The Attempt at a Solution
so when t=0, the capacitor is a closed circuit.
The Req of the circuit is (20k)(30k)/(20k+30k)= 12000 ohms.
U = I Req = (7.5x10-3A) (12000 ohms) = 90 Volts
so there is a voltage drop of 90 volts across the 20k resistor (R1), when t=0.
RC= Req C = (12000 ohms) (0.1 x10-6F) = 0.0012 sec
∴ UR1= Uo (e-t/RC) = UR1= 90V (e-t/0.0012)
Ok, so then the current flowing through the capacitor would be the same current flowing through the 30k resistor when t=0.
Using current division, 7.5x10-3A [20 / (20+30)] = 0.003 A
iC1 = io (e-t/RC) = (0.003A)(e-t/0.0012)
This is my final answer, I've erased my initial attempt as it was wrong. Could someone verify if this is correct?