Q.The total impedance of a circuit decreases when a capacitor is added in series with L and R. Explain why.
In a series L-R circuit (before the capacitor is added), the impedance is . When a capacitor is now added in series, the voltage phasors across L and across C are exactly out of phase with each other (one leads the current by , the other lags it by ), so instead of adding, their magnitudes SUBTRACT: the new impedance becomes . As long as is smaller than (or even if it slightly exceeds it, up to the point where the reduction in the reactive term outweighs any other change), the quantity is smaller than alone, so : the overall impedance decreases. Physically, the capacitor's voltage drop works to CANCEL part of the inductor's voltage drop (since they are exactly out of phase), so less net reactive voltage needs to be supplied by the source for the same current, which is exactly why adding C reduces the impedance. This cancellation is complete (impedance falls all the way to its minimum possible value, Z = R) exactly when , the resonance condition. [!ANSWER] Adding C in series reduces impedance because the capacitor's voltage is out of phase with the inductor's voltage, so the net reactance is smaller in magnitude than alone, shrinking .
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