Question

The voltage applied to the primary winding of an unloaded single phase transformer is given by
$v=400\cos \omega t+100\cos 3\omega t$
The primary has 500 turns and frequency of the fundamental component of applied voltage is 50 Hz. The maximum value of fundamental component flux is _______mWb.

• A. 2.546

Answer 1

The correct option is A 2.546
$\text{The applied voltage,}v=-e=N\frac{d\varphi }{dt}$
$\therefore$ $d\varphi =\frac{1}{N}\left(vdt\right)$

$\varphi =\frac{1}{N}\int vdt$

$\text{But}$ $v=400\cos \omega t+100\cos 3\omega t$
$\therefore \varphi =\frac{1}{500}\int (400\cos \omega t+100\cos 3\omega t)dt$

$=\frac{1}{5(2\pi \times 50)}[4\sin \omega t+\frac{1}{3}\sin 3\omega t]Wb$

$=\frac{1000}{500\pi }[4\sin \omega t+\frac{1}{3}\sin 3\omega t]mWb$

$\therefore \text{Maximum value}=\frac{1000\times 4}{500\pi }=2.546mWb$