The reading of the ammeter in the circuit is | vedclass

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(A) From the given circuit diagram, the capacitor $(X_C = 2 \, \Omega)$ and the inductor $(X_L = 2 \, \Omega)$ are connected in series with each other

Vedclass · Accepted Answer

$2$

Vedclass · Answer

(A) From the given circuit diagram, the capacitor $(X_C = 2 \, \Omega)$ and the inductor $(X_L = 2 \, \Omega)$ are connected in series with each other, and this combination is connected in parallel with the voltmeter. However, the ammeter is in series with the resistor $(R = 55 \, \Omega)$ and the $AC$ source $(110 \, V)$.Since the capacitor and inductor are in series, their net reactance is $X = X_L - X_C = 2 \, \Omega - 2 \, \Omega = 0 \, \Omega$.This means the branch containing the capacitor and inductor acts as a short circuit (zero impedance) for the $AC$ source.Therefore, the entire voltage of the source $(110 \, V)$ appears across the resistor $(R = 55 \, \Omega)$.The current $I$ measured by 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 is ............ $A$.

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