In the adjoining figure,the impedance of the | vedclass

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(C) The circuit consists of an inductor with inductive reactance $X_L = 30 \, \Omega$ and a capacitor with capacitive reactance $X_C = 20 \, \Omega$ c

Vedclass · Accepted Answer

$60$

Vedclass · Answer

(C) The circuit consists of an inductor with inductive reactance $X_L = 30 \, \Omega$ and a capacitor with capacitive reactance $X_C = 20 \, \Omega$ connected in parallel to an $AC$ source of $V = 90 \, V$.The current through the inductor is $i_L = \frac{V}{X_L} = \frac{90}{30} = 3 \, A$.The current through the capacitor is $i_C = \frac{V}{X_C} = \frac{90}{20} = 4.5 \, A$.Since the current in the inductor lags the voltage by $90^\circ$ and the current in the capacitor leads the voltage by $90^\circ$,the currents are in opposite phases.The net current through the circuit is $i = |i_C - i_L| = |4.5 - 3| = 1.5 \, A$.The impedance of the circuit is $Z = \frac{V}{i} = \frac{90}{1.5} = 60 \, \Omega$.

In the adjoining figure,the impedance of the circuit will be $... \Omega$.

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