Question

For a series $RLC$ circuit $R=X_{L}=2X_{C}$ . The impedance of the circuit and phase difference (between) $V$ and $i$ will be :-

• A. $\frac{\sqrt{5} R}{2},\left(tan\right)^{- 1}\left(2\right)$
• B. $\frac{\sqrt{5} R}{2},\left(tan\right)^{- 1}\left(\frac{1}{2}\right)$
• C. $\sqrt{5}X_{C},\left(tan\right)^{- 1}\left(2\right)$
• D. $\sqrt{5}R,\left(tan\right)^{- 1}\left(\frac{1}{2}\right)$

Answer 1

Answer: (B)

$X_{L}=R,X_{C}=R/2$
$\therefore tan\phi=\frac{X_{L} - X_{C}}{R}=\frac{R - \frac{R}{2}}{R}=\frac{1}{2}\Rightarrow \phi=\left(tan\right)^{- 1}\left(1 / 2\right)$
Also $Z=\sqrt{R^{2} + \left(X_{L} - X_{C}\right)^{2}}=\sqrt{R^{2} + \frac{R^{2}}{4}}=\frac{\sqrt{5}}{2}R$