A series LCR circuit is connected to a \( 230\,\mathrm{V} \), \( 50\,\mathrm{rad/s} \) source. Given \( L = 5\,\mathrm{H} \), \( C = 80\,\mu\mathrm{F} \) and \( R = 40\,\Omega \). Impedance of the circuit is
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \( 40\,\Omega \)
The impedance of a series LCR circuit is \( Z = \sqrt{R^2 + (X_L - X_C)^2} \), where \( X_L = \omega L \) and \( X_C = 1/\omega C \).
\( \displaystyle X_L = \omega L = 50 \times 5 = 250\,\Omega \)
\( \displaystyle X_C = \frac{1}{\omega C} = \frac{1}{50 \times 80\times10^{-6}} = \frac{1}{4\times10^{-3}} = 250\,\Omega \)
Here \( X_L = X_C \), so the circuit is at resonance and the reactances cancel.
\( \displaystyle Z = \sqrt{R^2 + 0} = R = 40\,\Omega \)