In the circuit shown below,the AC source has | vedclass

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(A) Given: Voltage $V = 20 \cos(\omega t)$,so peak voltage $V_0 = 20 \, V$. Angular frequency $\omega = 2000 \, rad/sec$.Inductive reactance $X_L = \o

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$2$

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(A) Given: Voltage $V = 20 \cos(\omega t)$,so peak voltage $V_0 = 20 \, V$. Angular frequency $\omega = 2000 \, rad/sec$.Inductive reactance $X_L = \omega L = 2000 	imes 5 	imes 10^{-3} = 10 \, \Omega$.Capacitive reactance $X_C = \frac{1}{\omega C} = \frac{1}{2000 	imes 50 	imes 10^{-6}} = \frac{1}{0.1} = 10 \, \Omega$.Since $X_L = X_C$,the circuit is in resonance.Total resistance of the circuit $R = R_L + R = 4 \, \Omega + 6 \, \Omega = 10 \, \Omega$.The impedance of the circuit at resonance is $Z = R = 10 \, \Omega$.The amplitude of the current $I_0 = \frac{V_0}{Z} = \frac{20}{10} = 2 \, A$.

In the circuit shown below,the $AC$ source has voltage $V = 20 \cos(\omega t)$ with $\omega = 2000 \, rad/sec$. The amplitude of the current will be nearest to.....$A$

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