# Capacitance - combining multiple capacitors for equivalent C

1. Feb 12, 2014

### MathewsMD

Capacitance -- combining multiple capacitors for equivalent C

In the given problem, Ceq = C1C2C3C4/(C1 + C2 + C3 + C4)

The answer says the equivalent capacitance is always less than C1 but I can't come up with thy. When I do this, I can't seem to prove that equivalent capacitance is always less than C1. For example, if I let:

C1 = 998 F, C2 = 999 F, C3 = 1000 F and C4 = 10001 F, then I get an answer (767119326 F) and this is much larger than C1. Any suggestions?

(I realize the capacitances used here are very large, but they are just used to make a point.)

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2. Feb 12, 2014

### BvU

Use Q=CV and discover that now four capacitors have to share the voltage that C1 would have had all on its own if C2,3,4 were absent!

3. Feb 12, 2014

### BvU

Perhaps your formula is questionable ?
When I do
$${1\over C_{eq}} = {1\over C_1} + {1\over C_2} + {1\over C_3} + {1\over C_4}$$
I get something completely different ! Which has the right dimension. Your $C_{eq}$ does not have the dimension of Farads ....

Hint: The $C_1 C_2 C_3 C_4$ in the numerator is correct.

4. Feb 12, 2014

### MathewsMD

Hmmm....
Well I know:

Ceq = Q(V1 + V2 + V3 + V4)/(V1V2V3V4)

I just don't quite see how this shows Ceq < C1...do you mind explaining further?

5. Feb 12, 2014

### BvU

You know this for two capacitors, because then it comes out right.
What you wrote here has the dimension Coulombs * Volts / Volts4 and that is not Farads ...
So: from where did you get this ?

6. Feb 12, 2014

### BvU

You mean you know that
$${1\over C_{eq}} = {V\over Q} = {V_1\over Q}+{V_2\over Q}+{V_3\over Q}+{V_4\over Q}$$

7. Feb 12, 2014

### MathewsMD

Sorry, yes.

Ceq = Q/(V1 + V2 + V3 + V4)

Also, Veq = V1 + V2 + V3 + V4

So since C1 = Q/V1 then Ceq must be smaller (i.e. it is divided by more positive numbers).

Thank you!

8. Feb 13, 2014

### BvU

Just in case you ever need to calculate the $C_{eq}$ for 4 capacitors in series: do you know how to work out
$${1\over C_{eq}} = {1\over C_1} + {1\over C_2} + {1\over C_3} + {1\over C_4}$$
to ${ C_{eq}} = {C_1 C_2 C_3 C_4 \over ???}$ ? And: can you guess why anyone would ever use capacitors in series in a circuit ?