AC Circuits & Phasors: JEE Main Complete Guide

Alternating current circuits contribute three to four questions every JEE Main session, and students who understand phasors solve them in 90 seconds flat while others waste five minutes on simultaneous equations. The phasor approach converts differential equations into simple geometry — once you see it, you cannot unsee it. This guide takes you from the definition of a phasor through the full LCR circuit, covering every question type the exam uses.

What Is a Phasor and Why It Works

A phasor is a rotating vector whose projection on the vertical axis gives the instantaneous value of a sinusoidal quantity. For voltage v = V₀ sin(ωt), the phasor has magnitude V₀ and rotates anticlockwise at angular frequency ω. The key insight: if two quantities have the same frequency but different phases, their phasors are fixed relative to each other. We can treat the entire circuit as a vector addition problem at one frozen instant.

Resistor: voltage and current are in phase (phasors parallel). Inductor: voltage leads current by 90° (V phasor is 90° ahead of I phasor). Capacitor: current leads voltage by 90° (I phasor is 90° ahead of V phasor). These three phase relationships are the foundation of every AC problem. Memorise them with the mnemonic CIVIL — in a Capacitor I leads V; in an Inductor V leads I. For the related chapter on AC generators and transformers, see our AC generator and transformer guide.

Impedance and the Phasor Triangle

For a series LCR circuit, the total voltage phasor is the vector sum of V_R (along I), V_L (90° ahead of I) and V_C (90° behind I). Since V_L and V_C are antiparallel, they partially cancel. Net reactive voltage = V_L − V_C = I(X_L − X_C) where X_L = ωL and X_C = 1/ωC are inductive and capacitive reactances respectively.

Impedance Z = √(R² + (X_L − X_C)²). Phase angle φ = tan⁻¹((X_L − X_C)/R). The phasor triangle has hypotenuse Z, base R, and height (X_L − X_C). JEE frequently asks you to find Z given R, L, C and ω, or to find the frequency at which φ takes a given value. Draw the phasor triangle every time — it takes three seconds and eliminates sign errors. Take a free AC circuits mock test to practise these calculations under time pressure.

Resonance: The Highest-Yield AC Topic

At resonance, X_L = X_C, so (X_L − X_C) = 0 and Z = R (minimum impedance, maximum current). Resonant frequency ω₀ = 1/√(LC) or f₀ = 1/(2π√(LC)). JEE tests three resonance ideas: (1) finding ω₀ or f₀ from L and C; (2) the quality factor Q = ω₀L/R = 1/(ω₀CR) which measures sharpness of resonance; (3) power at resonance P = V²rms/R (purely resistive). The Q factor appears increasingly in recent sessions — memorise Q = ω₀L/R and understand that higher Q means sharper, more selective resonance.

Average power in AC: P = V_rms × I_rms × cosφ where cosφ = R/Z is the power factor. For a purely reactive circuit (resistor absent), cosφ = 0 and average power is zero. For a resistive circuit, cosφ = 1. JEE asks power questions in about one in three AC problems. For deeper LCR content, see our LCR resonance guide and our LC oscillations guide.

RMS Values and Power Calculations

V_rms = V₀/√2 ≈ 0.707 V₀. Same for I_rms. These are the effective values — they produce the same heating as an equal DC value. Exam traps: (1) peak value problems disguised as RMS — always identify which is given; (2) instantaneous power p = vi = V₀I₀ sin(ωt) sin(ωt−φ) is not the same as average power P = ½V₀I₀ cosφ; (3) the RMS formula applies only to sinusoidal waveforms — for other waveforms you need to integrate. In multi-loop AC circuits, use Kirchhoff's laws in phasor form: phasor voltage drops sum to zero around any loop, and phasor currents sum to zero at any node. Apply this with the impedance of each element (R, jX_L, −jX_C) to find the unknown. For a full AC strategy see our alternating current guide.