# Instantaneous current after battery unplugged in RL circuit?

I've been racking my brain over this, and I can't find any clues in my textbook as to how to approach it.

I have the following circuit:

My goal is to find R such that, right after the switch is unplugged, the voltage between A and B is no more than 80V

I can easily apply Kirchoff's rules to find the currents after the switch has been closed a long time:

$$I_1- I_2 - I_3 = 0$$

$$12 - RI_3 = 0$$

$$10 + 7.5I_2 - RI_3 = 0$$

The result is:

$$I_3 = \frac{12}{R}$$

$$I_2 = \frac{4}{15}$$

$$I_1 = \frac{4R + 180}{15R}$$

Now, the switch is thrown open. The new circuit is described by a single loop. The thing I don't understand is the fact that $I_2$ is different than $I_3$, and yet the single loop must have a single constant current when the switch is thrown open. I don't know how to go about finding this new current. Furthermore, I would have to write down Kirchoff's loop rule for the new circuit, and that would require knowing the emf generated by the inductor, which would require $\frac{dI}{dt}$, which I also wouldn't quite know how to determine at the first instant.

Any guidance on this problem would be MUCH appreciated, I would really like to understand it and my textbook doesn't provide much to go on =\

Thanks!

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Because of the inductor, $i_2$ must be continuous across the switching time.

You've already calculated that $i_2(0-) = \frac{4}{15}A$ so, knowing that $i_2$ is continuous, you also have $i_2(0+) = \frac{4}{15}A$.

Now, since there is just a single loop after the switch opens, we have $i_3 = - i_2, t > 0$

This is all you need to complete the problem.

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Furthermore, I would have to write down Kirchoff's loop rule for the new circuit, and that would require knowing the emf generated by the inductor, which would require $\frac{dI}{dt}$, which I also wouldn't quite know how to determine at the first instant.
Think about this: in order to write down Kirchoff's loop rule for the original circuit in the late-time limit, you had to know the emf generated by the resistors. (Agreed?) And doing this requires $I_2$ and $I_3$. Did you have to know the values of $I_2$ and $I_3$ to write the loop rule? How did you deal with not having specific numeric values for $I_2$ and $I_3$? You can deal with not having $\frac{\mathrm{d}I_2}{\mathrm{d}t}$ the same way.
With regards to your first point, are you saying that the new loop will have current everywhere equal to $I_2$ after the switch is thrown open? Why $I_2$ and not $I_3$? Is it because the current "being lagged" is the one going through the inductor, and that's the only current remaining when the switch is thrown. Do I understand that correctly? – cemulate Nov 1 '12 at 19:24