero, because it responds to the average current, which is zero over a cycle. The heater still gets hot, because heating depends on i², which averages to $I^2R$. An ac ammeter reads the rms value instead.

Beyond the textbook. For the exam: $\text{rms} = \text{peak}/\sqrt{2}$, and meters read rms. In reality: many inexpensive multimeters do not compute a true rms. They measure the rectified average and multiply by a fixed constant calibrated for a sine wave. So they are accurate only for sinusoidal supplies and can read wrongly for square or distorted waveforms. Meters sold as "true rms" avoid this. Why it matters: rms is defined by heating, not by waveform shape, and the factor 1/√2 belongs to the sine wave only.

Phasors: seeing phase with a rotating arrow

Adding sine waves that are out of step is messy algebra. A phasor makes it a drawing. A phasor is a vector of length equal to the peak value that rotates anticlockwise with angular speed ω. Its vertical projection on the axis is the instantaneous value. Two quantities with a phase difference are two arrows with a fixed angle between them, and combining them is just vector addition.

In every ac circuit, series or parallel, the question you ask is simple: for each element, is the current phasor ahead of, behind or aligned with the voltage phasor? That is what the next three sections settle.

Resistor, inductor, capacitor: three personalities

Resistor. Voltage and current are in phase, and R does not depend on frequency. It converts electrical energy to heat in both half cycles.

Inductor. Apply $v = v_m \sin\omega t$ to a pure inductor L. The loop rule gives $v = L\,di/dt$. Integrating,

$$i = \dfrac{v_m}{\omega L}\sin(\omega t - \pi/2)$$

so the current lags the voltage by π/2 (a quarter cycle). Why: the inductor opposes any change in current by producing a back emf. The current can only catch up after the voltage has already pushed for a while. The amplitude is $i_m = v_m/\omega L$, which defines the inductive reactance

$X_L = \omega L = 2\pi\nu L$ (unit: ohm)

The faster the current tries to change, the harder the inductor fights it, so $X_L$ grows with frequency. At dc ($\omega = 0$) an inductor behaves like a plain wire. At very high frequency it blocks current almost completely.

Capacitor. Apply the same voltage to a pure capacitor C. Charge $q = Cv$, so $i = dq/dt = \omega C v_m \cos\omega t$:

$$i = i_m \sin(\omega t + \pi/2), \quad i_m = v_m/(1/\omega C)$$

The current leads the voltage by π/2. Why: current must flow first to deposit charge, and only then does the voltage across the plates build up. The capacitive reactance is

$$X_C = 1/(\omega C) = 1/(2\pi\nu C)$$ (unit: ohm)

A capacitor passes high frequencies easily and blocks dc, since at $\omega = 0$ the reactance is infinite.

Why is the average power zero for L and C? In one quarter cycle the element draws energy from the source and stores it (as magnetic field energy in L, electric field energy in C). In the next quarter cycle it hands the same energy back. Nothing is converted to heat.

Worked example: change the frequency. (a) A pure inductor of 25.0 mH is connected to a 220 V, 50 Hz source. Find $X_L$ and the rms current. (b) A 15.0 μF capacitor is on the same source. Find the rms current, then find it again after the frequency is doubled.

Common wrong step. Saying that the capacitor current halves when the frequency doubles, because "more frequency means more opposition". That is the inductor's rule. For a capacitor, higher frequency means less opposition and more current.

Pause and predict: a bulb is in series with an inductor on an ac source. You slide a soft iron rod into the inductor. Does the bulb get brighter or dimmer?

Check your answer: dimmer. The iron raises the inductance L, which raises $X_L = \omega L$, so the current falls. The same coil on dc would hardly change the brightness, because at dc only the wire's own small resistance limits the current.

The series LCR circuit

Now put R, L and C in series with the source. The same current i flows through all three, so we draw the current phasor as the reference. The voltage across R is in phase with it, across L it is π/2 ahead and across C it is π/2 behind. Because $V_L$ and $V_C$ point in opposite directions, they partly cancel:

$v_m^2 = v_R^2 + (v_C - v_L)^2$, so $Z = \sqrt{R^2 + (X_C - X_L)^2}$

Z is the impedance, the total opposition of the circuit, and $i_m = v_m/Z$. The phase angle of the current relative to the voltage is given by

$$\tan\varphi = \dfrac{X_C - X_L}{R}$$

with the convention $i = i_m \sin(\omega t + \varphi)$. The sign of φ tells you the nature of the circuit.

