I would like some guidance on this problem that I have:
This is the circuit:
And the question is:
Note: typo in the question, closes should be closed.
So my take on the question is:
There is two cases, one t<0 and one t=0. In order for me to find I1 I'm thinking that I need to calculate Vc. So when t<0 the switch has been closed for a long time and the capacitor should act like an open circuit, like this:
So I'm having a bit of a difficulty to go on because of the two voltage sources. In my mind I'm thinking that when the capacitor acts like an open circuit then VA should not be considered. However I won't be able to calculate Vc by only having VB, R2 and R1.
So my difficulty is what happens when the switch is closed. How am I supposed to find I1 when the switch is closed?
Any suggestions would be kind!
Best Answer
Checking the voltage loops you have:
Loop 1
$$2V - I_{R2}R2-I_{R1}R1=0$$
Loop 2
$$1V - V_{cap} + I_{R1}R1=0$$
At t=0, there's no current flowing into the capacitor because it is already charged, meaning that \$I_{R2}=I_{R1}\$. Rewriting the first loop equation and solving for the current gives:
$$I_{R1}=\dfrac{2V}{R_1+R_2} = 500\mu A$$
Replacing the current in the second equation and solving for the capacitor voltage yields
$$V_{cap} = 1V + 500\mu A \cdot 1k\Omega=1.5V$$
So basically your capacitor will discharge from 1.5V to 1V. The discharging curve can be computed using:
$$V_{CAP, t}=V_{CAP,t=0}e^{\dfrac{-t}{1k\Omega \cdot 500pF}}$$
The rest you can figure out.
Here is a small simulation just to double check it.