Capacitors connected in series.

In summary, the conversation discusses the charges of three capacitors, C1, C2, and C3, when connected in different series to a battery. C1 and C2 have a charge of 23.1 microC when connected in series, while C1 and C3 have a charge of 25.2 microC. The resulting charge when all three capacitors are connected in series to the same battery is unknown and can be determined using the equations Q=CV and \frac{1}{C_{eq}}=\frac{1}{C_1}+\frac{1}{C_2}+\frac{1}{C_3}. However, since the voltage is not given, it is not possible to calculate the
  • #1
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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 I've had no luck solving this problem.

There must be some rule (or trick) I am missing involving combing C12 and C13 that I am completely missing. Can anyone offer any insight into solving this problem?
 
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  • #2
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, let's see what's wrong with them.
 
  • #3
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
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:

1. How do capacitors behave when connected in series?

When capacitors are connected in series, the total capacitance decreases. This is because the effective area of the plates decreases, making it more difficult for the plates to store charge. As a result, the equivalent capacitance is always less than the smallest individual capacitor.

2. What happens to the voltage across each capacitor in a series circuit?

In a series circuit, the voltage across each capacitor is divided in proportion to the capacitance of each capacitor. This means that the capacitor with the largest capacitance will have the largest voltage drop, while the capacitor with the smallest capacitance will have the smallest voltage drop.

3. Can capacitors connected in series store more charge than a single capacitor?

No, the total charge stored in a series circuit of capacitors is the same as the charge stored in a single capacitor. This is because the same current flows through each capacitor, and the total charge is equal to the current multiplied by the total time.

4. How do capacitors in series affect the total energy stored?

The total energy stored in a series circuit of capacitors is less than the energy stored in a single capacitor. This is because the voltage across each capacitor is decreased, and energy is directly proportional to voltage. Therefore, the total energy stored in a series circuit is the sum of the individual energies stored in each capacitor.

5. What is the purpose of connecting capacitors in series?

Connecting capacitors in series can be used to achieve a specific capacitance value that is not available with a single capacitor. Additionally, it can be used to evenly distribute the voltage across multiple capacitors in a circuit, preventing any one capacitor from becoming overloaded.

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