Question

Derive an expression for average power of an AC (alternating current) circuit.

Answer 1

Hint: Use values of potential difference and current of an AC circuit and find average work first then find power from it.

Compete step by step solution:
Average power in an AC circuit:
Let V be the alternating potential difference in the AC circuit given as: -
${\text{V = }}{{\text{V}}_{\text{o}}}{\text{sin}}\omega {\text{t}}$ ….. (i)

Then the AC current developed will lag by a phase angle $\phi$ then,
${\text{I = }}{{\text{I}}_{\text{o}}}{\text{sin}}\left( {\omega {\text{t - }}\phi } \right)$ ….. (ii)

Now by using equations (i) and (ii) we will find the total work done over a complete cycle of AC
${\text{W = }}\int\limits_{\text{0}}^{\text{T}} {{\text{VI}} \cdot dt}$ …. (iii)

Put value of V and I from equations (i) and (ii) in the equation (iii)
$\therefore {\text{W = }}\int\limits_{\text{0}}^{\text{T}} {{{\text{V}}_{\text{o}}}} {\text{sin}}\omega t \cdot {{\text{I}}_{\text{o}}}{\text{sin}}\left( {\omega t{\text{ - }}\phi } \right) \cdot dt$
${\text{ = }}{{\text{V}}_{\text{o}}}{{\text{I}}_{\text{o}}}\int\limits_{\text{0}}^{\text{T}} {{\text{sin}}\omega t{\text{sin}}\left( {\omega t{\text{ - }}\phi } \right)} \cdot dt$
${\text{ = }}\dfrac{{{{\text{V}}_{\text{o}}}{{\text{I}}_{\text{o}}}}}{{\text{2}}}\int\limits_0^T {2\sin \omega t\sin \left( {\omega t - \phi } \right)} \cdot dt$
${\text{ = }}\dfrac{{{{\text{V}}_{\text{o}}}{{\text{I}}_{\text{o}}}}}{{\text{2}}}\int\limits_0^T {\cos \left( {\omega t - \omega t + \phi } \right)} - \cos \left( {\omega t + \omega t - \phi } \right) \cdot dt$ $\left[ {\because 2\sin {\text{A}}\sin {\text{B = }}\cos \left( {{\text{A - B}}} \right) - \cos \left( {{\text{A + B}}} \right)} \right]$

Thus,
${\text{W = }}\dfrac{{{{\text{V}}_{\text{o}}}{{\text{I}}_{\text{o}}}}}{{\text{2}}}\left[ {t\cos \phi - \sin \dfrac{{\left( {2\omega t - \phi } \right)}}{{2\omega }}} \right]_0^T$
${\text{W = }}\dfrac{{{{\text{V}}_{\text{o}}}{{\text{I}}_{\text{o}}}}}{{\text{2}}}\left[ {{\text{T}}\cos \phi } \right]$

So, ${\text{W = }}\dfrac{{{{\text{V}}_{\text{o}}}{{\text{I}}_{\text{o}}}}}{{\text{2}}} \cdot \cos \phi \cdot {\text{T}}$

$\therefore$ Average power in AC circuit over a complete cycle is given by: -
${\text{P = }}\dfrac{{\text{W}}}{{\text{T}}}$
${\text{ = }}\dfrac{{{{\text{V}}_{\text{o}}}{{\text{I}}_{\text{o}}}}}{{\text{2}}}\cos \phi$
${\text{ = }}\dfrac{{{{\text{V}}_{\text{o}}}}}{{\sqrt {\text{2}} }}\dfrac{{{{\text{I}}_{\text{o}}}}}{{\sqrt {\text{2}} }}\cos \phi$

$\therefore {\text{P = }}{{\text{V}}_{{\text{rms}}}}{{\text{I}}_{{\text{rms}}}}\cos \phi$

Note: Start a solution with the basic way to find the average work done by using its simple formula and be careful while using the properties and average value of trigonometric functions. At the end just convert work done into power.