AC Voltage Applied to a Pure Resistor

Setting up the circuit. A pure resistor $R$ (with no inductance or capacitance of its own) is connected directly across an AC source supplying $v=v_0\sin\omega t$.

Applying Ohm's law instantaneously. Ohm's law, $v=iR$, holds at EVERY instant for a resistor, not just for steady DC -- it makes no reference to how quickly $v$ might be changing. Substituting the source voltage directly gives the instantaneous current:

Voltage and current are in phase. Comparing this result with the applied voltage $v=v_0\sin\omega t$ shows that the current is ALSO a pure $\sin\omega t$ function, with exactly the SAME argument $\omega t$ -- there is no phase shift of any kind between them. Current and voltage rise to their peaks together, cross zero together, and reverse direction together; on a phasor diagram, their two phasors point in exactly the same direction at every instant, and on a waveform graph, the two sine curves rise and fall perfectly in step (both shown in the figure for this section). …