A resistor R and a capacitor C are connected in series to an ac source. Consider the following statements:
(A) The current in the circuit leads the applied voltage.
(B) The voltage across the capacitor is in phase with the current.
(C) If the frequency of the source is increased, the phase difference between the current and the applied voltage decreases.
(D) The average power consumed in the circuit is \( I^{2}R \), where I is the rms current.
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → (A), (C) and (D) only
A series RC circuit is a series LCR circuit with no inductor, so \( X_L=0 \) and the capacitive reactance is \( X_C=1/\omega C \). The phase angle between the current and the source voltage follows from \( \tan\phi=X_C/R \).
(A) The current in the circuit leads the applied voltage. — True. With \( X_C \gt X_L \) the circuit is capacitive, and the current leads the source voltage by an angle between 0 and \( \pi/2 \).
(B) The voltage across the capacitor is in phase with the current. — False. In a capacitor the current leads the voltage by \( \pi/2 \), so the capacitor voltage lags the current by a quarter of a cycle. It is the voltage across the resistor that is in phase with the current.
(C) If the frequency of the source is increased, the phase difference between the current and the applied voltage decreases. — True. The phase angle is set by \( \tan\phi=1/\omega CR \). A higher frequency lowers \( X_C \). So \( \tan\phi \) becomes smaller, and so does \( \phi \).
(D) The average power consumed in the circuit is \( I^{2}R \), where I is the rms current. — True. The capacitor takes no average power, and all the power is dissipated in the resistor. With \( \cos\phi=R/Z \), the average power \( I^{2}Z\cos\phi \) equals \( I^{2}R \).
NCERT (2026-27):
Phasor-diagram Solution – “If XC > XL, φ is positive and the circuit is predominantly capacitive. Consequently, the current in the circuit leads the source voltage.”
Summary – “4. … The current through the capacitor is π/2 ahead of the applied voltage. …”
Phasor-diagram Solution – “… we have”
\( \displaystyle \tan\phi=\frac{X_C-X_L}{R} \)\( \qquad (7.27) \)
Power in AC Circuit: the Power Factor – “… So, φ may be non-zero in a RL or RC or RCL circuit. Even in such cases, power is dissipated only in the resistor.”