A voltage \(v={v}_{0}\sin \omega t\) applied to a circuit drives a current \(i={i}_{0}\sin (\omega t+ϕ)\) in the circuit…

Explanation

To find the average power consumed in the circuit, we begin with the instantaneous power, which is the product of the voltage

\(v(t)\)and the current\(i(t)\)

:

\[p(t)=v(t) \cdot i(t)=\left(v_0 \sin (\omega t)\right) \cdot\left(i_0 \sin (\omega t+\phi)\right)\] \[\sin A \sin B=\frac{1}{2}[\cos (A-B)-\cos (A+B)]\] \[\sin (\omega t) \sin (\omega t+\phi)=\frac{1}{2}[\cos (-\phi)-\cos (2 \omega t+\phi)]\] \[\cos (-\phi)=\cos (\phi)\]

, thus:

\[p(t)=v_0 i_0 \frac{1}{2}[\cos (\phi)-\cos (2 \omega t+\phi)]\]

To find the average power over a cycle, integrate this expression over one period

\[T=\frac{2 \pi}{\omega}\]

, then divide by

\[T\]

:

\[P_{\mathrm{avg}}=\frac{1}{T} \int_0^T v_0 i_0 \frac{1}{2}[\cos (\phi)-\cos (2 \omega t+\phi)] d t\]

This separates into two integrals:

\[P_{\mathrm{avg}}=\frac{v_0 i_0}{2 T}\left[\int_0^T \cos (\phi) d t-\int_0^T \cos (2 \omega t+\phi) d t\right]\] \[P_{\mathrm{avg}}=\frac{v_0 i_0}{2 T}\left[\int_0^T \cos (\phi) d t-\int_0^T \cos (2 \omega t+\phi) d t\right]\]