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I'm just confused in general about what happens with charge, voltage, etc in parallel and series circuits with capacitors.

Anyways, I'm trying to find the total energy stored in $2$ equivalent capacitors in series vs in parallel, vs 1 capacitor alone. They're charged by a battery that has a constant voltage and current.

If I use $\rm P=\frac{1}{2}CV^2$ , I know using the equivalent capacitance for $2$ capacitors in series would be : $$\rm C_{eq}=\frac{1}{\frac{1}{C}+\frac{1}{C}}=\frac{1}{2}C$$

And for parallel it's just : $$\rm C_{eq}=2C$$

The equivalent capacitance for a single capacitor is simply $\rm C$.

My question(s) :

(a) What I do about the voltage? Is it the same for both in all $3$ cases ($2$ capacitors in series, $2$ in parallel, and 1 lone capacitor)?

(b) What if I wanted to instead use $P=\frac{1}{2}\frac{Q^2}{C}$? What happens with the charge in the $3$ cases?

Thanks for the help folks

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3 Answers 3

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(a) What I do about the voltage? Is it the same for both in all 3 cases (2 capacitors in series, 2 in parallel, and 1 lone capacitor)?

I'm assuming the battery charged the series, parallel, and single capacitors to the same battery voltage $V$.

For series capacitors the charge on each capacitor is the same regardless of the value of each capacitance, while the voltage on each depends on the capacitance. For two equally sized capacitors, the charge and voltage are the same and the voltage across each is one half the battery voltage. Therefore the total stored energy is

$$E_{series}=\frac{1}{2}C(V/2)^2+\frac{1}{2}C(V/2)^2=\frac{CV^2}{4}$$

Or, in terms of the single equivalent capacitance of $C/2$

$$E_{equiv}=\frac{1}{2}\frac{C}{2}V^{2}=\frac {CV^2}{4}$$

The capacitors in parallel have the same voltage across them and the charge depends on the capacitance. So the total stored energy for two equal parallel capacitors is

$$E_{parallel}=\frac{1}{2}CV^2+\frac{1}{2}CV^2=CV^2$$

Or, in terms of the single equivalent parallel capacitance of $2C$

$$E_{equiv}=\frac{1}{2}(2C)V^{2}=CV^2$$

Finally, for the single capacitor

$$E=\frac{1}{2}CV^2$$

(b) What if I wanted to instead use $P=\frac{1}{2}\frac{Q^2}{C}$? What happens with the charge in the $3$ cases?

Just plug in $\frac{Q}{C}$ for $V$ in each of the above equations.

Hope this helps.

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  • $\begingroup$ So in your $E_{series}$ equation, what is C? Do we not need to account for $1/C_{eq}$? $\endgroup$ Mar 4, 2022 at 17:11
  • $\begingroup$ @CottonHeadedNinnymuggins $C$ is each individual capacitor which, as I understand your question, are equal. The total energy stored on the equivalent series capacitance has to equal the sum of the stored energy on each capacitance, for conservation of energy. The stored energy on each capacitor is based on the voltage on each, which is half the voltage $V$.across the series combination, $\endgroup$
    – Bob D
    Mar 4, 2022 at 17:16
  • $\begingroup$ @CottonHeadedNinnymuggins I've updated my answer in terms of the equivalent capacitances. Hope that clarifies it for you. $\endgroup$
    – Bob D
    Mar 4, 2022 at 17:44
  • $\begingroup$ That makes perfect sense, thank you $\endgroup$ Mar 4, 2022 at 19:17
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In simple cases when the capacitors are connected in series the charge stored on each of the capacitors is the same and for capacitors in parallel the potential difference across each of them is the same.

In general stored electrical energy conservation does not work because if the charge needs to be redistributed this will mean that an electric current must flow in wires which have resistance and so some of the stored electrical energy will be converted into heat.
That is the reason for the discrepancy in the electrical energy stored by the capacitors in your post.

The Wikipedia article Two capacitor paradox and the Kirk T. McDonald paper A Capacitor Paradox explain fully what happens if the wire resistance is zero (very, very small).

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Here are some ways of thinking about the equivalent capacitance.

A "equivalent capacitance between terminals A and B" is a single capacitance that replaces a "network of only capacitances with terminals A and B" such that the rest of the circuit can't tell the difference between two cases. Specifically, this means that [in the context of charging and discharging the network with an external battery connected at A and B]

  • the voltage across A and B is the same in both cases
  • the [total] charge seen at terminal A is the same in both cases (and similarly for B)
  • the current into A (and out of B) is the same in both cases
  • the energy stored is the same in both cases

The bracketed condition means that the paradox mentioned by @Farcher involving one of the capacitors being charged while another is uncharged does not apply here.

So, if your three capacitor networks are connected to identical batteries, then their corresponding equivalent-capacitances are connected to those identical batteries.


For the case of two identical capacitors,

  • when connected in series, the two capacitors and its equivalent capacitance could be modeled as all having the same area of one plate, but the equivalent capacitance has twice the separation between the equivalent-plates compared to the separation between the plates of one of the original capacitors. (This is consistent with the equivalent capacitance being half of one original capacitance.)

  • when connected in parallel, the two capacitors and its equivalent capacitance could be modeled as all having the same separation between their plates, but the equivalent capacitance has twice the area of its equivalent-plate compared to the area of the plate of one of the original capacitors. (This is consistent with the equivalent capacitance being twice one original capacitance.)

The above could be generalized to different capacitances connected all-in-series or all-in-parallel.


For the general case of different capacitances,

  • For an all-in-parallel combination,
    the voltages across the various capacitors are equal to each other
    and are equal to the voltage across the equivalent capacitance.
    The charges on the left-plate of the equivalent-capacitance is equal to the sum of the charges on the left-plates of each of these capacitances.
  • For an all-in-series combination,
    the charge on the left-plates of each capacitor are equal to each other
    and are equal to the charge on the left-plate of the equivalent capacitance.
    The voltage across the equivalent-capacitance is equal to the sum of the voltages across each of these capacitances.

Hopefully, this gives you enough information to answer your questions.

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