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?*

^{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.

You should use balancing resistors across the capacitors, assuming it's a simple series connection of electrolytic capacitors.

The appropriate resistor to use across each capacitor is

R = \$ \frac {n V_M - V_b}{0.0015C V_b}\$

R is in M ohms

n is the number of capacitors

C in uF is the capacitance of each capacitor

\$V_M\$ is the voltage allowed on each capacitor

\$V_b\$ is the total voltage across the string

If we allow 400V across each cap and total voltage will not exceed 15kV
then we have a balancing resistor value of 555K ohms.

Each one needs a power rating suitable for the highest possible voltage,
so

\$P_R = \frac {V_M^2}{R} \$ = 0.29W for this example

The balancing resistors will suck 0.54mA in this example, which is 8.1W total,
a significant power loss, so you may not be able to get full voltage from your coil.

Reference: Cornell-Dublier *Aluminum Electrolytic Capacitor Application Guide*

## Best Answer

The answer to this comes from considering what is capacitance: it is the number of coulombs (C) of charge that we can store if we put a voltage (V) across the capacitor.

Effect 1:If we connect capacitors in series, we are making it harder to develop a voltage across the capacitors. For instance if we connect two capacitors in series to a 5V source, then each capacitor can only charge to about 2.5V. According to this effect alone, the charge (and thus capacitance) should be the same: we connect two capacitors in series, each one charges to just half the voltage, but we have twice the capacity since there are two: so break even, right? Wrong!Effect 2:The charges on the near plates of the two capacitors cancel each other. Only the outer-most plates carry charge. This effect cuts the storage in half.Consider the following diagram. In the parallel branch on the right, we have a single capacitor which is charged. Now imagine that if we add another one in series, to form the branch on the left. Since the connection between the capacitors is conductive, bringing the two plates to the same potential, the

`-----`

charges on the bottom plate of the top capacitor will annihilate the`+++++`

charges on the top plate of the bottom capacitor.So effectively we just have two plates providing the charge storage. Yet, the voltage has been cut in half.

Another way to understand this is that the two plates being charged are

farther apart. In free space, if we move plates farther apart, the capacitance is reduced, because the field strength is reduced. By connecting capacitors in series, we are virtually moving plates apart. Of course we can place the capacitors closer or farther on the circuit board, but we have now have two gaps instead of one between the top-most plate and the bottom-most plate. This reduces capacitance.