After the current in the circuit of Fig. P30.63 has reached its final, steady value with switch S1 closed and S2 open, switch S2 is closed, thus short-circuiting the inductor. (Switch S1 remains closed. See Problem 30.63 for numerical values of the circuit elements.) (a) Just after S2 is closed, what are Vac and Vcb , and what are the currents through R₀ , R, and S2 ? (b) A long time after S2 is closed, what are Vac and Vcb , and what are the currents through R₀ , R, and S2 ? (c) Derive expressions for the currents through R₀ , R, and S2 as functions of the time t that has elapsed
since S2 was closed. Your results should agree with part (a) when
t = 0 and with part (b) when$\mathit{t}\mathbf{\to }\mathbf{\infty }$. Graph these three currents
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.

$\mathbf{\sum }\mathit{V}\mathbf{=}\mathbf{0}$

Current in a R-L Circuit is given by

${\mathit{i}}_{\mathbf{R}}\left(t\right)\mathbf{=}\frac{\mathbf{\epsilon }}{{\mathbf{R}}_{\mathbf{t}\mathbf{o}\mathbf{t}\mathbf{a}\mathbf{l}}}\left(e^{-\frac{R_{total}}{L}}\right)$

Where $\mathit{\epsilon }$is the emf of the cell ${\mathit{R}}_{\mathbf{t}\mathbf{o}\mathbf{t}\mathbf{a}\mathbf{l}}$is the total resistance of circuit $\mathit{L}$ is the inductance of the circuit

So the current flows through the inductor after the switch S₂ is closed is$i=0.18A$and this current flows though the resistance R. Using Kirchhoff’s law we get

$\epsilon +V_L-iR-i_0{R}_0=0$

Plug the values of $i,\epsilon ,V_L,Rand{R}_0$into the equation to get value of$i_0$

$36V+\left(0.18A\right)\left(150\Omega \right)-\left(0.18A\right)\left(150\Omega \right)-i_0\left(50\Omega \right)=0$

Get

$i_0=0.72A$

Now get the voltage$V_{ac}$using ohm’s law

$$\begin{gathered} V_{ac}=\left(0.72A\right)\left(50\Omega \right) \\ {V}_{ac}=36V \end{gathered}$$

Just after the switch is closed the voltage across inductor is zero, therefore voltage for cb is zero

$V_{cb}=36V$

After a long time when the switch S₂ is closed the voltage across inductor is zero there for voltage between c and b is zero

$V_{cb}=0$

Use Kirchhoff’s loop rule

$$\begin{gathered} \epsilon -V_{ac}-V_{cb}=0 \\ 36V-V_{ac}-0=0 \\ {V}_{ac}=36V \end{gathered}$$

Now use Ohm’s law to get the current$i_0$

$$\begin{gathered} i_0=\frac{\epsilon }{R_0} \\ {i}_0=\frac{36V}{50\Omega } \\ {i}_0=0.72A \end{gathered}$$

The current through $R_0$the same for S₂

$I_{S2}=0.72A$

While the current is zero across R because it breaks the path

$I_R=0A$

We know, current in the R-L circuit is given by

$i_R\left(t\right)=i_0\left(e^{-\frac{R}{L}t}\right)$

Plug in the values we get

$i_R\left(t\right)=\left(0.18A\right)\left(e^{-12.5s^{-1}t}\right)$