Charging a Capacitor with Another Capacitor

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Homework Help Overview

The discussion revolves around a circuit involving two capacitors, specifically a 12 microfarad capacitor and a 4 microfarad capacitor, and the process of charging the latter through a switch mechanism. Participants are tasked with formulating a differential equation to describe the charge on the 4 microfarad capacitor over time after the initial charging phase.

Discussion Character

  • Exploratory, Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • The original poster attempts to apply Kirchhoff's Loop Rule to derive a differential equation for the charge on the 4 microfarad capacitor. Some participants question the meaning of a high resistance value mentioned in the context and whether it is a typographical error. Others discuss the configuration of the capacitors, debating whether they are in series or parallel based on their charge and voltage characteristics.

Discussion Status

Participants are actively engaging with the problem, offering insights into the differential equation and discussing the configuration of the capacitors. There is a recognition of differing interpretations regarding the series and parallel arrangement of the capacitors, with some guidance provided on the conditions that define these configurations.

Contextual Notes

There are indications of missing information, such as the voltage source value, and a potential typo in the resistance value mentioned. The discussion also reflects on the assumptions regarding the arrangement of the capacitors and their behavior in the circuit.

breez
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http://img230.imageshack.us/img230/8480/circuitbb5.th.gif

Switch 1 is turned on until the 12 microF capacitor is completely charged. Then Switch 1 is switched off, while switch 2 is switched on.

How do you write a differential equation that may be used to find charge on the 4 mircroF capacitor as a function of time?

My attempt:

Applying Kirchhoff's Loop Rule: [tex]0 = V_1 - R_1 \frac{dq}{dt} - R_2 \frac{dq}{dt} - \frac{q}{C_2}[/tex], where q is the charge on the 4 microF capacitor.

This can be rewritten as
[tex]0 = \frac{Q-q}{C_1} - \frac{dq}{dt} (R_1 + R_2)- \frac{q}{C_2}[/tex]

In this last step I assumed the charged discharged from the 12 microF capacitor is the charge that appears on the 4 microF capacitor.

(loops used are clockwise)
 
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What does the [tex]\mbox{20M} \Omega[/tex] refer to? How can a voltage source have such a high resistance? The voltage provided by the source isn't given as well. Looks like there's something wrong with the picture.
 
Yeah that's a typo in the book. It should be 20MV.
 
The exact question asks:

a. What is the charge on the 4 microF capacitor after switch 2 is closed for 1 minute?

b. What is the final charge on the 2 capacitors?
 
Hi breez,

I think your differential equation is okay; and you don't need the differential equation to solve for part b.
 
Oh thanks. Are these 2 capacitors in series or parallel? I think they should be in series right?

EDIT: Actually after solving this problem the capacitors end up with different charges and different voltages I find they are in fact parallel...

I guess as long as 2 capacitors are in a closed loop, they must be in parallel since their 2 ends are at the same 2 voltages.

This is kind of counterintuitive because they seem to be in series...

So another requirement for them to be in series is not only for them to flow into one another, but for their ends to be at a different voltage right?
 
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They are not in series; in series capacitors have the positive plate of one capacitor attached to the negative plate of another.

Normally they are also not in parallel, since their voltages are not the same. When the current goes to zero, however, the potential differences across the resistors will vanish and at that point they can be treated as if they are in parallel.
 

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