# Calculating the Charge of 3 Capacitors

1. Jul 18, 2013

### freddy13

1. The problem statement, all variables and given/known data

Three capacitors having capacitance C1 = 3.00 microF, C2 = 8.50 microF and C3 = 15.0 microF are connected to a 24.0 V battery as shown. Calculate the charge on C3.

2. Relevant equations

$$C_{eq}= \left ( \frac{1}{C_1+C_2} + \frac{1}{C_3}\right )^{-1}$$

Q = CΔV

3. The attempt at a solution

First I attempted to find the Capacitance saying that C1+C2= 11.5μC through the parallel addition of capacitors. From here, I plugged everything into the

Ceq=((1/C1+C2(1/C3))-1

equation and got my capacitance which was 2.74 μF. I took the result and plugged all of my values into the Q = CΔV equation, and got a final value of 57.096 μC, or 5.71*10-5 C.

I tried another way and got the same answer, so, any suggestions would be great. I think I messed up somewhere with the units.

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Last edited: Jul 18, 2013
2. Jul 18, 2013

### Staff: Mentor

Can you explain your formula for the equivalent capacitance? Are there some parentheses and an operator missing that might make the order of operations clear (and correct)?

3. Jul 18, 2013

### Simon Bridge

There are more "("'s than ")"'s.

Is that: $$C_{eq} = \left ( \frac{1}{C_1} + C_2 \left ( \frac{1}{C_3} \right ) \right )^{-1}$$
or maybe: $$C_{eq}= \left ( \frac{1}{C_1+C_2} + \frac{1}{C_3}\right )^{-1}$$

Neither of those equations nets me the same answer you got though.
I think you need to go over your arithmetic carefully.

4. Jul 18, 2013

### freddy13

It is the second one! Sorry for the miscommunication!

5. Jul 18, 2013

### freddy13

This one to be more exact!

6. Jul 18, 2013

### Staff: Mentor

Okay, can you run the calculation again using that formula? What value do you get for the equivalent capacitance? (The value you gave previously is not correct, although that formula is good. So... finger+calculator problems?)

7. Jul 18, 2013

### freddy13

I got 6.509 microF this time, does that sound correct?

8. Jul 18, 2013

### freddy13

Which, multiplied by the 24 V would give me 156.226 micro C, or 1.56*10^-4 C....is that right?

9. Jul 18, 2013

### Staff: Mentor

Yes. Much better

10. Jul 18, 2013

### freddy13

Thank you very much! Gotta watch those number entering errors!