# Capacitor with resistors

1. Mar 27, 2012

### NWNINA

1. The problem statement, all variables and given/known data

In the following circuit, the ideal battery has an electromotro force of 12V, the resistances R1=4 ohms and R2 = 6 ohms. The capacito of 6x10^-6 F can be found discharged initially. The switch closes at t=0. Calculate the potential difference in the capacitor when t=2$\tau$

2. Relevant equations

q=q_o e^(-t/RC)

$\tau$ =RC

3. The attempt at a solution

I have tried many things, but nothing seems to be working.

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2. Mar 27, 2012

### Staff: Mentor

Hi NWNINA! http://img96.imageshack.us/img96/5725/red5e5etimes5e5e45e5e25.gif [Broken]

What have you tried? Show your working.

Last edited by a moderator: May 5, 2017
3. Mar 27, 2012

### NWNINA

this is one of the things i tried:

Vo = (12V/10) * 6 = 7.2V

V=(7.2V)*e^(-2t/t)=.97V

ΔV=6.23V

Last edited by a moderator: May 5, 2017
4. Mar 28, 2012

### Staff: Mentor

If R1 were zero ohms, after the switch is closed what would be the final voltage across the capacitor? Trace the full path of current as it leaves the battery and charges the capacitor.

5. Mar 28, 2012

### theBEAST

I would like to know how to do this question as well. So I found the current using I=I_o*e^(-2t/t) and then I used V=IR to solve for the potential difference. Am I on the right track? If I am what would I use for R, the total resistance of the circuit or just the 4ohm resistor?

6. Mar 28, 2012

### Rayquesto

how would you find Potential Difference when a charge exists in a capacitor? Also, I believe that equation for charge after some time in the capacitor is too general. You must consider the initial charge in the emf and then take a special difference. The equation is derived from a differential equation that looks like this: q(t)=Q0(1-e^-t/RC)

7. Mar 28, 2012

### Staff: Mentor

Can you trace the closed-loop path that current would be taking?