The reading of the ammeter in the circuit sho | vedclass

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(B) In the given circuit,the inductor $(X_L = 5 \ \Omega)$ and the capacitor $(X_C = 5 \ \Omega)$ are connected in series with each other,and this com

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$2$

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(B) In the given circuit,the inductor $(X_L = 5 \ \Omega)$ and the capacitor $(X_C = 5 \ \Omega)$ are connected in series with each other,and this combination is connected in parallel with the voltmeter. Since $X_L = X_C$,the series combination of the inductor and capacitor acts as a short circuit because the net impedance $Z_{LC} = |X_L - X_C| = 0 \ \Omega$. Therefore,the voltage across the voltmeter is $0 \ V$. The entire source voltage of $110 \ V$ appears across the resistor $R = 55 \ \Omega$. The current $I$ through the ammeter is given by Ohm's law: $I = \frac{V}{R} = \frac{110 \ V}{55 \ \Omega} = 2 \ A$.

The reading of the ammeter in the circuit shown will be: (in $A$)

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