In the circuit shown in Fig. P30.61, E = 60.0 V,R1 = 40.0 Ω, R2 = 25.0 Ω, and L = 0.300 H. (a) Switch S is closed. At some time t afterward, the current in the inductor is increasing at a rate of di>dt = 50.0 A>s. At this instant, what are the current i1 through R1 and the current i2 through R2 ? (Hint:Analyze two separate loops: one containing E and R1 and the other containing E, R2 , and L.) (b) After the switch has been closed a long time, it is opened again. Just after it is opened, what is the current through R1 ?

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.

$\sum V=0$

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

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

Now apply Ohm’s law to get $i_1$

$$\begin{gathered} i_1=\frac{\epsilon }{R_1} \\ {i}_1=\frac{60V}{40\Omega }=1.5A \end{gathered}$$

And we get$i_2$

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

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

$$\begin{gathered} i_2=\frac{\epsilon }{R_1} \\ {i}_2=\frac{60V}{25\Omega }=2.4A \end{gathered}$$

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 $i_1=i_2=2.4A$.