# Deriving Voltage from the Inductor Equation

1. Homework Statement

L = 20mH = 20 x 10-3 H

i = 40 mA for t≤0
i = A1e-10,000t + A2e-40,000t A for t≥0

The voltage at t=0 is 28 V.

I have to find the equation for the voltage for t>0.
Then I have to find the time when power is zero.

2. Homework Equations

v(t) = L* di/dt
p(t) = L*i* di/dt

3. The Attempt at a Solution

I know that current through an inductor has to be continuous, so at t=0, I can say:
A1 + A2 = 40 x 10-3 A

I then found di/dt:
di/dt = -10,000A1e-10,000t - 40,000A2e-40,000t

Since v = L di/dt:
v = (20 x 10-3 H)*(-10,000A1e-10,000t - 40,000A2e-40,000t)
v = -200A1e-10,000t - 800A2e-40,000t

At t=0, v=28 V, so:
v(t=0) = -200A1 - 800A2 = 28 V

So I have the following two equations:
A1 + A2 = 40 x 10-3 A
-200A1 - 800A2 = 28 V

When I solve these using matrices on my calculator, I get:
A1 = 0.1
A2 = -0.06

When I plug those into the equation for v, I get:
v(t) = -20A1e-10,000t + 48A2e-40,000t

This answer is one of the multiple choice answers. However, the system tells me this answer is wrong and the correct answer is:
v(t) = 18A1e-10,000t + 10A2e-40,000t

I don't understand why my answer is wrong, so any explanation would be appreciated!
Once I understand that, I know that to find the time when power is zero, I just set p(t) = L*i* di/dt to zero and solve for t.

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kuruman
Homework Helper
Gold Member
I don't understand your solution or the solution that is reportedly correct. If you plug in numerical values for A1 and A2, the expression for v(t) should not have A1 and A2 in it. Actually $v=-L\frac{di}{dt}$ but that will only change the overall sign of v(t) without changing the values of the coefficients in front of the exponentials.

v(t) should not have A1 and A2 in it.
Yes, that's clearly a typo.
I like your answer better than theirs. But, I would suggest that you read the problem carefully and make sure you wrote it correctly at the start.

Yes, my last two equations for v(t) should not have A1 and A2 in them. Sorry for the typo.

I double checked that I copied everything over correctly and I can't find an obvious error.

kuruman
Homework Helper
Gold Member
The solution $V(t)=18e^{-10000t}+10e^{-40000t}$ is incompatible with the the initial current value of 40 mA. Assume that this solution is correct, then you can integrate to find $I(t)$ from$$I(t)=-\frac{1}{L}\int V(t)dt$$and finally calculate the initial current $I(0)$. I did that and found about 100 mA, but you should check it out for yourself. So it looks like that either the supposed answer is incorrect or the initial value of 40 mA is incorrect.

The solution $V(t)=18e^{-10000t}+10e^{-40000t}$ is incompatible with the the initial current value of 40 mA. Assume that this solution is correct, then you can integrate to find $I(t)$ from$$I(t)=-\frac{1}{L}\int V(t)dt$$and finally calculate the initial current $I(0)$. I did that and found about 100 mA, but you should check it out for yourself. So it looks like that either the supposed answer is incorrect or the initial value of 40 mA is incorrect.
I found the same, but then I remembered there is a constant added from the integration. The voltage tells you how the current changes, not what it's value is.
However, the values implied by the voltage equation for A1 & A2 do not give the correct initial current in the given current equation.

kuruman
Homework Helper
Gold Member
There's probably more to this than meets the eye because of the two different time constants. @Nicole D can you post the circuit diagram that goes with this and the original statement of the problem? Thanks.

This is a picture of the original problem statement. There wasn't a circuit diagram with it. It is possible that the given solution is wrong - I know other people are having trouble with this problem.

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kuruman
Homework Helper
Gold Member
Thank you for posting this. It looks like there are no hidden pitfalls. Frankly, I don't see how the supposedly correct solution is actually correct. However, I think your original solution is also incorrect. You should use $V(t)=-L\frac{di}{dt}$ to get the answer. I was wrong in post #2 when I claimed that the negative sign will just change the overall sign of $V(t)$. I also think you should bring to the attention of whoever assigned you this problem that there appears to be something wrong with it.

Thank you for posting this. It looks like there are no hidden pitfalls. Frankly, I don't see how the supposedly correct solution is actually correct. However, I think your original solution is also incorrect. You should use $V(t)=-L\frac{di}{dt}$ to get the answer. I was wrong in post #2 when I claimed that the negative sign will just change the overall sign of $V(t)$. I also think you should bring to the attention of whoever assigned you this problem that there appears to be something wrong with it.
"passive sign convention" is important here. It means that a positive inductor voltage increases the inductor current. $V(t)=L\frac{di}{dt}$

NascentOxygen
Staff Emeritus