The reading of the ammeter in the circuit sho | vedclass

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(A) In the given circuit,the capacitor $(X_C = 5 \ \Omega)$ and the inductor $(X_L = 5 \ \Omega)$ 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 capacitor $(X_C = 5 \ \Omega)$ and the inductor $(X_L = 5 \ \Omega)$ are connected in series with each other,and this combination is connected in parallel with the voltmeter. However,the entire series combination of the $LC$ circuit is in series with the resistor $(R = 55 \ \Omega)$ and the $AC$ source $(110 \ V)$.Since $X_L = X_C = 5 \ \Omega$,the net reactance of the $LC$ branch is $X = |X_L - X_C| = |5 - 5| = 0 \ \Omega$.This means the $LC$ branch acts as a short circuit (zero impedance) at resonance.The total impedance of the circuit is simply the resistance $R = 55 \ \Omega$.Using Ohm's law for the $AC$ circuit,the current $I$ is given by $I = \frac{V}{R} = \frac{110 \ V}{55 \ \Omega} = 2 \ A$.Therefore,the ammeter reading is $2 \ A$.

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

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