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# Capacitor bank charging in series & parallel

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#### B-BOMB7

##### New Member
Hey, I'm going to the Regional Science Fair in Washington for my work on a Thompson Coilgun, which is basically an electromagnet, but I want to impress the judges somewhat by using a capcitor bank. The only problem is I'm only 15 and I don't know the equations for how to find the voltage and capacitance for capacitors in series and parallel circuitry. Any help that would be given in Lamens Terms would be greatly appreciated! Thankyou VERY much!

Bryan[/i]

Capacitors in parallel add up i.e if C1 and C2 are to capacitors connected in parallel then the equivalent capacitance C would be C = C1 + C2
And in series, the equivalent capacitance become 1/C = (1/C1) + (1/C2)
Thus the capacitance will effectively reduce in series combination.

If you want a large capacitor bank then connect as many large capacitors in parallel as you want but take care of the charging voltage as a big cap. bank can give you a very tasty and nasty shock which may be much much higher than you can get from the wall socket. Never touch the terminals of very large capacitors even if it has been lying idle for years.

Kinjal, i think you mean a large voltage capacitor....wich is charged at a high voltage.
and yes they stay charged for a lot of time. i found one charged at around 150V in al old tv that wasnt pluged in for about 5 years. today in many electronics where they are used there is used a resistor in paralel with it so that it will discharge it. it is a large values, some take a few minutes to discharge, but it is better like this.
also, if you have many caps in paralel in different values and voltages, you should consider that the maximum voltage you can apply to the balk is equal the the minimum voltage on all the caps.

So how, if at all, is the voltage of the capacitors effected by joining the caps in parallel and series circuits. Do the voltages add together in parallel and decline in series, how does this work. Also, how do I know what type of power supply should be used for any given cap rating. Thanks for your help in my upcoming victory in the science fair.

think of something: in paralel all the voltage you apply is applied to all of them, so this voltage must not exeed any of the voltages on the caps, for example if you have 10u/20V, 100u/16V, 2200u/10V the smallest voltage is 10V so you mustnt put more that 10V to the bank.
in series it is a bit harder, because if you have different values. ill say it for 2 caps. if you have 2 caps, C1 and C2 and you apply a voltage to the series VT then the voltage accross them will be V1 for C1 and V2 for C2.
VT=V1+V2; V1/V2=C2/C1 and from here you get:
V1=VT*C2/(C1+C2) and V2=VT*C1/(C1+C2)

So it's not like with batteries where if you put three 1.5V batteries in parallel the voltages add together and the current stays the same and when you put them in series the voltage stays the same and the currents add together. Just confirming eveything to make sure I don't do anything wrong and end up paying for it.

Thanks,
Bryan

You have that backwards.

Series batteries - voltage adds up, current stays the same
Parallel batteries - voltage stays the same, current adds up

i think he wasnt paying attention to what he was saying. you cant get this wrong that much
here:
http://www.tpub.com/neets/book2/3e.htm
i think this should be enough.

Maybe, but whether he meant capacitors or batteries, he still had it backwards. It doesn't matter what components you are using, a series circuit behaves one way and a parallel circuit behaves another. The quantities mentioned were voltage and current, not total capacitance.

Hi B-BOMB7,

I will try to say this plain and easy.

If you put two 6mfd caps in parallel,
thats side by side, you get 12mfd.

If you put two 6mfd caps in series,
thats one after the other, you get 3mfd.

If you put three 6mfd caps in parallel,
you get 18mfd.

If you put three 6mfd caps in series,
you get 2mfd.

The voltage working is a different carry-on
with series, because the voltage given to
each is a function of its value,
the smaller values get more voltage.

The logic behind that is that the current
through the series chain applies to each,
and the smaller ones charge up quicker.

Ive tried to express this in simple terms
i know side by side doesnt really mean its
parallel .... its just an expression ok.

Best of luck with it,
John

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