Question

A resistor of resistance $100\Omega$ is connected to an AC source $ϵ=\left(12V\right)sin\left(250\pi s^{-1}\right)t$.

Find the power consumed by the bulb.

Answer 1

Supply voltage $E=12\sin \left(250t\right)$

Peak voltage $V_o=12$ volts

Peak current $I_o=\frac{V_o}{R}=\frac{12}{100}=0.12\Omega$

Power consumed $P=\frac{V_o{I}_o}{R}=\frac{12\times 0.12}{100}=0.0144$watt