For a series LCR circuit, \( R = X_L = 3X_C \). The impedance of the circuit and the phase difference between voltage \( V \) and current \( I \), respectively, will be
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \( \displaystyle\frac{R\sqrt{13}}{3}\,\Omega \) and \( \tan^{-1}\displaystyle\frac{2}{3} \)
For a series LCR circuit the impedance and the phase angle are
\( \displaystyle Z = \sqrt{R^{2} + (X_L - X_C)^{2}}, \qquad \tan\phi = \frac{X_L - X_C}{R} \)
Given \( X_L = R \) and \( X_C = R/3 \):
\( \displaystyle X_L - X_C = R - \frac{R}{3} = \frac{2R}{3} \)
\( \displaystyle Z = \sqrt{R^{2} + \frac{4R^{2}}{9}} = \sqrt{\frac{13R^{2}}{9}} = \frac{R\sqrt{13}}{3} \)
\( \displaystyle \tan\phi = \frac{2R/3}{R} = \frac{2}{3} \;\Rightarrow\; \phi = \tan^{-1}\frac{2}{3} \)
Since \( X_L \gt X_C \), the circuit is inductive and the voltage leads the current by this angle.