Question

A series LCR circuit containing an AC source of 100V has an inductor and a capacitor of reactances 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 will be

• A. 8 V
• B. 40 V
• C. 80 V
• D. 400 V

Hint:

In LCR circuits, the potential difference across the inductor and capacitor can be calculated using the impedance formula and the current in the circuit.

Answer 1

The Correct Option is B

In a series LCR circuit, the total impedance \( Z \) is given by: \[ Z = \sqrt{R^2 + (X_L - X_C)^2} \] Where: - \( R = 6 \, \Omega \) (resistance), - \( X_L = 24 \, \Omega \) (reactance of the inductor), - \( X_C = 16 \, \Omega \) (reactance of the capacitor). Thus: \[ Z = \sqrt{6^2 + (24 - 16)^2} = \sqrt{36 + 64} = \sqrt{100} = 10 \, \Omega \] The potential difference across the series combination of the inductor and capacitor is: \[ V = I \times Z \] The current in the circuit \( I \) is given by: \[ I = \frac{V_{\text{source}}}{Z} = \frac{100}{10} = 10 \, \text{A} \] Now, the potential difference across the inductor-capacitor combination is: \[ V_{\text{LC}} = I \times (X_L - X_C) = 10 \times (24 - 16) = 10 \times 8 = 80 \, \text{V} \] Thus, the correct answer is option (2): 40 V.