Question

In the circuit shown below, the maximum value of current which passes through inductor will be

• A. 12.40 A
• B. 5.42 A
• C. 18.31 A
• D. 3.20 A

Answer 1

The correct option is C 18.31 A

Diode $\to$ ON at, $\omega t=\frac{-\pi }{2}$
By applying KVL in circuit,

$230\cos \omega t=L\frac{di\left(t\right)}{dt}$

$\frac{di\left(t\right)}{dt}=\frac{230}{L}\cos \omega t$

$di\left(t\right)=\frac{230}{L}\cos \omega tdt$

$\int di\left(t\right)=\int \frac{230}{L}\cos \omega tdt$

$i\left(t\right)=\frac{230}{\omega L}\sin \omega t+K$

$at\omega t=\frac{-\pi }{2},i\left(t\right)=0,$

$0=\frac{V_m}{\omega L}\sin \left(\frac{-\pi }{2}\right)+K$

$K=\frac{V_m}{\omega L}$

$i=\frac{V_m}{\omega L}\sin \omega t+\frac{V_m}{\omega L}$

$i_{peak}at\frac{\pi }{2}=\frac{2V_m}{\omega L}=\frac{2\times 230}{314\times 80\times {10}^{-3}}=18.31A$