In the circuit shown in the figure,neglecting | vedclass

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(D) The circuit consists of a resistor $R = 30\,\Omega$,an inductor $X_L = 25\,\Omega$,and a capacitor $X_C = 25\,\Omega$ in series connected to a $24

Vedclass · Accepted Answer

$0\,V, 8\,A$

Vedclass · Answer

(D) The circuit consists of a resistor $R = 30\,\Omega$,an inductor $X_L = 25\,\Omega$,and a capacitor $X_C = 25\,\Omega$ in series connected to a $240\,V$ $AC$ source.Since $X_L = X_C = 25\,\Omega$,the circuit is in a state of resonance.In a series $LCR$ circuit,the voltage across the inductor is $V_L = I X_L$ and the voltage across the capacitor is $V_C = I X_C$. These voltages are $180^{\circ}$ out of phase.The voltmeter is connected across the series combination of the inductor and the capacitor. Therefore,the reading of the voltmeter is $V = |V_L - V_C|$.Since $X_L = X_C$,we have $V_L = V_C$,which implies $V = |V_L - V_L| = 0\,V$.The total impedance of the circuit at resonance is $Z = R = 30\,\Omega$.The current in the circuit is given by $I = \frac{E}{Z} = \frac{240\,V}{30\,\Omega} = 8\,A$.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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