# Current through inductor as function of time

1. Mar 1, 2014

### Nick O

1. The problem statement, all variables and given/known data
The L=100 mH inductor in the following figure has an initial current of Io=10 mA. If the voltage is, v(t)=1e−10t+2e−5t V, what is the current, i(t), through the inductor?

Express your answer as a function of time with units of mA.

The figure shows an independent voltage source connected to an inductor.

2. Relevant equations

v = L(di/dt)
i dt = (v dt)/L
$i = \int_{t_0}^{t} v dt + i_0$

3. The attempt at a solution

I solved the following equation:

$i(t) = \frac{1}{0.1 H}(\int_{t_0}^{t} (e^{-10T}+2e^{-5T}) dT)*\frac{10^3 mA}{1 A} + 10 mA$

and obtained the following:

$i(t) = (e^{-0.1} + 4e^{-0.05} - e^{-10t} - e^{-5t})*10^3 + 10 mA$

My homework software rejects this answer, saying that my "answer either contains an incorrect additive numerical constant or is missing one."

I can think of nothing that I might have omitted, and I know that this software is very picky. For example, when rounding, it rejects the "round 5 to even" rule that was instilled in me early on as completely incorrect, always expecting 5 to be rounded up even when followed only by zeroes. Given how picky the software is, I am not altogether convinced that my answer is actually wrong.

Does anyone see any obvious oversights in my work?

2. Mar 1, 2014

### SammyS

Staff Emeritus
What did you use for t0, and why are you not simply using an indefinite integral, then applying the initial condition?

3. Mar 1, 2014

### Nick O

That's a very good question, and tells me that I need to get some sleep. Somehow, my $t_0$ ended up being 0.01 (that is, $i_0$). What nonsense!

Thanks!