- #1
Saladsamurai
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Homework Statement
I am working on this transient problem. When the switch is open, current only flows in the right loop and so the initial current on the inductor is io+ = 4A flowinf downward through the inductor.
When the switch is thrown at t = 0, the other loop gets started. I reduced the circuit to a Thevenin equivalent as shown. Now, I know that I am supposed to be solving for the i(t) shown in the last diagram, but I know from the 6A current source in the second to last diagram that my the real current is flowing in the opposite direction. So instead, I will solve for the real current iR(t) flowing counter clockwise and then just negate it to yield the i(t) asked for.
Since I know the direction of the real current iR(t), I have labeled the polarities of my circuit elements accordingly. However, when I solve the loop equation:
1.
(12/5)*iR + 10*i'R = 72/5
using the Laplace transform, I get
2.
(12/5)*I(s)R + 10*sIR(s) - 10*4 = 72/5
3.
iR(t) = 6[1 - (1/3)*e-6t/25]
so that
4.
i(t) = 2*e-6t/25 - 6
But the answer in the book is
5.
i(t) = 10*e-6t/25 - 6
which I can rectify by messing with the signs in my loop equation. What am I doing wrong here?
Perhaps it has to do with the fact that my initial condition is a negative current with respect to iR, so instead of subtracting it in the Laplace domain (equation 2), I should have added it? Actually, I am quite confident that that is what it is. What do you think?
EDIT Solved: That was the mistake. Since the initial current through the inductor was 'downward,' it was directed opposite my assumed positive direction for iR(t). So when the Laplace calls for the IC, the sign has to be consistent.
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