91Ó°ÊÓ

Ohm’s law states that the voltage across a conductor is directly proportional to the current flowing through it, provided all physical conditions and temperatures remain constant

$V=IR$

According to Kirchhoff’s Lawthe sum of the voltages around the closed loop is equal to null.

$∑V=0$

The voltage drop across the inductor is given by

$V_L=L\frac{di}{dt}$

Input the values here to get $V_L$

$$\begin{gathered} V_L=\left(0.3H\right)\left(50A/s\right) \\ {V}_L=15V \end{gathered}$$

Apply Kirchhoff’s Loop rule to get the voltage drop across the resistance

$$\begin{gathered} V_R=\mathrm{Î\mu }-V_L \\ {V}_R=60V-15V \\ {V}_R=45V \end{gathered}$$

Now apply Ohm’s law to get $i_1$

$$\begin{gathered} i_1=\frac{\mathrm{Î\mu }}{R_1} \\ {i}_1=\frac{60V}{40\mathrm{Î\copyright }}=1.5A \end{gathered}$$

And we get$i_2$

$$\begin{gathered} i_2=\frac{V_L}{R_2} \\ {i}_1=\frac{45V}{25\mathrm{Î\copyright }}=1.8A \end{gathered}$$

The current through $R_{1}and{R}_2$is $i_1=1.5A$and $i_2=1.8A$respectively.

When the switch is closed for long time the current passes through$R_2$and the inductor blocks the current. So,$i_2$will be

$$\begin{gathered} i_2=\frac{\mathrm{Î\mu }}{R_1} \\ {i}_2=\frac{60V}{25\mathrm{Î\copyright }}=2.4A \end{gathered}$$

And after the switch is open , current flows through$R_1$will be same of$i_2$

$i_1=i_2=2.4A$

Hence The current through $R_{1}and{R}_2$is $i_1=i_2=2.4A$.