Chapter 33, Problem 58PQ

To determine

The expression for average power supplied.

Answer to Problem 58PQ

The expression for average power supplied is $P_{\text{avg}}={\epsilon }_{\text{rms}}\times I_{\text{rms}}\times \cos \phi$.

Explanation of Solution

Write the expression for average power.

    $P_{\text{avg}}=\frac{{\displaystyle {\int }_0^{\text{T}}dU}}{{\displaystyle {\int }_0^{\text{T}}dt}}$                                            (I)

Here, $P_{\text{avg}}$ is the average power, $U$ is the energy dissipated in time $t$ and $T$ is one complete time period.

Write the expression for current to the circuit connected to an alternating source of emf.

    $I=I_{\max }\sin \omega t$                                (II)

Here, $I$ is the current in the circuit, $I_{\max }$ is the peak value of the current and $\omega$ is the angular frequency.

Write the expression for emf in the circuit.

    $\epsilon ={\epsilon }_{\max }\sin \left(\omega t+\phi \right)$                       (III)

Here, ${\epsilon }_{\max }$ is the peak value of emf, $\epsilon$ is the induced emf, and $\phi$ is the phase difference.

Write the expression for energy dissipated in time $t$.

    $dU=\epsilon Idt$                      (IV)

Substitute ${\epsilon }_{\max }\sin \left(\omega t+\phi \right)$ for $\epsilon$ and $I_{\max }\sin \omega t$ for $I$ in the equation (IV) to solve for expression for $dU$.

    $\begin{array}{c}dU={\epsilon }_{\max }\sin \left(\omega t+\phi \right)I_{\max }\sin \omega tdt\\ ={\epsilon }_{\max }{I}_{\max }\left(\sin \left(\omega t+\phi \right)\sin \omega tdt\right)\\ ={\epsilon }_{\max }{I}_{\max }\left(\cos \left(\omega t-\omega t-\phi \right)+\cos \left(\omega t+\omega t+\phi \right)\right)dt\\ ={\epsilon }_{\max }{I}_{\max }\left(\cos \left(\phi \right)+\cos \left(2\omega t+\phi \right)\right)dt\end{array}$

Substitute ${\epsilon }_{\max }{I}_{\max }\left(\cos \left(\phi \right)+\cos \left(2\omega t+\phi \right)\right)dt$ for $dU$ in equation (I) to calculate the expression for $P_{\text{avg}}$.

    $P_{\text{avg}}=\frac{{\displaystyle {\int }_0^{\text{T}}{\epsilon }_{\max }{I}_{\max }\left(\cos \left(\phi \right)+\cos \left(2\omega t+\phi \right)\right)dt}}{{\displaystyle {\int }_0^{\text{T}}dt}}$

Integrate the  expression between the limits $0$ and $T$.

    $\begin{array}{c}{P}_{\text{avg}}=\frac{{\displaystyle {\int }_0^{\text{T}}\frac{{\epsilon }_{\max }{I}_{\max }}{2}\left(\cos \left(\phi \right)+\cos \left(2\omega t+\phi \right)\right)dt}}{{\displaystyle {\int }_0^{\text{T}}dt}}\\ =\frac{{\epsilon }_{\max }{I}_{\max }}{2}\left[\frac{\cos \left(\phi \right){\displaystyle {\int }_0^{\text{T}}dt}+{\displaystyle {\int }_0^{\text{T}}\cos \left(2\omega t+\phi \right)dt}}{{\displaystyle {\int }_0^{\text{T}}dt}}\right]\\ =\frac{{\epsilon }_{\max }{I}_{\max }}{2}\left[\frac{\cos \left(\phi \right){\left[t\right]}_0^{\text{T}}+\frac{1}{2\omega }{\left[\sin \left(2\omega t+\phi \right)\right]}_0^{\text{T}}}{{\left[t\right]}_0^{\text{T}}}\right]\\ =\frac{{\epsilon }_{\max }{I}_{\max }}{2}\left[\frac{\cos \left(\phi \right)T+\frac{1}{2\omega }\left[\sin \left(2\omega T+\phi \right)-\sin \phi \right]}{T}\right]\end{array}$                     (V)

Now,

    $\omega =\frac{2\pi }{T}$

Substitute, $\frac{2\pi }{T}$ for $\omega$ in the equation (V).

    $\begin{array}{c}{P}_{\text{avg}}=\frac{{\epsilon }_{\max }{I}_{\max }}{2}\left[\frac{\cos \left(\phi \right)T+\frac{1}{2\frac{2\text{π}}{T}}\left[\sin \left(2\frac{2\text{π}}{T}T+\phi \right)-\sin \phi \right]}{T}\right]\\ =\frac{{\epsilon }_{\max }{I}_{\max }}{2}\left[\frac{\cos \left(\phi \right)T+\frac{T}{4\text{π}}\left[\sin \left(4\text{π}+\phi \right)-\sin \phi \right]}{T}\right]\\ =\frac{{\epsilon }_{\max }{I}_{\max }}{2}\frac{T}{T}\left(\cos \left(\phi \right)+\frac{1}{4\text{π}}\left[\sin \left(\phi \right)-\sin \phi \right]\right)\\ =\frac{{\epsilon }_{\max }{I}_{\max }}{2}\cos \left(\phi \right)\end{array}$                            (VI)

The above expression can be rewritten as.

    $P_{\text{avg}}=\frac{{\epsilon }_{\max }}{\sqrt{2}}\times \frac{I_{\max }}{\sqrt{2}}\times \cos \phi$                                                                                    (VII)

Also, rms value of current is given by.

    $I_{\text{rms}}=\frac{I_{\max }}{\sqrt{2}}$

Here, $I_{\text{rms}}$ is the rms value of current.

And,

The rms value of emf is given by.

    ${\epsilon }_{\text{rms}}=\frac{{\epsilon }_{\max }}{\sqrt{2}}$

Here, ${\epsilon }_{\text{rms}}$ is the rms value of induced emf.

Substitute ${\epsilon }_{\text{rms}}$ for $\frac{{\epsilon }_{\max }}{\sqrt{2}}$ and $I_{\text{rms}}$ for $\frac{I_{\text{rms}}}{\sqrt{2}}$ in equation (VII).

    $P_{\text{avg}}={\epsilon }_{\text{rms}}{I}_{\text{rms}}\cos \phi$

Therefore, the expression for average power supplied is $P_{\text{avg}}={\epsilon }_{\text{rms}}{I}_{\text{rms}}\cos \phi$.