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Calculate the average power in an electrical circuit

by RoRi

Let e(t) and i(t) denote the current and voltage, respectively, in a circuit at time t. Define

    \[ e(t) = 160 \sin t, \qquad i(t) = 2 \sin \left( t - \frac{\pi}{6} \right). \]

Then, we define the average power by the integral equation

    \[ A = \frac{1}{T} \int_0^T e(t) i(t) \, dt, \]

where T is the period of the voltage and current. Find T and calculate the average power.

First, since the voltage is given by e(t) = 160 \sin t we know it is periodic with period 2 \pi, so T = 2 \pi. Then, we compute the average power,

    \begin{align*} A &= \frac{1}{2 \pi} \int_0^{2 \pi}320 \sin t \sin \left(t - \frac{\pi}{6} \right) \, dt \\ &= \frac{160}{\pi} \int_0^{2 \pi} \sin t\left( \sin t \cos \frac{\pi}{6} - \sin \frac{\pi}{6} \cos t \right) \, dt \\ &= \frac{160}{\pi} \left( \cos \frac{\pi}{6} \int_0^{2 \pi} \sin^2 t \, dt - \sin \frac{\pi}{6} \int_0^{2 \pi} \sin t \cos t \, dt \right) \\ &= \frac{160}{\pi} \left( \frac{\sqrt{3}}{2} \int_0^{2 \pi} (1 - \cos 2t) \, dt - \frac{1}{4} \int_0^{2 \pi} \sin 2t \, dt \right) \\ &= \frac{80 \sqrt{3}}{2 \pi} (2\pi - 0) - \frac{40}{\pi}\cdot 0 \\ &= 80 \sqrt{3}. \end{align*}

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