Calculating Potential Difference in a Capacitor with Resistors Circuit

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Homework Help Overview

The discussion revolves around calculating the potential difference across a capacitor in a circuit that includes resistors and an ideal battery. The circuit parameters include a battery voltage of 12V, resistances of 4 ohms and 6 ohms, and a capacitor with a capacitance of 6x10^-6 F, initially discharged.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants share various attempts to calculate the potential difference, with some using the formula for charge and others questioning the assumptions about resistance in the circuit. There are inquiries about the correct approach to find the final voltage across the capacitor and how to account for initial conditions.

Discussion Status

The discussion is active, with participants exploring different methods and questioning the validity of their approaches. Some have provided partial calculations, while others seek clarification on the underlying principles and assumptions involved in the problem.

Contextual Notes

Participants are considering the implications of varying resistance values and the initial conditions of the capacitor. There is also a mention of the need to trace the current path in the circuit, which suggests a focus on understanding the circuit dynamics.

NWNINA
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Homework Statement



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


Homework Equations



q=q_o e^(-t/RC)

\tau =RC

The Attempt at a Solution



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

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

What have you tried? Show your working.
 
Last edited by a moderator:
NascentOxygen said:
Hi NWNINA! http://img96.imageshack.us/img96/5725/red5e5etimes5e5e45e5e25.gif

What have you tried? Show your working.



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:
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.
 
NascentOxygen said:
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.

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?
 
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)
 
theBEAST said:
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?
Can you trace the closed-loop path that current would be taking?
 

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