Help with a Schaum's Outline Problem

Thread Starter

Jake1234

Joined Oct 14, 2007
19
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.
 

Ratch

Joined Mar 20, 2007
1,070
Jake1234,

If someone can verify my claim or assist in any errors I may have not seen I would appreciate it.
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
 

Thread Starter

Jake1234

Joined Oct 14, 2007
19
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

instead of,

p(t) = 40*e^-1000*t mW?
 
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