The reading of the ammeter in the circuit sho | vedclass

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(A) In the given circuit,the capacitor (with reactance $X_C = 5 \ \Omega$) and the inductor (with reactance $X_L = 5 \ \Omega$) are connected in serie

Vedclass · Accepted Answer

$2$

Vedclass · Answer

(A) In the given circuit,the capacitor (with reactance $X_C = 5 \ \Omega$) and the inductor (with reactance $X_L = 5 \ \Omega$) 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 branch containing the inductor and capacitor is in a state of series resonance. In a series $LC$ circuit at resonance,the net impedance $Z_{LC} = |X_L - X_C| = |5 \ \Omega - 5 \ \Omega| = 0 \ \Omega$. Therefore,the potential difference across the $LC$ combination is $V_{LC} = I 	imes Z_{LC} = I 	imes 0 = 0 \ V$. Since the voltmeter is connected across this $LC$ combination,it will read $0 \ V$. However,the ammeter is in series with the resistor $R = 55 \ \Omega$ and the $AC$ source of $110 \ V$. Since the $LC$ branch acts as a short circuit (zero impedance),the entire voltage of the source $110 \ V$ appears across the resistor $R$. Using Ohm's law,the current $I$ through the ammeter is $I = V / R = 110 \ V / 55 \ \Omega = 2 \ A$.

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

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