Concept:Rewriting both quantities as sine functions reveals their relative phase angle. If voltage leads current by \(90^\circ\), the circuit is purely inductive.
Formula / Reaction:$$V = 3 \cos(\omega t) = 3 \sin\left(\omega t + \frac{\pi}{2}\right), \quad i = 4 \sin(\omega t)$$
Solution:- Comparing phases: Voltage phase is \(\omega t + \pi/2\) and current phase is \(\omega t\).
- Voltage leads current by \(\frac{\pi}{2}\) (\(90^\circ\)).
- A \(90^\circ\) voltage lead is the characteristic signature of an inductive circuit.
Why other options are incorrect:- Opt_A: In a capacitive circuit, current leads voltage by \(90^\circ\).
- Opt_C: In a resistive circuit, voltage and current are in-phase.
- Opt_D: The phase relationship uniquely identifies an inductive circuit.
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