Question

An alternating voltage is given by $e=(6\sin \omega t+8\cos \omega t)$ $volt$. The peak value of voltage is given by:

Answer 1

$e=6\sin \omega t+8\cos \omega t$

$⟹e=6\cos (\omega t-90)+8\cos \omega t$

Taking maximum values

$⟹e=6\angle -90+8\angle 0$

$⟹e=6\cos 90-6j\sin 90+8$

$⟹e=8-6j$

$⟹e=10\angle -36.86$

$⟹e=10\cos (\omega t-36.86)$

So, maximum or peak value $=10$