Capacitance - combining multiple capacitors for equivalent C

[MathewsMD]

Capacitance -- combining multiple capacitors for equivalent C

In the given problem, C_eq = 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.)

 

[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!

 

[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.

 

[MathewsMD]

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!

Hmmm...
Well I know:

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

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

 

[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 ?

 

[BvU]

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

 

[MathewsMD]

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

Sorry, yes.

C_eq = Q/(V₁ + V₂ + V₃ + V₄)

Also, Veq = V₁ + V₂ + V₃ + V₄

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

Thank you!

 

[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 ?