# Why does a capacitor prevent the spark?

#### arcnets

I read this in Feynman's book 'Surely you are joking Mr.Feynman' but I didn't understand it.

'If you close an electrical switch, there will be a spark shortly before the contact is made. If you don't want any spark, just put a capacitor across the switch.'

My question: Why does a capacitor prevent the spark? There's still the same voltage on the switch, isn't it?

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#### chroot

Staff Emeritus
Gold Member
Capacitors force voltage changes to be continuous -- in other words, they ensure that the voltage won't go from 0V to +V instantaneously. The capacitor will have to charge from 0V to +V, which will take a period of time.

- Warren

#### Njorl

I don't think this helps for closing a switch unless it is arranged so that the capacitor is linked to the switch. For example, a capacitor with a long charge time is rigged so that it bridges the switch just before the switch is closed.

Njorl

#### arcnets

chroot - I think the capacitor is already charged before you close the switch. And if you short a charged capacitor, there will be a spark won't it?
Njorl - do you think Feynman is wrong here?

#### Njorl

I have "Surely you must be joking..." at home. What page is it on? I'm not about to say Feynman's wrong lightly. Then again, he'd turn over in his grave if I didn't accept the possibility.

Njorl

#### Integral

Staff Emeritus
Gold Member
If a cap is placed parallel to the switch contacts, it will create a alternate current path for transient currents. Rather than jumping the closing gap of the switch contacts, current will take the path of least resistance into the capasitor.

Consider that the cap will charge to circiut voltage while the switch is open, this provides a source of mobile electrons that are able to redistribute as the EM field changes around the closing switch contacts, essentially this will allow circiut current to flow BEFORE the switch contacts actually close. Thus eliminating any chance of a spark.

Such capasitors are in common use in such places as old fashioned auto distributors (across the points) and in nearly every high voltage contactor circiut.

Modern Solid State Relays (SSR) avoid this propblem in AC voltage switching by changing state only at a AC zero Volts crossing. So the switch changes state the circiut voltage is O, thus no arc is possible.

#### arcnets

Thanks! I understand now.

#### Njorl

Originally posted by Integral
If a cap is placed parallel to the switch contacts, it will create a alternate current path for transient currents. Rather than jumping the closing gap of the switch contacts, current will take the path of least resistance into the capasitor.

Consider that the cap will charge to circiut voltage while the switch is open, this provides a source of mobile electrons that are able to redistribute as the EM field changes around the closing switch contacts, essentially this will allow circiut current to flow BEFORE the switch contacts actually close. Thus eliminating any chance of a spark.

Such capasitors are in common use in such places as old fashioned auto distributors (across the points) and in nearly every high voltage contactor circiut.

Modern Solid State Relays (SSR) avoid this propblem in AC voltage switching by changing state only at a AC zero Volts crossing. So the switch changes state the circiut voltage is O, thus no arc is possible.
For a DC circuit, this just isn't going to work. The cap gets fully charged. For the transient pulse to go to the capacitor, it would be overcharging it, not discharging it. The electrons won't go that way.

Think of this. Charge a capacitor. Remove it from the charger. Put a switch across the contacts. Close the switch. You get a spark. By putting the cap across the switch of a DC circuit, you probably are making things worse.

It would work for an AC circuit, but the capacitor would have to be small, so as not to be an AC short.

Njorl

#### Integral

Staff Emeritus
Gold Member
OK, lets consider this.
Why does a switch arc? Essentially it can be treated as a capacitor with the contacts as the plate. When the contacts are far apart there is a very small capacitance, but as the switch closes the capacitance increases rapidly causing a significant change in the amount of charge stored in the "plates" this current surge is the source of the spark.
Now put in a parallel capacitor, since capacitance of contacts will be small in comparance to that of the capacitor it will make a very small contribution to the total capacitance. (Recall that caps in parallel add as series resistors) now when the switch is open the cap, including the switch contacts will charge to circiut voltage. As the switch closes there will be only a small increase in TOTAL capacistance so therefore a small surge current (if any). Thus no arc.

I think that an isolated charged capacitor should not be used as a model for this situation, here we have other circiutry which limits the total current flow, that is not true when a cap is shorted.

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