AC Voltage Across a Pure Resistor

Identify V₀ and R. Apply I₀ = V0/R. Write current with the same phase as voltage. Convert peak values to RMS only when the question asks for RMS or average power. State phi = 0 for a pure resistor.

Useful diagrams include the simple AC source-resistor circuit, sinusoidal V-t and I-t graphs on the same axes, and a phasor diagram with V and I along the same line. Labels should include V, I, R, time axis, amplitude, and phase difference zero.

Writing I = I₀ sin(omega t - pi/2) is wrong because that lag occurs for a pure inductor, not for a pure resistor.

When a sinusoidal AC voltage is applied across a pure resistor, the current is also sinusoidal and remains in phase with the applied voltage.

When a sinusoidal AC voltage is applied across a pure resistor, the current is also sinusoidal and remains in phase with the applied voltage. Instantaneous voltage: V = V₀ sin omega t. Instantaneous current: I = I₀ sin omega t, where I₀ = V0/R. RMS values: Vrms = V0/sqrt(2), Irms = I0/sqrt(2). R is resistance in ohm, V in volt, I in ampere, and omega in rad s^-1. Use these relations only for a pure resistor with negligible inductance and capacitance. If a 220 V RMS AC supply is connected to a 110 ohm resistor, the RMS current is I = V/R = 220/110 = 2 A. Voltage and current reach their peak values together.

For a resistor, Ohm's law applies at every instant. If the applied voltage is V = V₀ sin omega t, then the instantaneous current is I = V/R = (V0/R) sin omega t = I₀ sin omega t. Since voltage and current have the same sine factor, their maxima, minima, and zero values occur at the same instant. The phase difference between voltage and current is zero. RMS values are used because AC changes continuously; Vrms = V0/sqrt(2), Irms = I0/sqrt(2), and Vrms = Irms R. Instantaneous voltage: V = V₀ sin omega t. Instantaneous current: I = I₀ sin omega t, where I₀ = V0/R. RMS values: Vrms = V0/sqrt(2), Irms = I0/sqrt(2). R is resistance in ohm, V in volt, I in ampere, and omega in rad s^-1. Use these relations only for a pure resistor with negligible inductance and capacitance. A 50 ohm resistor is connected to V = 100 sin(314t) V. Here V₀ = 100 V and R = 50 ohm. I₀ = V0/R = 100/50 = 2 A. Therefore I = 2 sin(314t) A. Vrms = 100/sqrt(2) = 70.7 V and Irms = 2/sqrt(2) = 1.41 A. The current is in phase with voltage. Questions usually ask for the expression of current, RMS value, phase difference, average power, or a graph/phasor showing V and I in a resistor. Writing I = I₀ sin(omega t - pi/2) is wrong because that lag occurs for a pure inductor, not for a pure resistor.