Physics · Alternating Current · NEET
On DC the frequency is zero, so the inductor's reactance X_L = 2 pi f L becomes zero. It behaves like an ordinary wire and does nothing to limit current, so only R matters and current I = V/R is large. On AC, the inductor adds reactance X_L, making the total opposition Z = sqrt(R^2 + X_L^2) larger than R. A bigger opposition means a smaller current, so I = V/Z is smaller on AC.
No. A capacitor blocks steady DC. When you connect DC, the capacitor charges up until its voltage equals the source, and then the current stops (steady current = 0). Its DC reactance X_C = 1/(2 pi f C) becomes infinite at f = 0, like an open switch. But on AC the voltage keeps reversing, so the capacitor keeps charging and discharging and AC current does flow. This is the key clue in the NEET 2019 question.
Impedance is Z = sqrt(R^2 + (X_L - X_C)^2). Because the reactance term is squared and added under the root, Z can never be smaller than R. Z equals R only in two cases: a pure resistor, or at resonance where X_L = X_C cancel out. This is why the AC current I = V/Z is never larger than the DC current I = V/R for the same circuit and same voltage magnitude.
Check if steady DC current flows. If DC current flows (non-zero), there is NO capacitor in series, because a capacitor would block DC completely. If DC current is zero, a capacitor is present. Then compare: R = V_dc / I_dc gives the resistance, and Z = V_ac / I_ac gives the impedance. If Z > R and DC current flows, the extra element is an inductor (series LR).
A circuit when connected to an AC source of 12 V gives a current of 0.2 A. The same circuit when connected to a DC source of 12 V gives a current of 0.4 A. The circuit is:
Try the real previous-year questions from this chapter — each with the answer and a full solution.
On DC, current is limited only by resistance R (I = V/R), because the inductor's reactance is zero and the capacitor blocks steady DC. On AC, current is limited by the full impedance Z = sqrt(R^2 + (X_L - X_C)^2), so the same circuit usually carries less current on AC than on DC.
Inductive reactance is X_L = 2 pi f L. At DC the frequency f = 0, so X_L = 0 and the inductor is just a wire. As frequency rises, X_L rises, so it opposes AC more. At very high frequency an inductor almost blocks the current.
Capacitive reactance is X_C = 1/(2 pi f C). At DC (f = 0), X_C is infinite, so no steady current flows. On AC the capacitor keeps charging and discharging as the voltage reverses, so AC current flows, and X_C is smaller at higher frequency.
Yes, for the same circuit and same voltage magnitude, because Z is always greater than or equal to R. AC current equals DC current only when the circuit is a pure resistor or at resonance (X_L = X_C), where Z = R.
