Consider the circuit shown in Fig. P30.63.Let E = 36.0 V, R0 = 50.0 Ω, R = 150 Ω, and L = 4.00 H.
(a) Switch S1 is closed and switch S2 is left open. Just after S1 is closed, what are the current i0 through R0 and the potential differences vac and vcb ? (b) After S1 has been closed a long time (S2 is still open) so that the current has reached its final, steady value, what are i₀ , Vac , and Vcb ? (c) Find the expressions for i₀ , Vac , and Vcb as functions of the time t since S1 was closed. Your results should agree with part (a) when t = 0 and with part (b) when$\mathit{t}\mathbf{\to }\mathbf{\infty }$. Graph i₀ , Vac ,and Vcb versus time.

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.

$\sum _{}V=0$

When the switch is just closed the initial current is zero i.e$i_0=0A$ and in this case the voltage across ab will be zero $V_{ab}=0$.

And as no voltage drop occurs therefore the voltage across bc is the same of the emf of the battery$V_{bc}=36V$

Hence the voltmeters read$V_{ab}=0,V_{bc}=36V$ and the current is $i_0=0$.

After a long time of closing of the switch, the voltage across the inductor is zero, so use the ohm’s law to get$i_0$.

$\begin{array}{r}{i}_0=\frac{\epsilon }{R_{eq}}\\ i_0=\frac{36\mathrm{V}}{(150\mathrm{\Omega }+30\mathrm{\Omega })}\\ i_0=0.18\mathrm{A}\end{array}$

Now use this value to get the voltage across$V_{ab}$ and$V_{bc}$ by

$V_{ab}=i_0{R}_0=\left(0.18\mathrm{A}\right)\left(50\mathrm{\Omega }\right)=9\mathrm{V}$

And

$V_{bc}=i_{0}R=(0.18A)\left(150\Omega \right)=27V$

Hence The voltmeters read$V_{ab}=9,V_{bc}=27V$ and the current is $i_0=0.18A$.

Use Kirchhoff’s rule in the bottom loop and get the equation

$\begin{array}{r}\epsilon -V_{ac}-V_{cb}=0\\ \epsilon -i_0{R}_0-i_{0}R-L\frac{di}{dt}=0\end{array}$

Rearranging

$\begin{array}{r}L\frac{di}{dt}=\epsilon -i_0\left(R_0+R\right)\\ \frac{di}{-i+\frac{\epsilon }{\left(R_0+R\right)}}=\frac{\left(R_0+R\right)}{L}dt\end{array}$

Integrate both sides

$\ln \left(\frac{-i+\frac{\epsilon }{\left(R_0+R\right)}}{\frac{\epsilon }{\left(R_0+R\right)}}\right)=-\frac{\left(R_0+R\right)}{L}dt$

Take exponent on both sides

$\frac{-i+\frac{\epsilon }{\left(R_0+R\right)}}{\frac{\epsilon }{\left(R_0+R\right)}}=e^{-\frac{\left(R_0+R\right)}{L}dt}$

Rearrange to get

$i_{0=}\frac{\epsilon }{\left(R_0+R\right)}\left(1-e^{-\frac{\left(R_0+R\right)}{L}t}\right)$

Plug in the values$R,R_0,Land\epsilon$ of to get

$i_0=0.180A\left(1-e^{-\frac{t}{0.02s}}\right)$

Hence the current is$i_0=0.180A\left(1-e^{-\frac{t}{0.02s}}\right)$