Find the impedance of the circuit in Omega. | vedclass

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(C) The circuit consists of an inductor and a capacitor connected in parallel across an $AC$ source of $90 \; V$.For the inductor branch,the current i

Vedclass · Accepted Answer

$60$

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(C) The circuit consists of an inductor and a capacitor connected in parallel across an $AC$ source of $90 \; V$.For the inductor branch,the current is $I_L = \frac{V}{X_L} = \frac{90}{30} = 3 \; A$.For the capacitor branch,the current is $I_C = \frac{V}{X_C} = \frac{90}{20} = 4.5 \; A$.Since the currents in the inductor and capacitor are $180^{\circ}$ out of phase,the total current $I$ drawn from the source is $I = |I_C - I_L| = |4.5 - 3| = 1.5 \; A$.The impedance $Z$ of the parallel circuit is given by $Z = \frac{V}{I} = \frac{90}{1.5} = 60 \; \Omega$.

Find the impedance of the circuit in $\Omega$.

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