Question

A resistor of resistance 160 $\Omega$, an inductor of 280 mH, and a capacitor are connected in series to an AC source of 80 V, 50 Hz. If the circuit is in resonance, the potential difference across the capacitor is:

• A. 44 V
• B. 40 V
• C. 88 V
• D. 80 V

Hint:

In series RLC resonance: $I = V / R$.
Voltage across L or C can exceed supply voltage.
Check reactance formulas: $X_L = 2 \pi f L$, $X_C = 1/(2 \pi f C)$.
Resonance condition: $X_L = X_C$.

Answer 1

The Correct Option is C

• At resonance, the inductive reactance $X_L$ = capacitive reactance $X_C$, so $V_{RMS}$ across R is $I \cdot R$, and current $I = \frac{V_{source}}{R}$.
• Voltage across capacitor: $V_C = I \cdot X_C$.
• Series resonance voltage across L or C can exceed source voltage (series resonance phenomenon).
• Using $V_C = I \cdot X_C = V \frac{X_C}{R} = 80 \cdot \frac{X_C}{160}$.
• After calculating $X_C = \frac{1}{2 \pi f C}$ and plugging in, $V_C = 88$ V.
• Hence, voltage across capacitor = 88 V.