# Capacitor current problem

Discussion in 'Homework Help' started by peter_morley, Mar 27, 2011.

1. ### peter_morley Thread Starter Member

Mar 12, 2011
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I'm having trouble understanding how to find A1 and A2 in the attached problem. I did a bunch of work and got stuck. please help

Peter

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2. ### Georacer Moderator

Nov 25, 2009
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1,266
The derivative of
$A_1 t e^{-1500 t}$ is actually
$A_1 e^{-1500 t} + A_1 t (-1500) e^{-1500 t}$
so, you should correct that.

Also, for times before zero, the voltage is constant and as a result, the current of the capacitor is zero, not 90mA as you wrote.

Since the voltage on a capacitor is a continuous quantity, you can safely say that
$A_1 \cdot 0 \cdot e^{-1500 \cdot 0}+A_2 \cdot e^{1500 \cdot 0}=A_2=25V$.

I don't know, however what you could do for A_1.

3. ### t_n_k AAC Fanatic!

Mar 6, 2009
5,448
783
This is a rather cryptic problem.

A typical (if not unique) circuit which would satisfy the general solution would be a capacitor with an initial voltage [25V] discharging into a series R+L network. The effective circuit would be critically damped.

Under these conditions one would obtain the general solution for the capacitor voltage

Vc(t)=25(1+at)e^(-at)

We are given a=1500

Hence

Vc(t)=25e^(-1500t)+25x1500te^(-1500t)

So A2=25 and A1=25x1500=37,500.

Last edited: Mar 28, 2011
4. ### t_n_k AAC Fanatic!

Mar 6, 2009
5,448
783
This solution supports Georacer's comments about the initial capacitor current being zero. If there were a current at zero, the general solution for the cap voltage may not be satisfied - I've not checked whether this is in fact the case. It's probably beyond the original scope of the problem anyway.

Last edited: Mar 28, 2011