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

When V = V_0 sin(omegat) is applied across a pure resistor R, the current I = V/R = (V_0/R) sin(omegat) = I_0 sin(omegat). Voltage and current are in phase (phi = 0). The instantaneous power p = VI = V_0I_0sin^2(omegat) oscillates between 0 and V_0I_0. Average power P = V_rmsI_rms = V_0I_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.