Concept:In phasor diagrams, phasors rotate counterclockwise. A phasor positioned \(90^\circ\) ahead counterclockwise leads the other by \(90^\circ\).
Formula / Reaction:$$i(t) = I_0 \sin\left(\omega t + \frac{\pi}{2}\right), \quad v(t) = V_0 \sin(\omega t)$$
Solution:- The current phasor \(\vec{I}\) points along \(+y\) (\(90^\circ\)) while the voltage phasor \(\vec{V}\) points along \(+x\) (\(0^\circ\)).
- This shows that current leads voltage by \(90^\circ\) (or \(\pi/2\text{ rad}\)).
- This is the defining characteristic of a purely capacitive AC circuit.
Why other options are incorrect:- Opt_A: In a pure inductor, voltage leads current by \(90^\circ\) (current lags along \(-y\)).
- Opt_B: In a pure resistor, \(\vec{V}\) and \(\vec{I}\) are collinear (in phase, \(\phi = 0^\circ\)).
- Opt_D: DC does not have rotating phasor representations with phase angles.
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