A capacitor blocks DC and allows AC because its capacitive reactance is X_C = 1/(2 pi f C). For steady DC the frequency f = 0, so X_C is infinite and no current flows. For AC f is not zero, so X_C is finite and current flows. Memory hook: "zero frequency, infinite reactance" - the capacitor is a wall for DC and a door for AC.
The reactance-frequency curve: X_C shoots to infinity as f approaches 0 (so a capacitor blocks DC), and drops to a small value at high frequency (so it allows AC). This single graph is the whole concept.
Your doubts, answered
Does current really flow THROUGH the capacitor, or is it just charging?
No charge crosses the gap between the plates - the insulator (dielectric) blocks it. In DC, the capacitor charges up once, then charge stops moving, so steady current is zero. In AC the source reverses direction every half cycle, so the plates charge, discharge and re-charge with opposite sign over and over. This continuous back-and-forth of charge in the wires looks like a steady current in the circuit, even though no charge jumps the gap. The changing electric field between the plates is called displacement current, and it equals the wire current.
Why is capacitive reactance infinite for DC?
Reactance is X_C = 1/(2 pi f C). For pure DC the frequency f = 0. Put f = 0 into the formula and X_C = 1/0 = infinity. Infinite opposition means current I = V/X_C = V/infinity = 0. So a fully charged capacitor behaves like a broken wire (open circuit) for DC.
Why does a capacitor pass MORE current at higher frequency?
X_C = 1/(2 pi f C). Frequency f is in the bottom of the fraction, so a bigger f gives a smaller X_C. Smaller reactance means the capacitor opposes the current less, so I = V/X_C is larger. This is why a capacitor easily passes high-frequency AC but strongly opposes low-frequency AC and completely blocks DC (f = 0).
If no charge crosses the plates, how can we measure a current in the circuit?
An ammeter in the wire measures charge flowing in the wire per second, not charge crossing the gap. When AC drives charge on and off the plates repeatedly, real electrons move in the connecting wires the whole time. So the ammeter reads a real current even though the gap itself is never crossed. Only the average NET charge transferred is zero, but the moving charge (the current) is very real.
⚠️ The NEET trap ✗ AC current physically jumps across the gap between the two capacitor plates. ✓ No charge ever crosses the insulating gap. AC keeps charging and discharging the plates, so charge moves only in the wires; the changing field between the plates is displacement current, which equals the conduction current in the wires. 🧠 Plates never touch - the current is in the WIRES, the field is in the GAP.
Real NEET questions
2023
An AC source is connected to a capacitor C. Due to decrease in its operating frequency:
A · Capacitive reactance decreases
B · Displacement current increases
C · Displacement current decreases ✓
D · Capacitive reactance remains constant
Solution: For a capacitor, current I = E / X_C where X_C = 1/(2 pi f C), so I = E * 2 pi f C. Step 1: write the dependence - I is directly proportional to f. Step 2: frequency f decreases, so X_C = 1/(2 pi f C) increases. Step 3: since current is proportional to f, the current falls. The conduction current equals the displacement current, so the displacement current decreases. Answer: (C). This is exactly the DC limit taken one step further: push f all the way to 0 and the current would drop to zero - that is DC being blocked.
2020
A 40 microF 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: angular frequency omega = 2 pi f = 2 * 3.14 * 50 = 314 rad/s. Step 2: capacitive reactance X_C = 1/(omega C) = 1/(314 * 40e-6) = 1/(0.01256) = 79.6 ohm. Step 3: rms current I = V_rms / X_C = 200 / 79.6 = 2.51 A, which is nearly 2.5 A. Answer: (A). Notice current DOES flow through a capacitor on AC - because f is not zero, X_C is finite.
2024
A 10 microF 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: X_C = 1/(omega C) = 1/(2 pi f C) = 1/(2 * 3.14 * 50 * 10e-6) = 1000/pi ohm = 318.5 ohm. Step 2: peak voltage V_0 = sqrt(2) * V_rms = 1.414 * 210 = 297 V. Step 3: peak current i_0 = V_0 / X_C = 297 / 318.5 = 0.93 A. Answer: (A). Again the capacitor passes AC because the finite frequency gives a finite X_C.
Solved Alternating Current NEET PYQs
Try the real previous-year questions from this chapter — each with the answer and a full solution.
Yes, in the steady state. There is a brief charging current when you first connect DC, but once the capacitor is fully charged, X_C is effectively infinite (f = 0) and the steady DC current is zero. It behaves like an open circuit.
What is the formula that explains this behaviour?
Capacitive reactance X_C = 1/(2 pi f C) or X_C = 1/(omega C). Put f = 0 for DC to get infinite reactance (blocks DC); put a non-zero f for AC to get finite reactance (allows AC).
Why does a capacitor allow high-frequency AC more easily than low-frequency AC?
Because X_C is inversely proportional to f. High frequency gives small reactance, so more current flows. Low frequency gives large reactance, so less current flows. This makes a capacitor act like a high-pass filter.
What is displacement current in a capacitor?
It is the current-like effect of the changing electric field between the plates, given by I_d = epsilon_0 * (d(phi_E)/dt). Maxwell introduced it so that current stays continuous across the gap. In magnitude it equals the conduction current in the connecting wires.
How is this the opposite of an inductor?
An inductor has X_L = 2 pi f L, which is zero for DC (passes DC freely) and large for high-frequency AC (blocks AC). A capacitor is the mirror image: it blocks DC and passes AC. This contrast is a favourite NEET comparison.
