Question

In a series resonant $LCR$ circuit, the voltage across $R$ is $100V$ and $R=1k\Omega $ with $C=2μF$ . The resonant frequency $\omega $ is $200rads^{- 1}$ . At resonance the voltage across $\textit{L}$ is

• A. $4\times 10^{- 3}V$
• B. $2.5\times 10^{- 2}V$
• C. $40V$
• D. $250V$

Answer 1

Answer: (D)

Current, $I=\frac{E}{Z}$
where, $E=\sqrt{V_{R}^{2} + \left(V_{L} - V_{C}\right)^{2}}$
$Z=\sqrt{R^{2} + \left(X_{L} - X_{C}\right)^{2}}$

At resonance, $X_{L}=X_{C}$
$\therefore $ $Z=R$
Again at resonance, $V_{L}=V_{C}$
$\therefore $ $E=V_{R}$
$\therefore $ $I=\frac{V_{R}}{R}=\frac{100}{1 \times 10^{3}}=$ $0.1A$
$\therefore \quad V_L=I L \omega=\frac{I}{C \omega}=\frac{0.1}{\left(2 \times 10^{-6}\right) \times(200)}$
$\therefore $ $V_{L}=250V$ .