Question

Derive expression $P=V_{rms}{I}_{rms}cos\delta$ for an A.C. circuit.

Answer 1

Solution:-

Average power in LCR circuit:-

let $V=V_0\sin \omega tthenI=I_0\sin \omega t-\varphi$

where, $\varphi$ = phase angle

Total work done over complete cycle:-

$W={\int }_0^{T}VIdt$

$={\int }_0^T{V}_0\sin \omega t.I_0\sin \omega t-\varphi dt$

$=\frac{V_0{I}_0}{2}{\int }_0^{T}2\sin \omega t.\sin \omega t-\varphi dt$

Using this

$2sinA.sinB=\cos (A-B)-\cos (A+B)$

we get

W$=\frac{V_0{I}_0}{2}{\int }_0^T[\cos \varphi -cos(2\omega t-\varphi )]dt$

W$=\frac{V_0{I}_0}{2}\left[T\cos \varphi \right]$

$\frac{W}{T}=\frac{V_0}{\sqrt{2}}.\frac{I_0}{\sqrt{2}}.\cos \varphi$

P=$V_{rms}.I_{rms}.\cos \varphi$