# Help with a Schaum's Outline Problem

Discussion in 'Homework Help' started by Jake1234, Jul 22, 2008.

1. ### Jake1234 Thread Starter Member

Oct 14, 2007
19
0
Heh... This problem seems incredibly easy, but the answers they produce in the book seem wrong.

The problem is as follows,

A 10 uF capacitor discharges in an element such that its voltage is v = 2*e^-1000*t. Find the current and power delivered by the capacitor as a function of time.

Schaum's Outline Ans: i = 20*e^-1000*t mA
p = 40*e^-1000*t mW

I used the ICE equation to get the current delivered however when I took the derivative of the voltage function I ended up with this i = -20*e^-1000*t mA. My answer had a negative out in front because of the Chain Rule operation on the e^-1000*t portion.

Then when I computed power as v*i my answer was 40*e^-2000*t mW because multiplying two exponential functions results in adding the exponents.

I don't know if I'm making a sign error or what, but the answers Schaum's provides seem very wrong. If someone can verify my claim or assist in any errors I may have not seen I would appreciate it.

2. ### Ratch New Member

Mar 20, 2007
1,068
3
Jake1234,

Be appreciative.

v(t) = 2*e^-1000*t

v(0) = 2

Initial Energy in capacitor = (1/2)CE^2 = (1/2)*10E-6*(2^2) = 2E-5

E(t) = 2E-5 - (1/2)*10E-6*(2*e^-1000*t)^2 = (2E-5)*e^-2000t

P(t) = dE/dt = 0 - (2E-5)*-2000e^-2000 = 0.04e^-2000*t

i(t) = P(t)/v(t) = (0.04e^-2000*t)/2*e^-1000*t = .02e^-1000t

Ratch

3. ### Jake1234 Thread Starter Member

Oct 14, 2007
19
0
Thank you very much Ratch.

I was failing to see that their was initial energy in the capacitor at time t = 0.

However, this means that I was correct in saying the answers Schaum's Outline provided are wrong because the power function which should be

p(t) = 40*e^-2000*t mW

p(t) = 40*e^-1000*t mW?

4. ### Ratch New Member

Mar 20, 2007
1,068
3
Jake1234,

You are correct.

Ratch