I have read everything about the capacitors and still have a question or two.

1. A 40 uF from a small capacitor (used on a board) vs a large capacitor with the same 40uF. Now the volts are the same at 440 What is the difference in output. Am just trying to figure a problem out.

2. The capacitor(1) has a 40uF and a rating of 440 v while the other capacitor has a rating of 650v with 40 uF. Is this the voltage that is delivered once the capacitor is discharged from a full charge.

 

I am not sure about your first question but for the second.

The voltage rating of a cap is the maximum voltage which can be applied to the cap. The discharge voltage will be whatever voltage you have let the cap charge to. Capacitance is charge per volt, so they'll both charge at the same rate provided the RC Time constant for charging is the same. The equivalent series resistance of each cap will effect this time, but probably not as much as whatever external resistor you have limiting current to the cap.

 

Excellent answer Robert. As for the first question, all Q ratings being the same, a 40uf cap, physically large or small, will behave the same electronically regardless of size. The 40uf rating (within the devices design tolerance, typically +/- 5%) is the specific capacitance the device is designed to exhibit in an electronic circuit, and behaves the same regardless of physical dimensions or shape.

Now other ratings like Q factor, which is the Quality rating of the capacitor, takes into account the dielectric leakage as well as the plate resistance and inductance, among numerous other factors. A smaller high-Q tantalum cap will have better performance than a rolled dielectric aluminum/paper can cap even though the two have the same uF rating.

Hope this helps.

 

Ok--thanks to both of you.

what would be the difference in a 40uF and a 100Uf at the same volts input. say 115 VAC

 

Ok--thanks to both of you.

what would be the difference in a 40uF and a 100Uf at the same volts input. say 115 VAC

They may both explode!

A lot of higher capacitance values are electrolytic capacitors,which in their most common form are polarised capacitors,which have to be connected with the lead marked + to the positive side of a DC supply,& the - lead to the negative side of the supply.
If they are reversed,they will become extremely hot,& may explode.

As an AC supply continuously reverses in polarity,they will blow up on an AC supply also.

If both capacitors are designed for use with AC mains,the 40uF one will have a reactance at 60Hz,of 66.3Ω
By Ohm's Law,the current through that capacitor at 115 V will be 1.73A approx.

The 100uF cap will have a reactance of 26.5Ω
By Ohm's Law,the current through that capacitor at 115 V will be 4.34A approx.

Depending upon the circuit,this may make a lot of difference!

PS:I'm not sure if you have quite got it about the voltage ratings of capacitors.
An analogy is that I can put 250 km/h rated tyres on my car,but that doesn't enable it to reach 250 km/h.!

 

Your question displays a bias that you should correct, a capacitor does not have an input and an output. It is a 2 terminal device. The circuit you put it in may have an input and output however, but it is not intrinsic to the device. This sort of concept makes more sense for some 4 termial devices like transformer.

 

So the 115Volt. is the same while it is charging and while it is discharging --so to speak.

Why then do you get a large voltage? at discharge?

Still having a hard time with this.
Thanks

Bladerunner

 

So the 115Volt. is the same while it is charging and while it is discharging --so to speak.

Why then do you get a large voltage? at discharge?

Still having a hard time with this.
Thanks

Bladerunner

This is because you are not thinking things through!

First,lets take the case,where you are charging to 115V DC.
In this case,the capacitor will charge to 115V DC.

The reason you get a big spark if you then short the disconnected capacitor is not because the voltage has increased,but
because the internal resistance of a capacitor is very low,so you get a high current through your short circuit.

In an 115V AC circuit,the capacitor first charges in one direction,then the other,so during normal operation,there is no net DC charge.

What does happen,sometimes,is that the AC supply is switched off when the incoming waveform is at,or close to its peak voltage on one or the other half cycle.

Mechanical switches are not very fast,but they are often faster than the time for one cycle of a mains waveform,so unless they operate exactly at the zero crossing point,there will be some net charge remaining on the capacitor after the switch is turned off.

The worst,(but extremely unlikely) case,of a very fast switch turned off exactly on a voltage peak suggests a possible charge of around 162V DC.
NOTE: This is not due to any magic in the capacitor,but
simply because the peak voltage of an AC waveform is √2 times Volts r.m.s.

I never thought this could happen,but one day about 30 years ago,I was trying to fix my neighbour's washing machine.
I turned it off,unplugged it,& went to adjust something in the vicinity of a big AC rated capacitor,& got zapped.
I pulled the thing apart,looking for evidence of a DC supply,till the reason dawned on me.