In the circuit shown in the figure,neglecting | vedclass

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(D) The circuit is an $LCR$ series circuit where the voltmeter is connected across the series combination of the inductor and the capacitor.In an $LCR

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$0\,V, \,8\,A$

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(D) The circuit is an $LCR$ series circuit where the voltmeter is connected across the series combination of the inductor and the capacitor.In an $LCR$ series circuit,the voltage across the inductor is $V_L = I X_L$ and the voltage across the capacitor is $V_C = I X_C$.Since the inductor and capacitor are in series,the same current $I$ flows through both.The voltage across the combination is $V_{LC} = |V_L - V_C| = I |X_L - X_C|$.Given $X_L = 25\,\Omega$ and $X_C = 25\,\Omega$,we have $X_L - X_C = 0$,so the voltmeter reading is $0\,V$.The impedance of the circuit is $Z = \sqrt{R^2 + (X_L - X_C)^2} = \sqrt{30^2 + (25 - 25)^2} = 30\,\Omega$.The current in the circuit is $I = \frac{V}{Z} = \frac{240\,V}{30\,\Omega} = 8\,A$.Thus,the voltmeter reading is $0\,V$ and the ammeter reading is $8\,A$.

In the circuit shown in the figure,neglecting source resistance,the voltmeter and ammeter readings will respectively be:

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