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A typical AC question: in a series LCR circuit, the voltmeters across R, L and C read 240 V, 200 V and 100 V, yet the supply is only 260 V. Students who add the readings get 540 V and decide the question is wrong. It is not. In AC circuits, voltages are out of step with each other, and you have to add them with directions. This post shows how phasors make that easy, then works a full series LCR example.
What is a phasor, and why do we need one?
In an AC circuit, every voltage and current varies as a sine wave at the same frequency, but they do not all peak at the same moment. The NCERT Class 12 chapter Alternating Current represents each one as a phasor: a rotating arrow whose length is the peak value and whose angle shows the phase. The rules you need are:
In a resistor, voltage and current are in phase.
In an inductor, the current lags the voltage by π/2.
In a capacitor, the current leads the voltage by π/2.
Because the current is the same everywhere in a series circuit, draw the current phasor first. Then V_R lies along it, V_L points 90° ahead and V_C points 90° behind. V_L and V_C point in opposite directions, so they partly cancel.
How do you get impedance from the phasor diagram?
The supply voltage is the hypotenuse of a right triangle with sides V_R and (V_L − V_C). Dividing by the current gives the impedance:
Z = √(R² + (X_L − X_C)²), where X_L = ωL and X_C = 1/(ωC).
NCERT derives exactly this and adds the phase angle, tan φ = (X_C − X_L)/R in its sign convention. If X_C > X_L the circuit is capacitive and the current leads the source voltage. If X_L > X_C it is inductive and the current lags. Both reactance and impedance are in the JEE (Main) 2026 syllabus, along with resonance, power in AC circuits and wattless current, and the JEE (Advanced) 2026 brochure lists LCR circuits with AC sources. Check your own year's syllabus.
Worked example: a series LCR circuit
A 260 V (rms) supply at ω = 1000 rad/s drives R = 120 Ω, L = 0.10 H and C = 20 μF in series. Find the current, the voltage across each element, the phase angle, the power and the resonant frequency.
Reactances. X_L = ωL = 1000 × 0.10 = 100 Ω. X_C = 1/(ωC) = 1/(1000 × 20 × 10⁻⁶) = 50 Ω.
Impedance. Z = √(120² + (100 − 50)²) = √(14400 + 2500) = √16900 = 130 Ω.
Current. I = 260/130 = 2.0 A (rms).
Element voltages. V_R = 2.0 × 120 = 240 V, V_L = 2.0 × 100 = 200 V, V_C = 2.0 × 50 = 100 V. These are the readings from the opening paragraph. Check: √(240² + (200 − 100)²) = √(57600 + 10000) = √67600 = 260 V. It matches the supply.
Phase. X_L > X_C, so the circuit is inductive and the current lags the voltage. tan φ = 50/120 = 5/12, so cos φ = 12/13 ≈ 0.92 and φ ≈ 22.6°.
Power. P = V I cos φ = 260 × 2.0 × 12/13 = 480 W. Check with I²R = 2.0² × 120 = 480 W. Only the resistor dissipates power on average; the inductor and capacitor store and return energy each cycle.
Resonance. ω₀ = 1/√(LC) = 1/√(0.10 × 20 × 10⁻⁶) = 1/√(2 × 10⁻⁶) ≈ 707 rad/s. At this frequency X_L = X_C, Z = R = 120 Ω and the current would rise to 260/120 ≈ 2.2 A, its largest possible value.
Which mistakes come up again and again?
Mixing rms and peak values. For sine waves, peak = √2 × rms. Meters read rms. Use one or the other throughout.
Using f where ω is needed. X_L = ωL = 2πfL. Forgetting the 2π is a common way to lose a full question.
Thinking power is V × I. Average power is V I cos φ. In a purely inductive or capacitive circuit, cos φ = 0 and the current is called wattless.
Getting lead and lag backwards. Remember that in a capacitor the current leads, because charge must flow in before voltage builds up.
Assuming resonance means infinite current. Z falls to R, not zero, so current is limited by the resistance.
The reactance formulas come from the same Faraday's law you used for induction, so if X_L feels like a formula with no meaning, revisit our post on finding the direction of induced current. If the calculus behind "current lags by π/2" is shaky, our guide to the calculus you need for JEE physics helps.
When you are unsure where your phasor diagram went wrong, Fermi's AI tutor can draw the figure on the board and check your understanding as it goes, so you see which arrow you placed the wrong way.
A practice routine
Day 1: Draw phasor diagrams for R only, L only, C only, RL, RC and LCR, with no numbers.
Day 2: Three series LCR problems. For each, check that the phasor sum of element voltages equals the supply.
Day 3: Two resonance questions and two power factor questions.
Day 4: A timed mixed set of six AC questions from previous papers.
Tomorrow, take one series LCR problem and solve it twice: once with the phasor triangle drawn to scale and once by formula, and compare.
