1. The problem statement, all variables and given/known data 1) Determine the current in each resistor and charge on the capacitor after a long time immediately after the switch is closed 2) Determine the current in each resistor and charge on the capacitor after a long time After the switch has been closed for a long time it is re-opened. 3) Determine the current in each resistor and the charge on the capacitor immediately after the switch has been re-opened. 4) Determine the time constant of the capacitor discharge. 2. Relevant equations V = I Req Q = CV 3. The attempt at a solution I got #1 and got I6 = 1A, I8 = 1.5A, and I12 = .5A I'm not sure what to do about #2 though. What is the voltage across the capacitor? I thought it would be 6V, because that is the voltage across the parallel series. However, there is a voltage drop after the 1A current runs across the 6ohm resistor so the V across the capacitor is 0. For #3 and 4, does the 8ohm resistor come into play or no? Thanks for the help guys.
Is the V across the capacitor just 6V? If so I think I'm good. #2 I through the 6ohm resistor is 0, then I through the other two resistors is 18V/20ohms right? #3 I = 6V/18ohms Q= 28 E-6 * 6 #4 (18 * 28E-6) = RC
Voltage which charges the capacitor is the voltage across 12 ohm resistance. Calculate that voltage. When the capacitor is completely charged, the same voltage will be the voltage across it.