1. The problem statement, all variables and given/known data http://img88.imageshack.us/my.php?image=physicsxt0.jpg 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 3. 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!