When an AC voltage v = vm sin(omega.t) is applied to a pure capacitor, the current is i = im sin(omega.t + 90 degrees). So the current LEADS the voltage by 90 degrees (a quarter cycle). The peak current is im = vm / Xc, where the capacitive reactance is Xc = 1/(omega.C). Memory hook: in a Capacitor, Current Comes first (C = Current leads).
The red current curve peaks a quarter cycle BEFORE the blue voltage curve, showing the current leads voltage by 90 degrees in a pure capacitor. Peak current im = vm/Xc with Xc = 1/(omega.C), and the average power over a full cycle is zero.
Your doubts, answered
Does the current lead or lag the voltage in a capacitor?
The current LEADS the voltage by 90 degrees (pi/2 radians) in a pure capacitor. If v = vm sin(omega.t), then i = im sin(omega.t + 90 degrees). Reason: the current must first push charge onto the plates before the voltage across the capacitor can build up, so current peaks a quarter cycle earlier than voltage. Trick: 'CiVil' - in C (capacitor), i leads V.
Why does a capacitor allow AC but block DC?
With DC, the capacitor charges up once and then current stops (fully charged plates oppose more charge). With AC, the voltage keeps reversing every half cycle, so the capacitor is charged and discharged again and again. This continuous back-and-forth flow of charge means AC current keeps flowing. So a capacitor blocks DC but passes AC.
What is the formula for peak current through a capacitor?
im = vm / Xc, where Xc = 1/(omega.C) = 1/(2.pi.f.C) is the capacitive reactance. In rms terms, Irms = Vrms / Xc. Note current is NOT V/R here - a pure capacitor has no resistance, only reactance Xc measured in ohms.
Does a capacitor consume energy from the AC source?
No. Over one full cycle the average power in a pure capacitor is zero. During two quarter-cycles it stores energy (charging) and during the other two it returns that energy to the source (discharging). Because the phase difference is 90 degrees, cos(90) = 0, so average power = Vrms.Irms.cos(phi) = 0. This is called wattless current.
How does the current change if I increase the frequency?
Higher frequency means higher omega, so Xc = 1/(omega.C) DECREASES. Smaller reactance means MORE current flows. So current through a capacitor increases with frequency. A capacitor 'likes' high frequency and offers almost infinite reactance at f = 0 (DC), which is why it blocks DC.
⚠️ The NEET trap ✗ Using i = vm/R or saying current lags voltage by 90 degrees (mixing up capacitor with inductor). ✓ A pure capacitor has NO resistance. Use im = vm/Xc with Xc = 1/(omega.C), and remember current LEADS voltage by 90 degrees in a capacitor (it lags in an inductor). 🧠 Capacitor = Current Comes first (leads). Inductor = current Lags. Do not swap Xc = 1/(omega.C) with XL = omega.L.
Real NEET questions
2016
A small signal voltage V(t) = V0 sin(omega.t) is applied across an ideal capacitor C. Which statement is correct?
A · Current I(t) lags voltage V(t) by 90 degrees
B · Over a full cycle the capacitor C does not consume any energy from the voltage source ✓
C · Current I(t) is in phase with voltage V(t)
D · Current I(t) leads voltage V(t) by 180 degrees
Solution: In a pure capacitor, current LEADS voltage by 90 degrees, so options A, C and D are wrong (the lead is 90 degrees, not lag, not in-phase, not 180). Average power over one cycle = Vrms.Irms.cos(phi). Here phi = 90 degrees, so cos(90) = 0, giving zero average power. The capacitor stores energy while charging and returns it while discharging, so over a FULL cycle it consumes no net energy. Correct answer: B.
2020
A 40 microfarad capacitor is connected to a 200 V, 50 Hz AC supply. The rms value of the current in the circuit is, nearly:
A · 2.5 A ✓
B · 25.1 A
C · 1.7 A
D · 2.05 A
Solution: Step 1: Xc = 1/(2.pi.f.C) = 1/(2 x 3.14 x 50 x 40x10^-6). Step 2: Denominator = 2 x 3.14 x 50 x 40x10^-6 = 0.01256. So Xc = 1/0.01256 = 79.6 ohms (about 80 ohms). Step 3: Irms = Vrms/Xc = 200/79.6 = 2.51 A, which is nearly 2.5 A. Correct answer: A.
2024
A 10 microfarad capacitor is connected to a 210 V, 50 Hz source. The peak current in the circuit is nearly (pi = 3.14):
A · 0.93 A ✓
B · 1.20 A
C · 0.35 A
D · 0.58 A
Solution: Step 1: Xc = 1/(2.pi.f.C) = 1/(2 x 3.14 x 50 x 10x10^-6) = 1/(0.00314) = 318.5 ohms. Step 2: Irms = Vrms/Xc = 210/318.5 = 0.659 A. Step 3: Peak current im = sqrt(2) x Irms = 1.414 x 0.659 = 0.93 A. Correct answer: A.
Solved Alternating Current NEET PYQs
Try the real previous-year questions from this chapter — each with the answer and a full solution.
What is the phase difference between voltage and current in a capacitor?
Exactly 90 degrees (pi/2 radians), with the current leading the voltage. If v = vm sin(omega.t), then i = im sin(omega.t + pi/2).
What is capacitive reactance and its SI unit?
Capacitive reactance Xc = 1/(omega.C) = 1/(2.pi.f.C) is the opposition a capacitor offers to AC. Its SI unit is the ohm, the same as resistance.
Why is a capacitor called a wattless element in AC?
Because the phase angle is 90 degrees, so cos(phi) = 0 and average power = 0. The current is called wattless current - it flows but dissipates no net energy over a cycle.
Does capacitive reactance depend on frequency?
Yes. Xc = 1/(2.pi.f.C), so reactance is inversely proportional to frequency. Higher frequency gives lower Xc and more current; at DC (f = 0) reactance is infinite and no current flows.
Why is this concept important for NEET?
NEET regularly asks direct numericals on capacitor current (rms and peak) and conceptual questions on the 90-degree phase lead and zero power. It is also the building block for series LCR circuits, resonance and power factor.
