Hello,
I had a homework question recently which went like this:
find i(t) for time>0 for the attached diagram
Now, I'm having a bit of trouble grasping the concept that when t<0 , the capacitor is considered a short circuuit while when t>0 its considered a normal capacitor. To solve the problem I obtained a circuit for t>0 and t<0. For t>0 i shorted the capacitor and had a series circuit with 24V, 6 ohms which gives 4 Amp current across the whole thing. The t<0 circuit gace a series circuit with the capacitor and 4,5 ohm resistors respectively. I obtained an answer of i(t)=4e^(-t/3)...my time constant is correct but the current value is not. The correct answer is:
1.778e^(-t/3). Any help would be greatly appreciated. Thanks!

* the switch is closed at t<0 and open at t>=0*

 

What did you calculate for the voltage across the capacitor at t=0?

With a 4A current flowing (or more, while the capacitor is charging), there will be an 8 volt drop across the 2 ohm resistor at t<0.

 

well to be honest i'm quite lost from then on. i mean i know i'm supposed to get values for t<0 and t>0, other than that i'm not too sure on what to do. i mean, i calculated the time constant of 3 correctly by doing T=R/C, therefore T=9/(1/3)=9 seconds.