The most important consequence is that the voltages across individual elements can each be larger than the source voltage. They are phasors, not scalars, so you must add them as a vector triangle and never algebraically.

Worked example: the coil that behaves differently on dc and ac. A coil carries 1 A when connected to 100 V dc, and 0.5 A when connected to 100 V ac at 50 Hz. Find its impedance and inductance.

Common wrong step. Writing $Z = R + X_L$, which gives $X_L = 100\ \Omega$ and $L = 0.32$ H. Resistance and reactance are a quarter cycle apart in phase, so they combine like the two sides of a right triangle, not like two resistors in series.

Worked example: a full LCR calculation. A source $v = 141.4 \sin(1000t)$ volts drives $R = 100\ \Omega$, $L = 0.1$ H and $C = 5$ μF in series. Find Z, the rms current, the voltage across each element, the phase angle and the power.

Common wrong step. Adding the element voltages to get $70.7 + 70.7 + 141.4 = 283$ V, which is nearly three times the source. The check in step 5 is the habit worth building.

Resonance: when L and C cancel

As you change the source frequency, $X_L$ rises and $X_C$ falls. At one special frequency they are equal, $X_L = X_C$, and the two voltages cancel exactly. Then

$\omega_0 = 1/\sqrt{LC}$, $\nu_0 = 1/(2\pi\sqrt{LC})$

At resonance $Z = R$ (the minimum possible), the current is maximum, $i_m = v_m/R$, $\varphi = 0$ and the power factor is 1.

Why does it happen? The inductor and capacitor trade energy back and forth between a magnetic field and an electric field, a bit like a pendulum swapping kinetic and potential energy. At ω₀ their exchange is perfectly matched, so the source only has to top up the energy that R turns into heat.

Resonance needs both L and C. In an RL or RC circuit, $V_L$ (or $V_C$) has nothing opposite in phase to cancel against, so no resonance is possible. The resonant frequency depends only on L and C. Changing R does not move the peak, it changes its height and width.

The quality factor measures how sharp the resonance is:

$Q = \omega_0 L/R = 1/(\omega_0 C R) = \omega_0/\Delta\omega$, bandwidth $\Delta\omega = R/L$

The bandwidth $\Delta\omega$ is the full width of the current curve between the two frequencies where the current falls to $i_m(\text{max})/\sqrt{2}$. A high Q means a narrow peak.

A radio tuner is a resonant circuit. Turning the dial changes C, which moves ω₀ until it matches the station you want. The signal at that frequency drives a large current, and the neighbouring stations are weak because the circuit has a high Q.

Worked example: resonance and voltage magnification. A series circuit has $R = 10\ \Omega$, $L = 0.1$ H and $C = 10\ \mu\text{F}$, driven by a 5 V rms source of variable frequency. Find the resonant frequency, Q, the current at resonance, the voltage across the inductor and the power.

Common wrong step. Believing that no element can have more voltage than the source. Here a 5 V source puts 50 V across the inductor. That is not a violation of energy conservation, because $V_L$ and $V_C$ cancel in the loop (the source sees only the 5 V drop across R), and the extra voltage is stored field energy.

Beyond the textbook. For the exam: the series LCR circuit resonates at $\omega_0 = 1/\sqrt{LC}$ and a tuner selects a station by adjusting C. In reality: a real coil has its own resistance and a real capacitor has losses, so the effective R in Q is larger than the resistor you added. Also, the 50 V across the capacitor in an example like this is a genuine hazard in high voltage tuned circuits such as induction heaters. Why it matters: resonance multiplies voltage, not power, and that is both the trick behind radio reception and a safety concern in power electronics.

Power in an ac circuit

At each instant the power is vi, but only its average over a cycle matters:

$P = VI\cos\varphi$, power factor $\cos\varphi = R/Z$

VI is the apparent power. $\cos\varphi$ is the fraction of it that is actually dissipated. The rest sloshes back and forth between source and fields.

