Power absorbed by multiple elements

[rude man]

Sweet so I found the power absorbed by the resistor, obviously did not need to for the inductor. But what about the power source I guess that's obvious because it supply's and does not absorb.

Right. Power supplies don't absorb power, they provide it.

Remeber from now on, U(-t) sets up the initial conditions and U(t) starts the input voltage (or whatever) at t = 0+. If the input has neiter U(t) nor U(-t) multiplying it then that input is there from t = -∞ to +∞. Like the 20V in part b.

Also, remember that the input can be time-shifted from t = 0: U(t-τ) means the input is zero until t = τ. And U[-(t - τ)] means the input is zero starting at t = -τ.

 

[DODGEVIPER13]

Hey man thanks for all your help I have the problem right now. I really need to work a ton of theses, so I can get used to them. I will add thanks for all your posts later as my ipad won't allow it for some reason.

 

[rude man]

At 200mS I got 494.6176. I got 400 from 20^2 times R which is 1. Obviously since you have questioned this its not correct

494.61W is correct for P(200 ms). The 400 number makes no sense. The idea is, you take your expression for i(t) for which you have worked so hard, plug in t = 0.2s, then square and multiply by 1 ohm. That's for t=200ms. For t=0 you obviously let t=0 in the same i(t) expression.

 

[DODGEVIPER13]

Yah 400 was my number for t=0 which I believe to be 625?

 

[rude man]

Yah 400 was my number for t=0 which I believe to be 625?

Right. P(0) = 25^2 = 625. Current at t=0 was 25A.

 

[rude man]

Hey man thanks for all your help I have the problem right now. I really need to work a ton of theses, so I can get used to them. I will add thanks for all your posts later as my ipad won't allow it for some reason.

No prob. Yer' welcome.

 

[DODGEVIPER13]

Heh whoops (b) is actually 20+5u(t) so that would mean at t=0- I would get I(-0)= 20 A and power would be 400 W it would. Also be 400 W at t= 0 and 0+. So at 200 mS it should be 20exp(-.8)+25(1-exp(-.8)) = 22.75 A then P( 200 mS)= 517.71 W.

 

[rude man]

Heh whoops (b) is actually 20+5u(t) so that would mean at t=0- I would get I(-0)= 20 A and power would be 400 W it would. Also be 400 W at t= 0 and 0+. So at 200 mS it should be 20exp(-.8)+25(1-exp(-.8)) = 22.75 A then P( 200 mS)= 517.71 W.

All correct! Good going.