What will be the reading in the voltmeter and | vedclass

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(B) Given, $X_{L} = 4 \, \Omega$ and $X_{C} = 4 \, \Omega$.In a series $LC$ circuit, the voltage across the inductor $(V_{L})$ and the capacitor $(V_{

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$0, 2 \, A$

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(B) Given, $X_{L} = 4 \, \Omega$ and $X_{C} = 4 \, \Omega$.In a series $LC$ circuit, the voltage across the inductor $(V_{L})$ and the capacitor $(V_{C})$ are $180^{\circ}$ out of phase.Therefore, the net voltage across the series combination of the inductor and capacitor is $V_{	ext{net}} = |V_{L} - V_{C}|$.Since $X_{L} = X_{C}$, the magnitudes of the voltages are equal, i.e., $V_{L} = I X_{L}$ and $V_{C} = I X_{C}$.Thus, $V_{	ext{net}} = I(X_{L} - X_{C}) = I(4 - 4) = 0 \, V$.The voltmeter is connected across this series $LC$ combination, so its reading is $0 \, V$.The total impedance of the circuit is $Z = \sqrt{R^{2} + (X_{L} - X_{C})^{2}}$.Substituting the values, $Z = \sqrt{45^{2} + (4 - 4)^{2}} = \sqrt{45^{2}} = 45 \, \Omega$.The current in the circuit is $I = \frac{V}{Z} = \frac{90 \, V}{45 \, \Omega} = 2 \, A$.Therefore, the voltmeter reading is $0 \, V$ and the ammeter reading is $2 \, A$.

What will be the reading in the voltmeter and ammeter of the circuit shown?

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