# $RL$ Circuit current derivation

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:

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?

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.

• 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. – relayman357 May 1 at 17:01

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.