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If we have three capacitors in series, would the energy supplied to the system be the same as the energy that is contained in the equivalent capacitance of these three capacitors?

Urb
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Not Friedrich gauss
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    No, half of the source energy supplied goes to the capacitors, the other half to heat loss in the wire resistance. See my answer here, the chart at bottom should make it clear. – relayman357 Oct 19 '20 at 13:16
  • @relayman357 If the wires have finite resistance, and if the initial charge on the capacitor is zero, and if it is a simple RC circuit. – Roger V. Oct 22 '20 at 07:46
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    @Vadim The same holds for an RLC circuit as well - not restricted to simple RC circuit. I agree that initial charge on capacitor must be zero for this to hold. – relayman357 Oct 22 '20 at 15:31
  • Regardless of capacitor initial charge, the resistance will dissipate the following $E_R$ energy: $E_R = \frac{1}{2}C(Vf - Vi)^2 $ where Vf is final voltage on cap, and Vi is initial voltage on cap. While the capacitor energy goes from $E_C=\frac{1}{2}C{Vi}^2$ to $E_C=\frac{1}{2}C{Vf}^2$. This applies to both adding charge to cap or discharging cap. – relayman357 Oct 26 '20 at 19:48

2 Answers2

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The energy conservation law tells us that the energy stored in the capacitors will be the energy supplied by the buttery, minus the energy dissipated in the wires while charging the capacitors.

Roger V.
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    "minus the energy dissipated" will consume half of the energy. – Ralf Kleberhoff Oct 19 '20 at 15:35
  • @RalfKleberhoff There is no reason why it should be exactly half... unless you are refering to the common error of why the capacitor energy is $CV^2/2$ and not $CV^2$. – Roger V. Oct 22 '20 at 07:30
  • Please compute the integral of charging a capacitor C from a constant voltage V through a resistor R. When we did it at university, the result was a 50:50 distribution of energy between resistor and capacitor. And/or show me some references explaining why E=CV²/2 should be a "common error". – Ralf Kleberhoff Oct 22 '20 at 07:38
  • The question about $CV^2/2$ is frequently posed here. Without computing integral: what about the case of ideal wires? What about the initial conditions? – Roger V. Oct 22 '20 at 07:40
  • I understand the initial condition to be uncharged capacitors (although it wasn't explicitely stated in the question, it's the only interpretation where the question makes sense). And ideal wires will result in an infinite current, meaning that you can only tackle it mathematically as the limit of resistance approaching zero. – Ralf Kleberhoff Oct 22 '20 at 07:54
  • @RalfKleberhoff Actually, one assumes ideal wires when defining the charging energy of a capacitor, well before one considers an RC circuit. And the question never specifies what the circuit is like - it may include something other than a simple resistance: inductances, nonlinear elements, etc. I can guarantee you that the energy conservation will always work, but I am not so sure about 50:50 rule. – Roger V. Oct 22 '20 at 08:00
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    I am sure about the 50:50 rule. Period. – Ralf Kleberhoff Oct 22 '20 at 08:05
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    @RalfKleberhoff if you are sure that this rule applies to any circuit, I would be ineterested to see the proof. – Roger V. Oct 22 '20 at 08:22
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TL;DR: The battery has to supply twice the energy that gets stored in the capacitors.

The characteristics of your power source matter.

The "three capacitors in series" don't make a difference, they behave just the same as one capacitor of appropriate value C.

In your question's title (but not in its body), you mention the power source to be a battery, and a battery is typically best described as constant-voltage source (voltage U) with a series resistor (resistance R).

The battery will supply current to the capacitor until the capacitor's voltage equals the battery voltage. During this charging process, the voltage difference between the battery and the partially-charged capacitor is the voltage drop of the resistor R, resulting in heat dissipation = energy loss.

Finally (after "infinite" time), the capacitor's voltage reaches the battery voltage. The charge necessary for this is

Q = C * U

and the energy stored in the capacitor is

Ecap = 0.5 * Q * U

The energy supplied by the battery is

Ebatt = Q * U

So, half of the battery energy goes into the capacitor, the other half gets dissipated in the resistor (wires, internal resistance of the battery or the capacitors.

If you had a controllable voltage source, capable of producing a slow voltage ramp rising from 0 to U, there would be nearly no voltage drop on the resistor, and nearly all the energy would reach the capacitor. But a battery doesn't work like that.

Ralf Kleberhoff
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