Consider the L-C-R circuit given below. The c | vedclass

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(A) Given: Peak voltage $V_0 = 220 \,V$,frequency $f = 50 \,Hz$,resistance $R = 400 \,\Omega$,capacitance $C = 200 \,\mu F = 200 	imes 10^{-6} \,F$,a

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$0.12$

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(A) Given: Peak voltage $V_0 = 220 \,V$,frequency $f = 50 \,Hz$,resistance $R = 400 \,\Omega$,capacitance $C = 200 \,\mu F = 200 	imes 10^{-6} \,F$,and inductance $L = 6 \,H$.Angular frequency $\omega = 2\pi f = 2 	imes \pi 	imes 50 = 100\pi \,rad/s$.Inductive reactance $X_L = \omega L = 100\pi 	imes 6 = 600\pi \,\Omega \approx 1884.96 \,\Omega$.Capacitive reactance $X_C = \frac{1}{\omega C} = \frac{1}{100\pi 	imes 200 	imes 10^{-6}} = \frac{1}{0.02\pi} = \frac{50}{\pi} \,\Omega \approx 15.92 \,\Omega$.Impedance $Z = \sqrt{R^2 + (X_L - X_C)^2} = \sqrt{400^2 + (1884.96 - 15.92)^2} = \sqrt{160000 + (1869.04)^2} \approx \sqrt{160000 + 3493310} \approx \sqrt{3653310} \approx 1911.36 \,\Omega$.Maximum current $I_0 = \frac{V_0}{Z} = \frac{220}{1911.36} \approx 0.115 \,A$.Rounding to the nearest value,the maximum current is approximately $0.12 \,A$.

Consider the $L-C-R$ circuit given below. The circuit is driven by a $50 \,Hz, AC$ source with peak voltage $220 \,V$. If $R=400 \,\Omega, C=200 \,\mu F$ and $L=6 \,H$,the maximum current in the circuit is closest to ............ $A$.

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