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livewire1717

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## Homework Statement

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

## 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**

^{-5}CPart e) U=QV/2, U=(6x10

^{-5})(20)/2=

**6x10**

^{-4}JI think part b might be wrong, which screwed up the rest.. Can anyone either confirm this or correct me? Thanks!

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