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Capacitors connected in series. 
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#1
Jul406, 06:52 PM

P: 36

Ive tried so many different ways using the formula for series capacitors and Q=CV and Ive had no luck solving this problem. There must be some rule (or trick) Im missing involving combing C12 and C13 that im completely missing. Can anyone offer any insight into solving this problem? 


#2
Jul406, 07:02 PM

P: 400

what do you mean treid all sorts of ways using the series capacitors and Q=VC?
did you use [tex]\frac{1}{C_{eq}}=\frac{1}{C_1}+\frac{1}{C_2}[/tex]? did you get all the three equations? (so you could calculate the forth) write them here, lets see whats wrong with them. 


#3
Jul406, 07:07 PM

P: 36

Well you would have
[tex]\frac{1}{C_{eq}}=\frac{1}{C_1}+\frac{1}{C_2}[/tex] and [tex]\frac{1}{C_{eq}}=\frac{1}{C_1}+\frac{1}{C_3}[/tex] and finally [tex]\frac{1}{C_{eq}}=\frac{1}{C_1}+\frac{1}{C_2}+\frac{1}{C_3}[/tex] but how could you find the final charge without using a voltage? 


#4
Jul406, 07:25 PM

P: 400

Capacitors connected in series.
right, so you got [tex]30.8*10^{6}=C_1V[/tex]
and [tex]23.1*10^{6}=C_{eq_1}V[/tex] and [tex]25.2*10^{6}=C_{eq_2}V[/tex] you should be able to find [tex]C_1[/tex], [tex]C_2[/tex], and [tex]C_3[/tex] with these equations (they all depend on the voltage, but because the fourth eq. should use the same voltage, it cancels out...) 


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