The reading of the ammeter in the circuit sho | vedclass

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(A) 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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(A) 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 $X_L = X_C = 5 \ \Omega$,the series combination of $L$ and $C$ is in a state of resonance.The impedance of the $LC$ branch is $Z_{LC} = |X_L - X_C| = |5 - 5| = 0 \ \Omega$.Because the $LC$ branch has zero impedance,it acts as a short circuit.Therefore,the entire voltage of the source $(110 \ V)$ appears across the resistor $(R = 55 \ \Omega)$.The current $I$ flowing 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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