Concept:In an AC circuit containing only a pure ohmic resistor, the opposition to current flow is constant and does not depend on time or rate of change of current.
Formula / Reaction:$$v = V_0 \sin(\omega t), \quad i = \frac{v}{R} = \frac{V_0}{R} \sin(\omega t) = I_0 \sin(\omega t)$$
Solution:- Both voltage and current waveforms cross zero, increase, and reach their positive and negative maximums at the exact same instants.
- The phase difference \(\phi = 0^\circ\), meaning current and voltage are strictly in-phase.
Why other options are incorrect:- Opt_A: Leading by one-half cycle corresponds to \(\phi = 180^\circ\).
- Opt_B: Leading by one-fourth cycle corresponds to \(\phi = 90^\circ\), which occurs in a pure capacitor.
- Opt_C: Leading by \(1.5\) cycles corresponds to \(540^\circ\).
Quality & Fidelity Assurance:
Every question on BeambePrep is rigorously curated against the official PMDC syllabus with zero filler, zero out-of-syllabus content, and zero typos. When an authentic past paper originally contained a historical mistake or ambiguity from the examining board (such as UHS or NUMS), BeambePrep faithfully reflects the original paper while detailing the nuance and scientific consensus in the autopsy above.