A parallel combination of a pure inductor and | vedclass

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(A) In a parallel resonant circuit,the capacitive reactance $(X_{C})$ and inductive reactance $(X_{L})$ are equal $(X_{L} = X_{C})$.Since the inductor

Vedclass · Accepted Answer

$i = 0, i_{L} = i_{C} 
eq 0$

Vedclass · Answer

(A) In a parallel resonant circuit,the capacitive reactance $(X_{C})$ and inductive reactance $(X_{L})$ are equal $(X_{L} = X_{C})$.Since the inductor and capacitor are in parallel,the voltage across them is the same.Therefore,the magnitudes of the currents are equal: $i_{L} = i_{C} = \frac{e}{X_{L}} = \frac{e}{X_{C}}$.However,the current through the inductor lags the voltage by $90^{\circ}$,and the current through the capacitor leads the voltage by $90^{\circ}$.Thus,the currents $i_{L}$ and $i_{C}$ are $180^{\circ}$ out of phase with each other.The net current $i$ in the circuit is the phasor sum of $i_{L}$ and $i_{C}$,which is $i = |i_{L} - i_{C}| = 0$.Hence,the condition is $i = 0$ and $i_{L} = i_{C} 
eq 0$.

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