A current that flows but delivers no net power, because it is a quarter cycle out of step with the voltage, is called wattless current. In any LCR circuit the power is always $I^2R$: the reactances never dissipate anything.

Why it matters. To deliver a fixed power P at voltage V, the current needed is $I = P/(V\cos\varphi)$. A 2.2 kW motor on 220 V needs 10 A if its power factor is 1, but 20 A if the power factor is 0.5. Since line loss is $I^2R$, the poor power factor quadruples the loss in the wires. The usual remedy is to connect a suitable capacitor in parallel with an inductive load such as a motor. The capacitor supplies the lagging wattless current locally, so the supply line carries less current.

Beyond the textbook. For the exam: a capacitor connected in parallel improves the power factor of an inductive load. In reality: electricity suppliers bill large industrial users partly on apparent power (kVA) or penalise low power factor, so factories install capacitor banks to save money. Why it matters: the current in the supply cable depends on $VI$, not on the useful watts.

Transformers

A transformer is two coils wound on a common soft iron core. An ac voltage across the primary ($N_p$ turns) sets up an alternating flux. The core guides that flux through the secondary ($N_s$ turns), where mutual induction creates an emf. It does not work on dc, because a steady current produces a steady flux and induces nothing.

For an ideal transformer (no losses):

$V_s/V_p = N_s/N_p$, $V_pI_p = V_sI_s$, so $I_s/I_p = N_p/N_s$

The flux per turn is the same in both coils, so each turn picks up the same emf, which makes the voltage ratio equal the turns ratio. Since power in equals power out, whatever you gain in voltage you pay for in current.

A transformer changes voltage and current but never power. It cannot make energy.

Real transformers lose some energy in four ways, each with a design remedy:

Worked example: why we transmit at high voltage. A village needs 44 kW. The supply line has a total resistance of 0.5 Ω. Compare the line loss if the power is sent at 220 V and at 4400 V, using a 1 : 20 step-up transformer.

Common wrong step. Calculating line loss as $V^2/R$ with the transmission voltage, which would suggest that higher voltage means more loss. The transmission voltage is not the voltage across the line. Only the small voltage drop IR appears across the wires, which is why the loss formula is $I^2R$.

Efficiency. A real transformer has efficiency $\eta = \text{output power}/\text{input power}$. Suppose a transformer steps 4400 V down to 220 V and supplies a 4.4 kW load. The secondary current is $4400/220 = 20$ A. An ideal one with 2000 primary turns needs $2000 \times 220/4400 = 100$ secondary turns and draws $20 \times 100/2000 = 1.0$ A from the 4400 V line. If it is only 90% efficient, the input power is $4400/0.9 = 4889$ W and the primary current rises to $4889/4400 = 1.11$ A. A common slip is to flip the turns ratio for current, which gives 400 A. Current is inversely proportional to turns.

Pause and predict: a transformer is connected to a 12 V battery instead of an ac source. What is the secondary voltage after the first instant?

Check your answer: zero. Once the dc current is steady, the flux is steady and there is no changing flux to induce a secondary emf. Only at the moment of switching on or off is there a brief pulse. Worse, a primary designed for ac has low resistance and little reactance at dc, so it can draw a damaging current.

Beyond the textbook. For the exam: a transformer works only on ac, and it hums because of core vibration. In reality: the hum is at twice the mains frequency (100 Hz in India), because the magnetic force on the core depends on $B^2$, which peaks twice per cycle. Large power transformers are over 98% efficient, one of the most efficient machines ever built. For very long distances and undersea cables, modern systems often use high voltage dc, converted at each end by power electronics, because it avoids some losses that affect long ac lines. Why it matters: ac won the early contest because of the transformer, and the engineering has since moved on in specific cases.

Exam traps

Formula sheet

60-second revision

Where this chapter connects

• Previous: Electromagnetic Induction supplies self and mutual induction, the two ideas behind inductors and transformers.
• Next: Electromagnetic Waves shows how an oscillating LC circuit can radiate, the principle behind every radio transmitter.
• Class XI link: Oscillations is the mechanical twin of an LC circuit. Mass compares with L, spring constant with 1/C, and the same $\omega_0 = 1/\sqrt{LC}$ form appears.
• Later link: Semiconductor Electronics shows how diodes rectify ac into dc.