Alternating Current — Class 12 Physics Notes & Practice

What You’ll Learn

Your Game Plan

1. Fix the vocabulary first. Peak, rms and average are three different numbers for the same wave. Get them straight on day one or every later answer is off by a factor of √2.
2. Learn the three personalities one at a time. R alone, L alone, C alone. For each: what is the opposition, what is the phase, how much power is used?
3. Draw phasors, do not solve equations. The series LCR circuit is a drawing problem. Practise the diagram until it takes thirty seconds.
4. Make resonance a picture, not a formula. X_L climbs with frequency, X_C falls with it — they must cross somewhere, and that crossing is ω₀.
5. Finish with power, generator and transformer. These carry the reliable 2- and 3-mark questions and are the easiest marks in the chapter.
6. Then do numericals until units stop being a problem. Henry, farad, microfarad, ohm, hertz, rad s⁻¹ — sloppy units lose more marks here than sloppy physics.

Study Notes

What Makes a Current “Alternating”

An alternating current is one whose direction reverses periodically and whose magnitude varies with time. In every circuit you meet at this level the variation is sinusoidal, so the source voltage is written

v = V₀ sin ωt

and the current that results is i = I₀ sin(ωt + φ), where that little φ is the entire subject of this chapter. Reading such an equation is a guaranteed 1-mark question, so decode each symbol properly:

• V₀, I₀ — peak (amplitude) values. The largest value reached, in volts and amperes.
• ω — angular frequency, in rad s⁻¹. It is linked to ordinary frequency by ω = 2πf, and to the period by T = 1/f = 2π/ω.
• φ — phase angle, in radians or degrees. A positive φ in the current means the current is ahead; a negative φ means it is behind.

Why sinusoidal, and why 50 Hz? Because AC is generated by spinning a coil in a magnetic field. Uniform rotation gives a flux that varies as cos ωt, so by Faraday’s law the induced emf varies as sin ωt — a sine wave falls out of the geometry of a rotation. Indian mains runs at f = 50 Hz, so ω = 2π(50) = 100π ≈ 314 rad s⁻¹ and T = 0.02 s. The current reverses twice per cycle, so 100 reversals per second.

Why bother with AC at all? One reason above all others: AC voltage can be stepped up and down almost losslessly by a transformer, and only AC can, because a transformer needs a changing flux. That is why power stations generate AC and why the last section of this chapter matters as much as the first.

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Peak, Average and RMS Values of AC

Here is a fair question: your meter says the socket supplies “220 volts”, but the voltage is a sine wave that spends its life changing. Which 220 volts? Not the peak — the peak is 311 V. Not the average — averaged over a full cycle it is exactly zero, because the wave is above the axis exactly as much as below it.

So we need an honest way to quote a number for something that keeps changing sign. The trick is to stop measuring the current and measure what it does.

Why heating, and why squaring? Because heating does not care about direction. Power dissipated in a resistor is p = i²R, and i² is positive whichever way the current flows. Squaring is the mathematical way of saying “ignore the sign, keep the effect” — the same idea as electrical power in Current Electricity, now applied to a wave. So: square the current, average the square over a full cycle, take the square root. Root of the mean of the square: rms.

Do that for i = I₀ sin ωt. Over a complete cycle the average of sin²ωt is exactly ½ (the sin² curve oscillates between 0 and 1 about the midline ½). Therefore

I_rms = I₀/√2 = 0.707 I₀   and   V_rms = V₀/√2 = 0.707 V₀

And the average value? Over a full cycle it is zero and tells you nothing. Over a half cycle it is genuinely useful (it is what a moving-coil meter with a rectifier responds to):

I_av (half cycle) = 2I₀/π = 0.637 I₀

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AC Through a Pure Resistor

Start with the honest character. Connect v = V₀ sin ωt across a resistance R. A resistor has no memory and no storage — it does not care whether the voltage is rising or falling, only how big it is right now. So Ohm’s law applies at every instant:

i = v/R = (V₀/R) sin ωt = I₀ sin ωt,   with   I₀ = V₀/R

Notice what is absent: there is no phase term. The current equation has the same sin ωt as the voltage, so voltage and current are in phase — φ = 0. They peak together, they cross zero together, they reverse together. Dividing by √2 gives the rms version, I_rms = V_rms/R, which is just Ohm’s law again.

Power in a resistor. Instantaneous power is p = i²R = I₀²R sin²ωt. Since sin² is never negative, the resistor absorbs energy throughout the cycle — it never gives any back. Averaging sin² to ½:

P_av = ½ I₀²R = I_rms²R = V_rmsI_rms = V_rms²/R

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AC Through a Pure Inductor — Inductive Reactance

Now the slow one. An inductor’s whole personality is in Lenz’s law: it opposes change. When you try to increase the current through a coil, the changing flux induces a back-emf that pushes against you (this is self-induction, straight out of Electromagnetic Induction).

