Alternating current circuits for JEE Main

Alternating current questions in JEE Main are mostly one circuit, the series LCR, asked in different ways: find the impedance, the current, the phase, the power, or the frequency at which everything lines up. The mathematics is Pythagoras on a phasor diagram. The traps are in the phases and in the fact that voltages across parts of an AC circuit do not simply add.

The official 2026 syllabus covers peak and rms values, reactance and impedance, the LCR series circuit, resonance, power in AC circuits, wattless current, the AC generator and the transformer. It builds directly on electromagnetic induction.

Peak, rms and average values

For v = V₀ sin ωt:

• V_rms = V₀/√2 ≈ 0.707 V₀ (the steady DC value that heats a resistor at the same rate)
• average over a full cycle = 0
• average over a half cycle = 2V₀/π ≈ 0.637 V₀

AC meters read rms values. Indian household supply of 220 V is an rms figure, so its peak is 220√2 ≈ 311 V. The √2 rule holds only for sinusoidal waves.

R, L and C on their own

X_L rises with frequency, so an inductor blocks high frequencies. X_C falls with frequency, so a capacitor blocks DC (ω = 0) completely and passes high frequencies easily.

The series LCR circuit

The same current flows through all three, so draw the current as the reference phasor. V_R is along it, V_L is 90° ahead and V_C is 90° behind. Because V_L and V_C point in opposite directions:

• V = √[V_R² + (V_L − V_C)²]
• Z = √[R² + (X_L − X_C)²]
• tan φ = (X_L − X_C)/R, where φ is how far the voltage leads the current

If X_L > X_C the circuit is inductive (current lags). If X_C > X_L it is capacitive (current leads).

Resonance

When X_L = X_C, the reactances cancel:

• ω₀ = 1/√(LC), so f₀ = 1/(2π√(LC))
• Z = R, its smallest value, so the current is largest
• φ = 0 and the power factor is 1

At resonance V_L and V_C are equal and opposite, and each can be many times the supply voltage. Their ratio to the supply is the quality factor, Q = ω₀L/R. A higher Q means a sharper resonance, which is what a radio tuner needs.

Power, power factor and wattless current

Only the resistor absorbs energy on average. The average power is

P = V_rms I_rms cos φ = I_rms² R, with power factor cos φ = R/Z.

Split the current into a part in phase with the voltage (I_rms cos φ) and a part 90° out of phase (I_rms sin φ). The second part does no average work, so it is called the wattless current. In a purely inductive or capacitive circuit, all the current is wattless.

A low power factor forces the supply to carry more current for the same useful power, which raises losses in transmission lines. That is why industries add capacitors to raise the power factor.

AC generator and transformer

Generator. A coil of N turns and area A rotating at ω in a field B gives e = NBAω sin ωt, with peak emf NBAω.

Transformer. Two coils share a core. For an ideal transformer:

V_s/V_p = N_s/N_p = I_p/I_s

A step-up transformer raises voltage and lowers current; step-down does the reverse. Power is transmitted over long distances at high voltage because, for the same power, the current is smaller and the I²R loss in the lines falls as the square of the current.

Real transformers lose energy through:

• copper losses (heating of the windings)
• eddy currents (reduced by laminating the core)
• hysteresis in the core (reduced by using soft iron)
• flux leakage between the coils

A transformer does not work on steady DC, because it needs a changing flux.

Worked problems

Problem 1: LCR circuit. R = 6 Ω, X_L = 12 Ω and X_C = 4 Ω are in series across 100 V rms. Find Z, the current, the power factor, the power and the voltage across each element.

Z = √(6² + 8²) = 10 Ω, so I_rms = 10 A.
cos φ = 6/10 = 0.6 (inductive, current lagging). P = 100 × 10 × 0.6 = 600 W; check I²R = 100 × 6.
V_R = 60 V, V_L = 120 V, V_C = 40 V. They add as √(60² + 80²) = 100 V, not 220 V.

Problem 2: resonance. L = 1 mH, C = 10 μF and R = 2 Ω in series across 20 V rms. Find ω₀, the current at resonance and the voltage across the inductor.

LC = 10⁻³ × 10⁻⁵ = 10⁻⁸, so ω₀ = 10⁴ rad/s (f₀ ≈ 1.6 kHz).
At resonance Z = R, so I = 20/2 = 10 A.
X_L = ω₀L = 10 Ω, so V_L = 100 V, five times the supply. Here Q = 10/2 = 5.

Problem 3: transformer. A transformer steps 220 V down to 11 V. The primary has 2,000 turns and the secondary delivers 4 A. Find the secondary turns and the primary current if it is (a) ideal and (b) 80% efficient.

N_s = 2,000 × 11/220 = 100 turns.
(a) I_p = 4 × 11/220 = 0.2 A.
(b) Output is 44 W, so input = 44/0.8 = 55 W and I_p = 55/220 = 0.25 A.

Problem 4 (numerical answer): capacitor on the mains. A capacitor of (100/π) μF is connected to a 220 V, 50 Hz supply. Find the rms current in amperes.

X_C = 1/(2π × 50 × (100/π) × 10⁻⁶) = 1/(100 × 10⁻⁴) = 100 Ω.
I_rms = 220/100 = 2.2 A. The average power is zero.

Practice set

1. What is the power factor of a purely inductive circuit?
2. An AC has I_rms = 5 A. Find its peak value.
3. An inductor of (0.5/π) H is on a 50 Hz supply. Find its reactance.
4. A series LCR circuit with R = 10 Ω is driven at resonance by 50 V rms. Find the current and the power.
5. A circuit takes 5 A rms from 200 V rms with the current lagging by 60°. Find the power and the wattless current.
6. In a series LC circuit, C is made four times larger. What happens to the resonant frequency?
7. A 100-turn coil of area 0.02 m² rotates at 100 rad/s in a 0.5 T field. Find the peak emf.
8. Why does a transformer not work on a steady DC supply?

Answers

1. Zero (φ = 90°).
2. 5√2 ≈ 7.07 A.
3. X_L = 2π × 50 × 0.5/π = 50 Ω.
4. Z = R, so I = 5 A and P = 5² × 10 = 250 W.
5. P = 200 × 5 × cos 60° = 500 W; wattless current = 5 sin 60° ≈ 4.33 A.
6. f₀ ∝ 1/√C, so it halves.
7. NBAω = 100 × 0.5 × 0.02 × 100 = 100 V.
8. Steady DC gives no change in flux, so no emf is induced in the secondary.

What to do next

• Draw the phasor diagram for inductive, capacitive and resonant LCR circuits.
• Solve 15 LCR problems, finding Z, I, φ and P each time, and check P against I²R.
• Practise transformer problems with efficiency, and resonance problems where V_L exceeds the supply.
• Revise electromagnetic induction, then use the physics formula sheet to review the electricity units together.