Q130 / 150 · MHT CET 2026 (16 April Shift 2) · Physics
Series LCR circuit containing an a.c. source of 100V has a inductor and a capacitor of reactance $24\ \Omega$ and $16\ \Omega$ respectively. If a resistance of $6\ \Omega$ is connected in series, then the potential difference across the series combination of inductor and capacitor only is
- Correct answerA$80\text{V}$
- B$8\text{V}$
- C$40\text{V}$
- D$20\text{V}$
Solution
Given $V = 100\text{V}$, $X_L = 24\ \Omega$, $X_C = 16\ \Omega$, and $R = 6\ \Omega$.
The net reactance of the circuit is $X = X_L - X_C = 24 - 16 = 8\ \Omega$.
The impedance of the series LCR circuit is given by $Z = \sqrt{R^2 + (X_L - X_C)^2}$.
Substituting the values, $Z = \sqrt{6^2 + 8^2} = \sqrt{36 + 64} = 10\ \Omega$.
The rms current in the circuit is $I = \dfrac{V}{Z} = \dfrac{100}{10} = 10\text{A}$.
The potential difference across the series combination of the inductor and capacitor is $V_{LC} = I \times |X_L - X_C|$.
$V_{LC} = 10 \times 8 = 80\text{V}$.
The net reactance of the circuit is $X = X_L - X_C = 24 - 16 = 8\ \Omega$.
The impedance of the series LCR circuit is given by $Z = \sqrt{R^2 + (X_L - X_C)^2}$.
Substituting the values, $Z = \sqrt{6^2 + 8^2} = \sqrt{36 + 64} = 10\ \Omega$.
The rms current in the circuit is $I = \dfrac{V}{Z} = \dfrac{100}{10} = 10\text{A}$.
The potential difference across the series combination of the inductor and capacitor is $V_{LC} = I \times |X_L - X_C|$.
$V_{LC} = 10 \times 8 = 80\text{V}$.
Hence, the correct answer is option: 'A'.