How can I accurately calculate the power absorbed in a circuit?

[karaandnick]

I wish someone could tell me what I am doing wrong. I have tried using P=VI then integrating P from t1 to t2 where I get 225uJ but it keeps telling me I am wrong. Seems like a rather simple problem.

Homework Equations

The Attempt at a Solution

 

[berkeman]

I wish someone could tell me what I am doing wrong. I have tried using P=VI then integrating P from t1 to t2 where I get 225uJ but it keeps telling me I am wrong. Seems like a rather simple problem.

Homework Equations

The Attempt at a Solution

Welcome to the PF.

I think you left out a factor of 1/2 in the first part of the integration. I get half of your answer.

 

[uart]

I think he's out by a factor of 6.

If you normalize everything it just comes down to $\int_0^1 x(1-x) dx = 1/6$

Doing it properly with V, I and tr in the equation you get $$\frac{V \, I \, t_r}{6}$$

 

[berkeman]

I think he's out by a factor of 6.

If you normalize everything it just comes down to $\int_0^1 x(1-x) dx = 1/6$

Doing it properly with V, I and tr in the equation you get $$\frac{V \, I \, t_r}{6}$$

Interesting approach. Did you try it just the standard way, though? The number for t is so small that the t^3 term after the integration is negligible (if I did it right), and only the t^2 term contributes to the answer...

 

[karaandnick]

I ended up figuring it out. I decided to write equations for the slopes of each line. So

for the voltage

-30/250x + 30

For current

30/250x
$\int$(-30/250x+30)(30/250x)dx from 0 to 250

Then multiply by 10^-3 to get in micro jewel

Comes to 37.5uJ

Crazy little problem but I am ME so not used to this sort of thing.

 

[karaandnick]

I just noticed everyone's approach worked, geniuses. Thanks for everything, its nice to know there are still people out there willing to help others. I will definitely write these approaches down in my book for later use.