So picture the sequence. The supply voltage shoots up first; the coil digs in its heels; the current only builds slowly, and does not reach its own maximum until the voltage has already come back down to zero. Work it through and the lateness is exactly a quarter of a cycle:

i = I₀ sin(ωt − π/2)  —  the current LAGS the voltage by 90°

(Formally: the applied voltage must balance the back-emf, V₀ sin ωt = L di/dt, and integrating a sine gives a negative cosine — which is a sine shifted back by π/2.)

How much does it oppose? The peak current works out as I₀ = V₀/(ωL). The quantity in the denominator plays the role R played before, so we name it the inductive reactance:

X_L = ωL = 2πfL  (unit: ohm, Ω)

Power in a pure inductor: exactly zero. This is the strange and beautiful bit. Because the phase difference is 90°, the current is positive while the voltage is negative for a quarter of every cycle. So for half of each cycle the coil stores energy in its magnetic field, and for the other half it returns every joule to the source. Averaged over a full cycle, P_av = 0 — an ideal inductor consumes no energy at all. It merely borrows.

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AC Through a Pure Capacitor — Capacitive Reactance

Now the eager one. A capacitor stores charge, q = Cv (revise the definition in Electrostatic Potential and Capacitance if it is rusty). Since the applied voltage is v = V₀ sin ωt, the charge on the plates must follow q = CV₀ sin ωt — and current is nothing but the rate at which charge arrives, i = dq/dt.

Differentiating a sine gives a cosine, so

i = ωCV₀ cos ωt = I₀ sin(ωt + π/2)  —  the current LEADS the voltage by 90°

Why leading makes physical sense. Think about an empty capacitor at the instant the voltage starts from zero. Empty plates offer no opposition whatsoever, so charge rushes in at maximum rate — the current is at its peak while the voltage is still zero. As the plates fill they start pushing back, and by the time the voltage reaches its own peak the plates are full and the current has fallen to zero. The current does everything first. That is ICE: I leads E.

How much does it oppose? Comparing I₀ = ωCV₀ with I₀ = V₀/X gives the capacitive reactance:

X_C = 1/(ωC) = 1/(2πfC)  (unit: ohm, Ω)

Note the inversion: “capacitors hate slow.” As f falls, X_C blows up — and at f = 0 (pure DC) X_C is infinite, which is the mathematics telling you the obvious truth that DC cannot cross the gap between the plates. Raise the frequency and X_C collapses; at high frequency a capacitor is practically a short circuit.

Power in a pure capacitor: also exactly zero. Same argument as the inductor. For a quarter-cycle the capacitor charges and stores energy in its electric field; for the next quarter-cycle it discharges and hands it all back. φ = 90°, so cos φ = 0 and P_av = 0.

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The Series LCR Circuit by Phasors

Now put all three in series across one AC source, and the tug-of-war begins. The CBSE 2026–27 syllabus asks for this circuit by phasors only, which is a gift: you get to solve it with a drawing instead of a differential equation.

What a phasor is. A phasor is an arrow whose length is the amplitude of a sinusoidal quantity and whose direction shows its phase. Two arrows at 90° mean two quantities a quarter-cycle apart. The magic is that adding sinusoids of the same frequency — normally an unpleasant trigonometry exercise — becomes ordinary vector addition of arrows.

The four-step construction. Learn it as a ritual:

1. Draw I horizontally (to the right). This is the reference; everything is measured against it.
2. V_R = IR along I. The honest one is in phase, so its arrow lies right on top of the current.
3. V_L = IX_L straight up, V_C = IX_C straight down. By ELI the inductor voltage is 90° ahead (rotate anticlockwise, i.e. up); by ICE the capacitor voltage is 90° behind (down). They point in opposite directions — that is the tug-of-war made visible.
4. Combine and take the resultant. V_L and V_C are antiparallel, so they simply subtract to a single vertical arrow of length |V_L − V_C|. Now add that to the horizontal V_R using Pythagoras.

V = √[ V_R² + (V_L − V_C)² ]     tan φ = (V_L − V_C)/V_R

Substituting V_R = IR, V_L = IX_L, V_C = IX_C and cancelling the common I turns the voltage triangle into the impedance triangle of the next section. Same triangle, different scale.

