91Ó°ÊÓ

We have given that 100 mH inductor whose windings have a resistance of 4.0 Ω is connected across a 12 V battery having an internal resistance of 2.0 Ω.

To get the value of stored energy, we have to first calculate the value of current.

The current can be determined by using Ohm's law.

$I=\frac{∆V}{R}=\frac{{\mathrm{Î\mu }}_{battery}}{R_{inductor}+r}$

Here ${\mathrm{Î\mu }}_{battery}$= Battery emf

$R_{inductor=}$Resistance of inductor

$r$= internal resistance of battery

Substituting the values in formula :

localid="1651037129243" $I=\frac{12V}{(4.0\mathrm{Î\copyright }+2.0\mathrm{Î\copyright })}$

$I=2.0$A.

Now we will determine the stored energy.

The formula of stored energy is :

$U_L=\frac{1}{2}L{I}^2$

Substituting the values in formula :

localid="1651037260134" $U_L=\frac{1}{2}(100×{10}^{-3}H){(2.0A)}^2$

$U_L=0.20J$

Energy stored in inductor in 0.20 J.