The circuit shown above includes a switch S, which can be closed to connect the 3-microfarad capacitor in parallel with the 10-ohm resistor or opened to disconnect the capacitor from the circuit.
Case I. Switch S is open. The capacitor is not connected. Under these conditions determine:
a) The current in the battery
b) the current in the 10-ohm resistor
c) the potential difference across the 10-ohm resistor
Case II. Switch S is closed. The capacitor is connected. After some time, the currents reach constant values. Under these conditions determine:
d) The charge on the capacitor
e) The energy stored in the capacitor
The Attempt at a Solution
part a)I found the total resistance to be 9 ohms: 6+ 1/(1/4+1/(10+2)) so I=72/9=8 amps
Part b) I thought that 1/4 of the current would go through the 10 ohm resistor (12 ohms total in that branch, 4 in the parallel branch) so 1/4 of 8 is 2 amps
Part c) V=IR, V=(2 amps)(10 ohms)=20 V
Part d) The voltage drop should be the same as in the 10 ohm resistor, Q=(3x10-6)(20)=6x10-5C
Part e) U=QV/2, U=(6x10-5)(20)/2=6x10-4J
I think part b might be wrong, which screwed up the rest.. Can anyone either confirm this or correct me? Thanks!