Reading the winner off the diagram. Three cases, and they are the whole story:

• X_L > X_C — the coil wins. The resultant tilts above the current, so V leads I: the circuit is inductive and the current lags. φ is positive.
• X_C > X_L — the capacitor wins. The resultant tilts below the current, so I leads V: the circuit is capacitive and the current leads. φ is negative.
• X_L = X_C — a draw. The vertical arrows cancel completely and V lies flat along I. φ = 0. That is resonance.

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Impedance, Phase Angle and the Impedance Triangle

Divide every side of the voltage phasor triangle by the common current I and the arrows stop being volts and become ohms. What you get is the impedance triangle — the single most useful picture in the chapter.

Impedance Z is the total opposition an AC circuit offers, combining the dissipating kind (resistance) with the timing kind (reactance). Because R and X act a quarter-cycle apart, they are perpendicular sides of a right triangle and combine by Pythagoras, never by addition:

Z = √[ R² + (X_L − X_C)² ]    and    I_rms = V_rms/Z

The phase angle φ is the angle between the hypotenuse Z and the base R, and the triangle hands you three equivalent expressions for it — use whichever the data fits:

tan φ = (X_L − X_C)/R    cos φ = R/Z    sin φ = (X_L − X_C)/Z

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Resonance in a Series LCR Circuit

Everything so far has been building to this, and if the two memory devices have stuck then resonance needs no memorising at all. Recall them:

• Inductors hate fast: X_L = ωL starts at 0 and climbs without limit as frequency rises.
• Capacitors hate slow: X_C = 1/ωC starts infinite and dies away as frequency rises.

One quantity climbing from zero, another falling from infinity, both continuous. They must cross. And at the crossing frequency the two bullies are pulling with exactly equal strength in exactly opposite directions, so they cancel completely. The tug-of-war ends in a draw, the reactive side of the impedance triangle collapses to nothing, and the circuit behaves as if L and C were not there at all. That crossing is resonance — not a formula to be remembered, but the unavoidable consequence of two opposite frequency habits.

Set X_L = X_C to find where the crossing happens:

ω₀L = 1/(ω₀C)  →  ω₀² = 1/LC  →  ω₀ = 1/√(LC)   and   f₀ = 1/(2π√(LC))

Sharpness, and the quality factor Q. Look again at the two curves above: same L, same C, same peak position — but the R = 20 Ω circuit gives a tall narrow spike while R = 60 Ω gives a low broad hump. A sharp circuit is selective: it responds to its own frequency and ignores its neighbours, which is exactly what tuning a radio requires. Sharpness is measured by the quality factor:

Q = ω₀L/R = 1/(ω₀CR) = (1/R)√(L/C) = ω₀/Δω

All four forms are the same number — a useful self-check in an exam. Note that R sits in the denominator of every one of them: resistance is the enemy of sharpness, because it is the only element that actually drains energy from the oscillation. Δω = R/L is the bandwidth, the width of the peak between the points where the power has fallen to half its maximum.

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Power in AC Circuits, Power Factor and Wattless Current

In a DC circuit, power is P = VI and there is nothing more to say. In AC that formula is almost always wrong, because V and I are not doing the same thing at the same time. When the current is positive and the voltage negative, energy is flowing back to the source, and any honest power calculation must subtract it.

Averaging the instantaneous product vi over a full cycle gives the result you must know:

P_av = V_rms I_rms cos φ

The product V_rmsI_rms is called the apparent power — what the circuit looks like it is using. P_av is the true power — what it actually uses. The bridge between them is:

Splitting the current in two. Resolve the current phasor into a part along the voltage and a part perpendicular to it, and the accounting becomes obvious:

• Active (wattful) component, I_rms cos φ — in phase with the voltage. This is the part that transfers real energy, and it is the part the meter charges you for.
• Wattless (reactive) component, I_rms sin φ — 90° out of phase with the voltage. Over a full cycle it delivers exactly zero net energy.

So what is wattless current? It is current that shows up to work but does no work. It genuinely flows — an ammeter reads it, and it heats the transmission wires on its way — yet it carries no net energy to the load, because it hands back in one half-cycle exactly what it took in the other. In a pure inductor or pure capacitor φ = 90°, so sin φ = 1 and the entire current is wattless.

This is not an academic curiosity. A factory full of motors (large inductance) runs at a poor power factor, so it must draw a large current to obtain the real power it needs — and that oversized current wastes energy as I²R heat in every cable feeding it. Which is why utilities charge industrial users for poor power factor, and why engineers wire capacitor banks across inductive loads: the capacitor’s leading current cancels the coil’s lagging current and pulls cos φ back towards 1.

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AC Generator

Where does the sine wave come from in the first place? From a coil being spun in a magnetic field. The AC generator (or alternator) converts mechanical energy into electrical energy, and it is nothing but Faraday’s law given a crank handle.

