Why two 5v 10uf capacitors in series 5uf 10v and not 10uf 10v like batteries?

Question

Why two 5v 10uf capacitors in series become 10v 5uf

and not 10v 10uf like how batteries behave?

How does the capacitance get halved when?

In parallel capacitance adds up the same way batteries do.

but in series capacitance halves. Not the same behavior as batteries. why do batteries stay the same capacity in series but higher voltage.

but capacitors get half the capacity at higher voltage?is there some way that its not as simple as this??

? · Answer

Ceq = C1 // C2 = 10 // 10 = 5 μF 
Since charge Q = Ceq*Veq  = C*V , then :

Q = C*V = 10*5 = 50 μCo
Veq = Q/Ceq = 50 μCo / 5 μF = 10 V

Anonymous · Answer

Because not like batteries or resistors.

Philomel · Answer

Yes 5u 10V. to understand you need to view what is happening to them with currents etc.
There are books in the library to explain it.

oyubir · Answer

Like batteries? Sure?

Sure, if you but in series 2 two batteries of 1.2V, 2200mAh, you get a 2.4V, 2200mAh battery. Not 2.4V, 1100 mAh, indeed.

But F is not a unit similar to Ah.
1 F = 1 Coulomb per volt. (Hence the formula Cu=q)
And 1 Amp = 1 Coulomb per second.
So 1 Amp×hour (Ah) = 1 Coulomb per second × 3600 seconds = 3600 coulombs.
So Ah and Coulombs are both units of the same thing (with a 3600 conversion ratio)
And 1 Farad is that, per volt.

To be more accurate
1C=1 As = 1/3600 Ah
1 Farad = 1 C/v = (1/3600) Ah/v

So if you want to convert, in Farad, my 2× 1.2V, 2200 mAh battery example, what you have is
two 1.2V, 2.2 Ah = 1.2V,7920 C = 1.2V,7920/1.2=6600 F
(6600 F × 1.2V is indeed 7920 C=2200 mAh)

Which turns into 2.4 V, 2.2 Ah = 2.4V,7920 C = 2.4V,7920/2.4=3300 F.

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Said otherwise, back, to your example,
5v, 10μF = 50 μC = 13.9 nAh
10v, 5μF = 50 μC = 13.9 nAh

So what you have is "two 5v, 13.9 nAh capacitors in series give one 10v, 13.9 nAh". Exactly like batteries.

oldschool · Answer

We know Q = CV and we know caps in series have the same Q
Let the equivalent capacitance = Ceq
Ceq = Q/V = Q/(V1+V2)= C1V1/(V1+V2) but V2 = C1V1/C2
V1 + V2 = V1(1+C1/C2) = V1(C2+C1)/C2
Q/V = C1V1/(V1+V2) = C1V1/[V1(C2+C1)/C2] = C1C2/(C1+C2) = Ceq
Hope this helps.

billrussell42 · Answer

voltages add in batteries, and they add in caps, or any series components, like resistors.

two caps each with 5 volts on them, in series become 10 volts.

the capacitance value in series uses the formula 1/C = 1/C₁ + 1/C₂. That is due to the physical nature of the caps. Look at two equal caps in series, the equiv is the same plate size and twice the spacing, which means half the value.

Why two 5v 10uf capacitors in series 5uf 10v and not 10uf 10v like batteries?

6 Answers