In an a.c. LCR circuit,a capacitor and an ind | vedclass

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(A) Given that $X_L = 4\,\Omega$ and $X_C = 4\,\Omega$.Since $X_L = X_C$,the circuit is in a state of electrical resonance.In resonance,the total impe

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$I = 2\,A$ and $V = 0\,V$

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(A) Given that $X_L = 4\,\Omega$ and $X_C = 4\,\Omega$.Since $X_L = X_C$,the circuit is in a state of electrical resonance.In resonance,the total impedance of the circuit $Z$ is equal to the resistance $R$,so $Z = R = 45\,\Omega$.The current in the circuit is given by $I = \frac{V_{source}}{Z} = \frac{90\,V}{45\,\Omega} = 2\,A$.The voltage across the series combination of the inductor and capacitor is $V_{LC} = I 	imes |X_L - X_C|$.Since $X_L = X_C$,we have $V_{LC} = 2\,A 	imes |4\,\Omega - 4\,\Omega| = 2\,A 	imes 0\,\Omega = 0\,V$.Therefore,the ammeter reading is $2\,A$ and the voltmeter reading is $0\,V$.

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