Capacitors connected in series.

by ness9660
Tags: capacitors, connected, series
ness9660 is offline
Jul4-06, 06:52 PM
P: 36
Suppose we have 3 capacitors, C1, C2, and C3.

If C1 is connected to a battery it's charge is 30.8 microC. C1 is disconnected and discharged.

If C1 and C2 are connected in series to the same battery, their charge is 23.1 microC. Both are disconnected and discharged.

If C1 and C3 are connected in series their charge is 25.2 microC. All 3 are disconnected and discharged.

Now, if all 3 are connected in series to the same what is the resulting charge?

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?
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fargoth is offline
Jul4-06, 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.
ness9660 is offline
Jul4-06, 07:07 PM
P: 36
Well you would have


and finally

but how could you find the final charge without using a voltage?

fargoth is offline
Jul4-06, 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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