• Tags: resistor, in-phase, power
• Difficulty: Foundation

When V = $V_0$ sin(omegat) is applied across a pure resistor R, the current I = $\frac{V}{R}$ = $\frac{V_0}{R}$ sin(omegat) = $I_0$ sin(omegat). Voltage and current are in phase (phi = 0). The instantaneous power p = VI = $V_0$$I_0$$sin^2$(omegat) oscillates between 0 and $V_0$$I_0$. Average power P = $V_{rms}$$I_{rms}$ = $V_0$$I_0$/2 = $I_{rms}^2$*R = $V_{rms}^2$/R. The power is always positive — a resistor always absorbs energy. The phasor diagram shows $V_R$ and I along the same direction. The impedance of a pure resistor is simply R, independent of frequency.