Question

Switch $S$ is closed at $t=0$, in the circuit shown. The change in flux in the inductor ($L=500mH$) from $t=0$ to an instant when it reaches steady state is :

• A. $2Wb$
• B. $1.5Wb$
• C. $0Wb$
• D. None of the above

Answer 1

The correct option is B $1.5Wb$

Before closing the switch, inductor acts as a short circuit as it is in steady state.

Hence, Initial current $I_i=\frac{10}{10}=1A$

Initial flux, ${\varphi }_i=L\left(I_i\right)=500mWb$

$=0.5Wb$

At steady state,

Final current $I_f=\frac{20}{5}=4A$
Final flux, ${\varphi }_f=L\left(I_f\right)=\left(0.5\right)\times 4=2Wb$

$\therefore$ $\Delta \varphi =1.5Wb$