The parts, and what each is for.

• Armature — a coil of N turns, area A, wound on a soft-iron core (the core concentrates the flux, raising the output).
• Field magnet — supplies a strong, steady, uniform field B.
• Slip rings — two rings, each permanently joined to one end of the coil. This is the crucial detail: because each ring keeps the same end of the coil forever, the output reverses every half turn and you get AC. (Replace them with a split-ring commutator and you get DC instead — that one component is the whole difference.)
• Carbon brushes — press against the rings to draw the current out to the external circuit.

The derivation, in three lines. Spin the coil with angular velocity ω, so at time t its normal makes an angle θ = ωt with B. The flux through N turns is

Φ = NBA cos ωt

By Faraday’s law the induced emf is e = −dΦ/dt, and differentiating the cosine gives

e = NBAω sin ωt = e₀ sin ωt,   with peak emf   e₀ = NBAω

A sine wave, purely because uniform rotation was a cosine of time. To raise the output you can increase N, B, A or ω — but ω is fixed at 50 Hz by the grid, so real machines increase N and B.

Where in the turn is the emf largest? Not where the flux is largest — where the flux is changing fastest. When the coil’s plane is parallel to B the flux through it is momentarily zero, but it is sweeping through the field lines at maximum rate, so e is at its peak. A quarter turn later the plane is perpendicular to B, the flux is maximum, and the emf is zero. Students get this backwards constantly. The forces on the coil sides here are the same current-in-a-field forces you met in Moving Charges and Magnetism, now being fought against by the engine that turns the shaft.

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Transformer

A transformer changes an alternating voltage from one value to another, at almost no cost in energy. It is the reason the entire world’s electricity is AC, and it works on one principle only: mutual induction.

Construction. Two coils — primary (N_p turns, connected to the input) and secondary (N_s turns, connected to the load) — wound on a closed core of laminated soft iron. The two coils are not electrically joined at all. The core’s job is to guide essentially all the magnetic flux from one coil through the other.

How it works, step by step. The alternating primary current sets up an alternating flux in the core. Because the core is closed, that same changing flux Φ threads every turn of both coils. By Faraday’s law each coil develops an emf proportional to its own number of turns:

e_p = −N_p dΦ/dt      e_s = −N_s dΦ/dt

Divide one by the other and dΦ/dt cancels, leaving the transformer equation:

V_s/V_p = N_s/N_p = k     and for an ideal transformer     I_s/I_p = N_p/N_s = 1/k

Why the grid steps voltage up to send it far. Power lost as heat in a transmission line is I²R. Transmit the same power P at 20 times the voltage and the current falls to 1/20, so the loss falls to (1/20)² = 1/400 of what it was. That single factor is why 400 kV lines exist, and why the transformer — not the generator — is the component that made national power grids possible.

Why it will not work on DC. A steady direct current gives a steady flux, and a steady flux means dΦ/dt = 0, so no emf appears in the secondary at all. Worse, the primary is a low-resistance coil, so DC would simply burn it out. A transformer needs change, which is precisely what AC provides.

Where real transformers lose energy (and what engineers do about it):

• Copper loss — I²R heating in the windings. Countered with thick, low-resistance wire.
• Flux leakage — not all the flux reaches the secondary. Countered with a closed core and interleaved windings.
• Hysteresis loss — energy spent repeatedly re-magnetising the core. Countered by choosing a soft magnetic material with a narrow hysteresis loop.
• Eddy-current loss — circulating currents induced in the core itself. Countered by laminating the core: thin sheets separated by insulating varnish break up the current loops.

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Alternating Current Class 12 Physics Numericals With Solutions

These are the mixed-skill problems that decide your marks — the ones where you must choose the right route yourself. Work each one on paper before reading the solution, and get into the habit of cross-checking the power two ways (V I cos φ and I²R). If the two disagree, something upstream is wrong.

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Formula List and Quick Revision

Everything worth carrying into the exam hall, on one screen. Every symbol in SI units: L in henry (H), C in farad (F), R, X and Z in ohm (Ω), f in hertz (Hz), ω in rad s⁻¹.

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Practice Worksheet

Ten original questions, in rising order of difficulty, covering every part of the chapter — the kind of alternating current class 12 physics important questions that actually appear. Write your answer out fully first, then open the accordion. Every answer below has been computed and cross-checked.

Kaizen: one phasor diagram a day beats twenty the night before the exam. Draw the current arrow, hang V_R, V_L and V_C on it, close the triangle — and in a week the drawing that takes you ten minutes today will take thirty seconds. Small, boring, daily improvement is how this chapter stops being frightening.