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I have 2 150F 2.7V SuperCaps. I know if I put them in series, and charge them, I will get a result of only 75 farads. I am wondering if I charge them each separately to full capacity before putting them in series, I will get 150 farads at 5.4 volts out of it.

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No, when you put capacitors in series, the capacitance is reduced. Here's one reason why. The energy \$W\$ in a capacitor is:

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

For your two 150F capacitors, each charged to 2.7V, the total stored energy is:

$$ \frac{1}{2} 150F (2.7V)^2 \cdot 2= 1093.5J $$

If you put them in series, and you somehow now had a 150F capacitor at 5.4V, then you would have the energy:

$$ \frac{1}{2} 150F (5.4V)^2 = 2187J $$

Somehow, you have created energy, which I'm told, violates some fundamental properties of the universe as we understand it. We know what actually happens with two identical capacitors in series is that the capacitance is halved. In that case, the energy you have after connecting them in series is:

$$ \frac{1}{2} 75F (5.4V)^2 = 1093.5J $$

You see, the capacitance is halved, but the stored energy is the same as the two capacitors charged separately, as you'd expect.

Here's where you are confused: capacitance is not capacity. Capacitance is a measure of how much electric charge it takes to make a change in voltage:

$$ C = Q/V $$ $$ V = Q/C $$ $$ VC = Q $$

Thus, a farad is a coulomb per volt. (A coulomb is an ampere-second). So say you have a \$1F\$ capacitor, and you move \$1A\$ for \$1s\$ through it. You have moved \$1C\$ of charge, and the voltage across the capacitor will have changed \$1V\$:

$$ 1F = 1C/V $$ $$ V = 1C/1F $$ $$ V = 1V $$

So why does putting two capacitors in series halve the capacitance?

schematic

simulate this circuit – Schematic created using CircuitLab

Consider circuit A. If \$10C\$ of charge moves through I1, then this same charge must flow through C1 (where else would it go?), and the voltage will change by:

$$ 10C/150F \approx 66mV $$

Now consider circuit B. If \$10C\$ of charge moves through I2, that charge flows through C2 and C3. Looking at each capacitor individually, the same charge has moved as before, and the voltage of each will change again by \$66mV\$. But, they are in series, so the voltage of the two considered together has changed by \$133mV\$. What's the capacitance?

$$ 10C/133mV \approx 75F $$

For every bit of charge you move through I2, you charge two capacitors at once, so the voltage changes twice as fast, so the capacitance is halved. But, to get that charge to move you had to apply a higher voltage (because the voltage increased faster), so the stored energy is the same as if you had charged them separately.

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  • \$\begingroup\$ Thanks for the explanation of capacitance in series / parallel, there's always lots of questions who miss this. \$\endgroup\$ – woliveirajr Jun 12 '13 at 12:26
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The unit of capacitance, the Farad, is the ratio of (the magnitude of the) charge (on one plate) to the voltage (across the capacitor):

\$C = Q/V \$

(1) If you charge two identical capacitors to identical voltages and then put them in series, the voltage across the series connected capacitors is simply twice the individual voltages, i.e., the voltage across the pair is twice the voltage across the individual capacitors.

(2) Since the voltage has doubled but the charge on either "external" plate has not changed, it is necessarily the case that the capacitance is 1/2 the individual capacitance.

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  • \$\begingroup\$ When starting from a zero state, yes. But the asker has an additional stipulation which this analysis doesn't cover. \$\endgroup\$ – Ignacio Vazquez-Abrams Jun 12 '13 at 3:03
  • \$\begingroup\$ @IgnacioVazquez-Abramsm, please elaborate on what you believe hasn't been covered. This seems to be an elementary matter. One does not "get 150 Farads at 5.4V" out of something. This, it seems to me, is an elementary misunderstanding of basic physical principles. \$\endgroup\$ – Alfred Centauri Jun 12 '13 at 3:09
  • \$\begingroup\$ The situation that the asker gives where the plates have already been charged. It is (almost) impossible to remove any of the charge, and it is impossible to add any more, hence the capacitance should ideally be 0 and the resistance infinite. \$\endgroup\$ – Ignacio Vazquez-Abrams Jun 12 '13 at 3:13
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    \$\begingroup\$ @IgnacioVazquez-Abrams, your response seems to be, at best, confused, and more likely nonsense. \$\endgroup\$ – Alfred Centauri Jun 12 '13 at 3:16

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