Capacitor in Series and Parallel

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To calculate the charge on each capacitor in a circuit with capacitors in series and parallel, the equation Q = CV is essential, where C is capacitance and V is voltage. The equivalent capacitance must first be determined, which allows for the calculation of total charge. Charge on capacitor plates is equal but opposite, meaning the net charge is zero, but the charge on each plate is not. The process involves working backwards from the equivalent capacitance to find the charge and voltage for each capacitor step by step. Understanding the rules for series and parallel configurations is crucial for accurate calculations.
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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:

<br /> \begin{array}{|c|c|c|c|}<br /> \hline<br /> &amp; C &amp; V &amp; Q \\<br /> \hline<br /> 1 &amp;\qquad &amp;\qquad &amp;\qquad \\<br /> 2&amp;&amp;&amp;\\<br /> 3&amp;&amp;&amp;\\<br /> 4&amp;&amp;&amp;\\<br /> eq&amp;&amp;&amp;\\<br /> 1,2&amp;&amp;&amp;\\<br /> 1,2,3&amp;&amp;&amp;\\<br /> \hline<br /> \end{array}<br />
 
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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