Current through inductor as function of time

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SUMMARY

The discussion focuses on calculating the current through a 100 mH inductor with an initial current of 10 mA, given the voltage function v(t) = 1e^(-10t) + 2e^(-5t) V. The participant derived the equation i(t) = (e^(-0.1) + 4e^(-0.05) - e^(-10t) - e^(-5t))*10^3 + 10 mA but faced rejection from homework software due to potential numerical constant errors. The participant questioned their use of the initial condition and the choice of t_0, ultimately realizing a mistake in setting t_0 to 0.01 instead of 0.

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Nick O
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Homework Statement


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.

inductor.PNG

Homework Equations



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

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?
 
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Nick O said:

Homework Statement


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.

View attachment 67149

Homework Equations



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

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

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