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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:

enter image description here

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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2 Answers 2

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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Let me address your points individually:

The thing I don't understand is the fact that I2 is different than I3, and yet the single loop must have a single constant current when the switch is thrown open.

Good question. If there were no inductor, the current would instantly change at the moment the switch is opened. But the inductor slows down that change in current so that it happens gradually, not instantaneously. So at the moment the switch is opened, the current through the inductor is the same as it was with the switch closed.

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.

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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? –  Chase Meadors Nov 1 '12 at 19:24
1  
I'm not sure if "being lagged" is the best way to describe it, but I think you may have the right idea. As I said, the inductor prevents the current through itself from changing instantaneously. So it is the current through the inductor that is forced to stay the same. Outside the inductor, there is no obstacle to the current changing instantaneously. –  David Z Nov 1 '12 at 19:26

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