In Figure, $\mathit{\epsilon }\mathbf{=}\mathbf{100}\mathit{V}\mathbf{,}\mathbf{}{\mathit{R}}_{\mathbf{1}}\mathbf{=}\mathbf{10}\mathbf{.}\mathbf{0}\mathbf{\Omega }\mathbf{,}\mathbf{}{\mathit{R}}_{\mathbf{2}}\mathbf{=}\mathbf{20}\mathbf{.}\mathbf{0}\mathbf{\Omega }\mathbf{,}\mathbf{}{\mathit{R}}_{\mathbf{3}}\mathbf{=}\mathbf{30}\mathbf{.}\mathbf{0}\mathbf{\Omega }\mathbf{,}\mathbf{}\mathbf{and}\mathbf{}\mathit{L}\mathbf{=}\mathbf{2}\mathbf{.}\mathbf{00}\mathbf{}\mathbf{H}$.Immediately after switch S is closed, (a) what is i₁? (b) what is i₂? (Let currents in the indicated directions have positive values and currents in the opposite directions have negative values.) A long time later, (c) what is i₁? (d) what is i₂? The switch is then reopened. Just then, (e) what is i₁? (f) what is i₂? A long time later, (g) what is i₁? (h) what is i₂?

1. Voltage is$\epsilon =100\mathrm{V}$
2. Resistance$R_1=10.0\mathrm{\Omega }$
3. Resistance$R_2=20.0\mathrm{\Omega }$
4. Resistance$R_3=30.0\mathrm{\Omega }$
5. Inductance is$L=2.00\mathrm{H}$
6. Figure 30-62 is the RL circuit

We use the concept of Ohm’s law and Kirchhoff’s voltage law. Using the equations, we can find the currents through resistors. For part c), we can write two equations from Kirchhoff’s voltage law, and by solving them simultaneously, we can find the current.

Formulae:

$i=\frac{\epsilon }{R}$

Current$i_1$immediately after the switch is closed:

When switch is closed, immediately the current through inductor is zero. We can write

$$\begin{gathered} i_1=\frac{\epsilon }{R_1+R_2} \\ {i}_1=\frac{100}{10.0+20.0} \\ {i}_1=3.33A \end{gathered}$$

Current$i_2$immediately after the switch is closed:

Fromthecircuit diagram, we can see that the current through$R_2$is also same as$R_1$

$i_2=3.33A$

Current $i_1$a long time later:

After a long time, the current will become constant, so the emf across the inductor becomes zero. We can write the current through resistance $R_3$as

$i_3=i_1-i_2$

We can apply Kirchhoff’s voltage law:

$$\begin{gathered} \epsilon -i_1{R}_1-i_2{R}_2=0.............................................................\left(1\right) \\ \epsilon -i_1{R}_1-\left(i_1-i_2\right)R_3=0......................................................\left(2\right) \end{gathered}$$

Rearranging equation (1) for we get

role="math" localid="1661850480735" $i_2=\frac{\left(\epsilon -i_1{R}_1\right)}{R_2}...................................................................\left(3\right)$

Plugging this value in equation (2), we get

$$\begin{gathered} \epsilon -i_1{R}_1-\left(i_1-\frac{\left(\epsilon -i_1{R}_1\right)}{R_2}\right)R_3=0 \\ \epsilon -i_1{R}_1-i_1{R}_3+\frac{\left(\epsilon -i_1{R}_1\right)}{R_2}{R}_3=0 \\ \epsilon -i_1{R}_1-i_1{R}_3+\epsilon \frac{R_3}{R_2}-\frac{i_1{R}_1}{R_2}{R}_3=0 \\ \epsilon \left(\frac{R_2+R_3}{R_2}\right)-\frac{i_1\left(R_1{R}_2+R_2{R}_3+R_1{R}_3\right)}{R_2}=0 \\ \epsilon \left(R_2+R_3\right)=i_1\left(R_1{R}_2+R_2{R}_3+R_1{R}_3\right) \\ {i}_1=\frac{\epsilon \left(R_2+R_3\right)}{\left(R_1{R}_2+R_2{R}_3+R_1{R}_3\right)} \end{gathered}$$

Plugging the values, we get

$$\begin{gathered} i_1=\frac{100\left(20.0+30.0\right)}{10.0\times 20.0+20.0\times 30.0+10.0\times 30.0} \\ {i}_1=\frac{5000}{1100} \\ {i}_1=4.55A \end{gathered}$$

Current$i_2$a long time later:

Plugging the value of$i_1$in equation (3), we get

$$\begin{gathered} i_2=\frac{\left(\epsilon -i_1{R}_1\right)}{R_2} \\ {i}_2=\frac{\left(100-4.55\left(10.0\right)\right)}{20.0} \\ {i}_2=2.73A \end{gathered}$$

When switch is reopened, the current through resistor will be zero.

$i_1=0A$

Because of inductor, the current through the resistor$R_3$changes slowly, so just astheswitch is opened, the current through it isthesame, that is,$i_3=i_1-i_2$

$i_3=4.55-2.73=1.82A$.

Fromtheright loop, the current through the resistor$R_2$will bethesame as$R_3$but in opposite direction. We get

$i_2=-1.82A$.

After a long time, later when switch is opened, the current$i_1=0A$

After long time later when switch is opened the current$i_2=0A$