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In an AC circuit with a pure INDUCTOR, the current
Aleads voltage by 90 degrees
Bis in phase with voltage
Cleads voltage by 180 degrees
Dlags voltage by 90 degrees
Answer & Solution
Correct answer: D. lags voltage by 90 degrees
1. Inductor: $v_L = L\,di/dt$.
2. If $i(t) = I_m\sin(\omega t)$, then $v_L(t) = \omega L I_m \cos(\omega t) = \omega L I_m \sin(\omega t + 90^\circ)$.
3. So VOLTAGE LEADS CURRENT by 90° — equivalently, current LAGS voltage by 90°.
4. Mnemonic: 'ELI' — in an inductor (L), voltage (E) precedes current (I).
5. Option A is the CAPACITOR behavior. Options B, C are wrong.
_Source: Tony Kuphaldt, "Lessons in Electric Circuits — AC", Vol II, Ch 3 (Inductor phase)._
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