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In the circuit shown below, the ac source has voltage $V=20 \cos (\omega t)$ volts with $\omega=2000 \mathrm{rad} / \mathrm{sec}$. the amplitude of the current will be nearest to
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$2 A$
$\begin{aligned} & R=6+4=10 \Omega \\ & X_L=\omega L=2000 \times 5 \times 10^{-3}=10 \Omega \\ & X_C=\frac{1}{\omega C}=\frac{1}{2000 \times 50 \times 10^{-6}}=10 \Omega \\ & \therefore Z=\sqrt{R^2+\left(X_L-X_C\right)^2}=10 \Omega \\ & \text { Amplitude of current }=i_0=\frac{V_0}{Z}=\frac{20}{10}=2 \mathrm{~A}\end{aligned}$
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