In the circuit shown in the figure,neglecting | vedclass

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(D) The circuit is an $LCR$ series circuit where the inductor and capacitor are connected in series with each other,and this combination is in series

Vedclass · Accepted Answer

$0\,V, 8\,A$

Vedclass · Answer

(D) The circuit is an $LCR$ series circuit where the inductor and capacitor are connected in series with each other,and this combination is in series with the resistor $R$.The voltmeter is connected across the series combination of the inductor $(X_L)$ and the capacitor $(X_C)$.The voltage across the inductor is $V_L = I X_L$ and the voltage across the capacitor is $V_C = I X_C$.Since these voltages are $180^{\circ}$ out of phase,the net 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 $V_{LC} = I 	imes 0 = 0\, V$.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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