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Recently I have been thinking about a circuits problem that caused me a lot of confusion.

First of all we know from thermodynamics that energy is conserved. I wanted to show a problem that seemed to contradict itself.

So the problem is:

Consider an ideal 8\$\mu\$F capacitor that we place in an RC charging circuit and charged it up till it stores a voltage of 25V.

8uF capacitor So we can say that the energy stored in the 8\$\mu\$F capacitor to be: $$W = \frac{1}{2}*8\mu*25^2 = 2500 \mu J$$

Now let's take a second ideal 2\$\mu\$ capacitor that we place in an RC charging circuit and charged it up till it stores a voltage of 5V.

2uF capacitor

So we can say that the energy stored in the 2\$\mu\$F capacitor to be: $$W = \frac{1}{2}*2\mu*5^2 = 25\mu J$$

So we can say that the total energy in both capacitors is \$2525 \mu J\$.

Now let's say after we charged the capacitors we placed in series in a circuit like this but with reversed polarity (the first capacitor is placed in the opposite polarity of the second capacitor)

circuit

Now we can take the equivalent capacitance of the both series capacitors, we get a \$1.6 \mu F\$, and the initial voltage in that capacitor will be 20V (Since the 25 and 5 are in opposite directions).

So we can say that the energy stored in the equivalent capacitor is

$$W = \frac{1}{2}*1.6\mu*20^2 = 320\mu J$$

Now this is extremely confusing. Where did the other energy go.

I thought to myself that this energy might just be trapped inside the capacitor, however if you put this on a SPICE simulator with the initial conditions and a load of \$R = 25k\$, you will see that it fully discharges.

Any clue on what could be the mistake that caused this contradiction?

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  • \$\begingroup\$ Your 2 μF capacitor is reverse-charged and possibly dead. You used the symbol for a polarized capacitor where you probably should be using a non-polarized one. \$\endgroup\$
    – Hearth
    Oct 5, 2023 at 12:14
  • \$\begingroup\$ @Hearth the polarity markings indicate the measurement, not the component type. \$\endgroup\$ Oct 5, 2023 at 16:05
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    \$\begingroup\$ @TimWilliams The + and - do. The | and ( on the other hand indicate a polarized capacitor. \$\endgroup\$
    – Hearth
    Oct 5, 2023 at 18:15
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    \$\begingroup\$ I've seen/used this symbol for decades, and always seen polarized with a + on the flat side. If anything the asymmetry suggests outer foil (analogous to the dot on an inductor) \$\endgroup\$ Oct 5, 2023 at 18:16

2 Answers 2

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however if you put this on a SPICE simulator with the initial conditions and a load of R=25k, you will see that it fully discharges.

No, it does not fully discharge, depending on what you call the 'it'.

The composite capacitor fully discharges. However, each individual capacitor is still charged somewhat, storing energy.

The easiest way to see this is to take two capacitors charged to equal voltage, and put them back to back. The terminal voltage of this composite capacitor is zero. It stores zero energy. Put a resistor across its terminals, and no energy will be lost to the resistor. The capacitors are still charged though, storing energy, which is inaccessible to those two terminals.

The complication of different voltages and capacitances in your question simply serves to obfuscate the fact that inaccessible charge still remains.

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  • \$\begingroup\$ Thank you so much for clarification! \$\endgroup\$
    – Mora
    Oct 5, 2023 at 9:41
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Any connection between capacitors will always have some resistance and/or inductance. If two capacitors which are charged to different potentials are connected to each other, charge will move between them so that it balances, possibly leaving the caps in a lower energy state than they had been previously. Any energy that is lost by the caps will be either dissipated by the resistance or stored into the inductance, regardless of how big or small the resistance might be. If the resistance is large, the energy will be dissipated slowly, and if it's small, the energy will be dissipated quickly, but if the caps are left connected until the system reaches equilibrium, the final charge state will be unaffected by the resistance and inductance.

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