Capacitor in Series and Parallel

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Homework Help Overview

The discussion revolves around a problem involving capacitors in series and parallel configurations, specifically focusing on calculating the charge on each capacitor given their capacitance and voltage. The original poster attempts to apply the formula Q = CV and considers the behavior of capacitors in different configurations.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the need to find equivalent capacitance and total charge before determining the charge on individual capacitors. There are questions about the interpretation of charge on capacitors, particularly regarding the concept of net charge versus charge on individual plates.

Discussion Status

Participants are actively exploring different methods to approach the problem, including the use of tables to organize information and the step-by-step breakdown of the circuit. Some guidance has been offered regarding the treatment of equivalent capacitance and the implications for charge calculations.

Contextual Notes

There is mention of a specific point in the circuit diagram that may be causing confusion, and participants are clarifying the relevance of this point in the context of the problem. The original poster is working within the constraints of a homework assignment, which may limit the information available for discussion.

Heat
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Homework Statement


In the figure View Figure , each capacitor has 4.00 micro F and V = 34.0 V.

YF-24-15.jpg

Calculate the charge on each capacitor.

Homework Equations



Q= CV

The Attempt at a Solution



I know that to get the charge of a capacitor we need to use Q = CV, also that capacitor in parallel add, and capacitor in series add reciprocals.

This is what I thought it would be with charge on each of 0.

6htk6s.jpg
 
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I don't know what you mean by having 0 charge, but try using a table:

[tex] \begin{array}{|c|c|c|c|}<br /> \hline<br /> & C & V & Q \\<br /> \hline<br /> 1 &\qquad &\qquad &\qquad \\<br /> 2&&&\\<br /> 3&&&\\<br /> 4&&&\\<br /> eq&&&\\<br /> 1,2&&&\\<br /> 1,2,3&&&\\<br /> \hline<br /> \end{array}[/tex]
 
Last edited:
It seems that the best way to work out this problem would be to find the equivalent capacitance and then the total charge. Once you have this, work your way back to the original circuit step by step and in the process find the charge. This is analogous to a series-parallel resistor network.

The thing is that charge on the plates of the capacitor are equal but opposite. That gives a total net charge of zero. Unfortunately the answer to your problem is not zero. When they ask to find charge, they don't mean net charge. Instead they want the charge on either plate, -Q or +Q. Quantity is the same, but their signs are opposite.
 
Start out with this:
YF-24-15.jpg


So, I got the C equiv for the parallel as shown below:
2eumjoy.jpg


and I got the series as shown below:
2jezj9x.jpg


but how do I simplify this even more, because I notice there is a d point there, and don't know what to do with it.

(I edited the last two. o:) )
 
Heat said:
but how do I simplify this even more, because I notice there is a d point there, and don't know what to do with it.

Just ignore it. All it is is a reference to a specific spot on the wire. It doesn't represent any physical object.

EDIT:
oh, and what program did you use to draw the diagrams?
 
The first image was given by the problem..the rest I just edited in photoshop. so I treat the circuit as a series now? If so, what do I do with the equivalent capacitance?
 
With the equivalent capacitance you can find the total charge. Then work your way back to the original circuit one step at a time, while finding the charge and voltage in each step. Its like working backwards from the equivalent capacitance to the original circuit. Remember the same rules that apply for series and parallel networks (in terms of voltage and current or charge in this case) will work here.
 

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