Capacitors connected in series.

  • Thread starter ness9660
  • Start date
  • #1
35
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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?
 

Answers and Replies

  • #2
320
6
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
35
0
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
320
6
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...)
 
Last edited:

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