
#1
Jan2013, 09:09 PM

P: 9

1. The problem statement, all variables and given/known data
Derive a relationship for Cs for a two capacitor series circuit with a resistor. Start by showing that the same charge separation q is present across the capacitor and each of the capacitors in series and that the voltage across Cs is equal to the sum of the potential differences across each capacitor. 2. Relevant equations 1/Cs=(1/C1)+(1/C2) 3. The attempt at a solution not sure how to approach this 



#2
Jan2113, 03:07 AM

P: 2,861

If they are all in series what does that say about the current?
What is the definition of current? 



#3
Jan2113, 07:57 AM

P: 9

I tried to work it out and i was wondering if this was the correct way to show wht the question is asking for
q1=C1V and q2=C2V q=q1+q2=(C1+C2)V CsV=q=(C1+C2)V Cs=C1+C2 



#4
Jan2113, 08:32 AM

Mentor
P: 10,766

Capacitor EquationThe approach CWatters suggested gives you the correct equations for the charge. 



#5
Jan2113, 08:33 AM

Mentor
P: 11,403

First you should argue why the charge separation should be the same on both capacitors. Then you can use that fact and the q = C*V relationship to work out the rest. {Note: If you happen to have a knowledge of calculus you could write the KVL equation for the circuit (Integral version) and answer all the questions of the problem from its characteristics} 



#6
Jan2113, 09:11 AM

P: 9

To show that they are equal charges q1=q2
And q=Cs^1V q=(C1^1+C2^1)V (C1^1+C2^1)V=Cs^1V Cs^1=C1^1+C2^1 



#7
Jan2113, 09:14 AM

Mentor
P: 10,766

Your solution will need V1, V2 in some way... 



#8
Jan2113, 10:03 AM

P: 9

Ok
V=q/C V=V1+V2 q/C=q/C1+q/C2 C^1=C1^1+C2^1 



#9
Jan2113, 10:48 AM

Mentor
P: 10,766

That is correct. You can argue that q1=q2 to satisfy charge conservation, or something similar.




#10
Jan2113, 11:38 AM

P: 9

Ok thank you



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