Current through inductor as function of time

Click For Summary
The discussion centers on finding the current through a 100 mH inductor given an initial current of 10 mA and a specific voltage function. The user attempted to solve the equation using integration but received a rejection from their homework software, which indicated a potential error in their additive constant. They expressed frustration over the software's strict requirements, particularly regarding rounding rules. A suggestion was made to clarify the limits of integration and confirm the initial condition used in the calculations. The user acknowledged a mistake in their choice of t_0, realizing it should be 0 instead of 0.01.
Nick O
Messages
158
Reaction score
8

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?
 
Physics news on Phys.org
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?
 
  • Like
Likes 1 person
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!
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

Similar threads

  • · Replies 10 ·
Replies
10
Views
880
Replies
12
Views
2K
Replies
3
Views
2K
Replies
34
Views
7K
  • · Replies 19 ·
Replies
19
Views
3K
Replies
18
Views
3K
  • · Replies 3 ·
Replies
3
Views
1K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 2 ·
Replies
2
Views
4K
  • · Replies 7 ·
Replies
7
Views
2K