You are right in that you get the same energy no matter how you arrange the capacitors. Putting capacitors in series is problematic, so all else being equal, I'd want to put all the capacitors in parallel too.

The tradeoff between voltage and current can be largely compensated for by adjusting the winding. Take a look at a datasheet for a family of relays or solenoids, for example. You will usually see what is otherwise the same product offered in different voltage and current combinations. The only difference is the coil.

Fortunately, the total size of the coil and the total copper used stays the same for a range of current/voltage tradeoffs that come out to the same power. For example, consider starting with a coil that draws 50 mA at 12 V. Now we want a 24 V version while keeping the overall device geometry the same. If we changed nothing, applying 24 V would cause twice the current, and therefore twice the magnetic field, and four times the power dissipation. Now imagine we make the wire cross-section half the area but double its length. That results in 4x the resistance, so half the current flows thru the coil at 24 V. Half the current around each turn in the winding is made up by twice as many turns, so the magnetic field stays the same. Half the area and twice the length is still the same amount of copper, just arranged differently. Twice the voltage and half the current is still the same power, so same heat to get rid of.

The same tradeoffs apply to your coil gun coil. To use a lower voltage, use thicker wire but less of it. To go from 1000 V to 200 V, use wire that is 5x shorter, but also has 5x area in cross-section. That means its diameter will be sqrt(5) larger.

Eventually the currents get so high that the feed lines start becoming significant. However, as long as you still have a reasonable number of turns, all should be OK. If the original used 50 turns, then 10 turns of the thicker wire should be fine. If the original only used 5 turns, then there isn't room left to scale down the coil to lower voltage and higher current. If so, this is probably why the original went to such awkward means to get a higher voltage.

## Best Answer

Capacitor banks designed for power factor correction are rated in kVAr (kilo-volt-ampere reactive) because it's convenient. One will typically know the reactive power required by some load, then it's simply a matter of selecting a capacitor of the equal but negative reactive power to improve the power factor.

Reactive power \$Q\$ for a purely reactive load (such as a capacitor) is calculated by:

$$ Q = {|V|^2 \over X} $$

Where \$V\$ is the voltage, and \$X\$ is the reactance which can be calculated by:

$$ X = {-1 \over 2 \pi f C} $$

where \$C\$ is the capacitance. Putting those together, the relationship between reactive power (kilo-volt-ampere reactive, \$Q\$) and capacitance (farad, \$C\$) is:

$$ Q = - 2 \pi\, f C\, |V|^2 $$

Since the frequency and voltage of a power distribution system are typically fixed, specifying the capacity in kVAr instead of F eliminates some of the mundane calculation required.

Since the reactive power is proportional to the square of the voltages, converting to kVAr at one voltage can be converted to another voltage by examining the ratio of the squares of the voltages. For example:

$$ {208^2 \over 240^2} = 0.7511 $$

So to convert a kVAr rating for 240 VAC to one for 208 VAC, multiply by 0.75.

Similarly adjustment is required if the frequency is other than specified.