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An ac voltage $V=220\sin(2\times10^3t)$ Volt is applied to a series LCR circuit. Then the current amplitude in this circuit is: (Given: L = 10 mH, C = 25 $\mu$F, R = 100 $\Omega$)
Detailed Solution
$X_L=\omega L=2\times10^3\times10\times10^{-3}=20\ \Omega$ and $X_C=\frac{1}{\omega C}=\frac{10^6}{2\times10^3\times25}=20\ \Omega$. Since $X_L=X_C$, the circuit is at resonance and Z = R. $I_0=\frac{V_0}{R}=\frac{220}{100}=2.2$ A.