In a series LCR circuit, the current is \( I=I_0\sin\omega t \). Which of the following is true?
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → All the above
Option (1) The current through the capacitor is \( I=I_0\sin\omega t \) — Incorrect. The statement is true, because the capacitor is in series and carries the circuit current. But it is only one of the three true statements, so this option alone is incomplete.
Option (2) The voltage across the inductor is \( V=V_0\sin\left(\omega t+\frac{\pi}{2}\right) \) — Incorrect. The statement is true, because the inductor voltage leads the current by \( \pi/2 \). But it is only one of the three true statements, so this option alone is incomplete.
Option (3) The current through the resistor is \( I=I_0\sin\omega t \) — Incorrect. The statement is true, because the resistor also carries the common series current. But it is only one of the three true statements, so this option alone is incomplete.
Option (4) All the above — Correct. All three statements hold:
In a series LCR circuit the elements lie on one path. So the same current flows through each element, with the same amplitude and phase. The capacitor and the resistor both carry \( I_0\sin\omega t \), so (1) and (3) hold.
The voltage across the inductor has amplitude \( V_0=I_0X_L \) and leads the current by \( \pi/2 \), so it is \( V_0\sin\left(\omega t+\frac{\pi}{2}\right) \) and (2) holds.
NCERT (2026-27):
Phasor-Diagram Solution – “… we see that the resistor, inductor and capacitor are in series. Therefore, the ac current in each element is the same at any time, having the same amplitude and phase.”
Phasor-Diagram Solution – “… let VL, VR, VC, and V represent the voltage across the inductor, resistor, capacitor and the source, respectively. From previous section, we know that VR is parallel to I, VC is π/2 behind I and VL is π/2 ahead of I.”