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I'm having trouble solving an RL circuit using the Laplace Transform.

There's just a 2H inductor in series with a 5M ohm resistor. The inductor is initially charged to 1.25A.

So... Here's what I've done: $$ v_L(t) = v_R(t)\\ 2i_L'(t) = 5Mi_L(t) $$ Now taking the Laplace Transform, $$ 2(I_L(s)s - 1.25) = 5M I_L(s)\\ 2.5 = 2I_L(s)s - 5M I_L(s)\\ 2.5 = I_L(s)(2s - 5M)\\ I_L(s) = \frac{2.5}{2s-5M}\\ I_L(s) = \frac{1.25}{s-2.5M} $$ Now taking the inverse Laplace transform, $$ i_L(t) = 1.25e^{2.5Mt} $$ This answer is wrong though, and there should be negative $$ i_L(t) = 1.25e^{-2.5Mt} $$

For the life of me, I can't work out where that negative comes from in the maths.

I'd really appreciate some help.


Oh, I should add that the $M$ is Mega (so $10^6$).

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For the system you wrote at the beginning your solution is correct. –  Artem Jun 7 '12 at 5:15

1 Answer 1

up vote 1 down vote accepted

The problem is with your conventions.

Either take $v_R = v_L$, in which case $i_R = -i_L$. This would be the usual convention (ie, current 'enters' the device through the $+$ terminal).

Or you could take $i_R = i_L$, in which case you will need to take $v_R = -v_L$.

In your problem above, you have, in effect, taken a negative resistance (or a negative inductance, of course). It might help to draw the circuit graph when conventions are not obvious.

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