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I am having trouble with the following question:

"After the current in the circuit of Fig. P30.63 has reached its final, steady value with switch $S_1$ closed and $S_2$ opened, switch $S_2$ is closed, thus short-circuiting the inductor. (Switch $S_1$ remains closed. See Problem 30.63 for numerical values of the circuit elements.) Derive expressions for the currents through $R_0$, $R$, and $S_2$ as functions of the time $t$ has elapsed since $S_2$ was closed."

Here is the figure given in the question: enter image description here

I understood that the current through $R_0$ remains constant, but I am having trouble deriving the expression for the current through $R$. The answer seems to be $$i_R(t)=\frac{\mathcal{E}}{R_\text{eq}}e^{-(R/L)t} \quad,$$ but I don't see how to obtain this expression. Can someone please show me the derivation of this with steps and explanations?

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In the small loop, after $S_2$ is closed: iR = - L(di/dt). (Negative because, i, is decreasing with time.) Experience shows that this type of equation has a solution of the form: i =A$e^{αt}$. Find (di/dt) and solve the equation for α. The constant, A, will be the initial current.

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  • $\begingroup$ Agree with R.W., also, notice that the short between c and b by S2 effectively takes those elements right out of the rest of the circuit. That will help you find current through R0 easy enough. $\endgroup$ – relayman357 May 1 at 17:01
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Since this is a homework-like question, I will only give the steps, but leave out the calculational details.

Choose your time scale such that the switch $S_2$ closes when $t=0$. Consider first the situation when switch $S_2$ is still open, i.e. for $t_0\le 0$.

  • Because switch $S_2$ is open, $R_0$, $R$ and $L$ are in series and the currents through these are all the same.
  • Because the system is in steady state, the current through $R_0$, $R$ and $L$ doesn't change with time, and hence the voltage across $L$ is zero.
  • Because the resistors $R_0$ and $R$ are in series, you should know how to calculate their equivalent resistance $R_\text{eq}$ from $R_0$ and $R$.
  • From Ohm's law you can calculate the current by $$i_R=\frac{\mathcal{E}}{R_\text{eq}} \tag{1}$$

And now consider the situation after $S_2$ closes, i.e. for $t\ge 0$:

  • From Ohm's law the voltage accross resistor $R$ is $U_R(t)=Ri_R(t)$.
  • From the definition of inductance the voltage across inductor $L$ is $U_L(t)=L\frac{di_L(t)}{dt}$.
  • Because $R$ and $L$ are in series, the currents through these are the same: $i_R(t)=i_L(t)$.
  • Because switch $S_2$ is closed the voltage between $c$ and $b$ is zero: $U_R(t)+U_L(t)=0$.
  • Putting together the previous 4 facts, you will get a differential equation for $i_R(t)$.
  • From (1) you have the starting condition for $i_R(0)$.
  • Now find the solution $i_R(t)$ of the differential equation with the starting condition.
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