What is the charge on the capacitor?
change in V =q/c
The Attempt at a Solution
I want to use Kirchoff rule but I don't know if this is the right way to go.
When the capacitor is completely charges, there is no current in that branch. The voltage across the capacitor is the potential difference across 40 ohm resistor.
t = 2 μF.Ya, I was able to figure out that the charge on the resistor would be 80 microCoulombs . The voltage difference between the plates of the capacitor was 40 volts and from there it was plugging in. Now I face another problem where I can't find how long it will take the capacitor to discharge to 25% of 80 microCoulombs. I derived that it should be -RC*ln(1/4)=t. I plug in R= 50, c = 2 and get the wrong answer :/
Sorry.It is typo. I mean C = 2 μF. Your R is correct. You didn't mention your answer.How does t = 2 MICROFARADS? Don't you mean seconds or milliseconds? I don't quite understand your logic though. Doesn't the capacitor discharge through 2 resistors that become 50 ohms?