I need help with capacitors...

[nati levia]

Hey guys,I need some help here...
I have two 1000μf 5v capacitors in series and a 100F 2.7 supercapacitor in parallel to them,
The question: Would it perform or be like a 100F supercapacitor but with 12.7v?

How are reading this secret message? Did you hack it or found it by accident?

 

[BobK]

The answer is no. It would be practically indistinguishable from a 100F 2.7V supercapacitor.

Bob

 

[nati levia]

The answer is no. It would be practically indistinguishable from a 100F 2.7V supercapacitor.

Bob

What i meant to type i would it be like a 10v capacitor but with 100F?.

 

[BobK]

And the answer is still no. A 2.7V capacitor does not act like a 10V capacitor. You cannot put 10V on the 2.7V capacitor just because there is a 10V capacitor in parallel.

You can make a 10V 100F capacitor by putting 4 400F 2.7V capacitors in series, but you need extra circuitry to ensure that they are each charged to less than 2.7V.

Bob

 

[Harald Kapp]

A bit simplified, but good enough here:
The max. voltage of a parallel circuit is the smallest max. voltage of any of the components.
The max. voltage of a series circuit is the sum of the max voltages of the components.

The max. voltage of your 2*1000 μF, 5 V capacitors is 10 V as these are in series.
The max. voltage of the 100 F, 2.7 V capacitor is 2.7 V. Therefore the total max. voltage is also 2.7 V, as the 100F capacitor is in parallel to the 2*1000 μF capacitors.

The total capacitance is Ctotal = (1000 μF in series with 1000 μF) parallel to 100 F
2 * 1000 μF in series is equal to 500 µF.
500 µF parallel to 100F is 100.5 F

That's abaout all the math that there is to this problem.

Look up series and parallel circuits for more detailed information.

 

[nati levia]

And the answer is still no. A 2.7V capacitor does not act like a 10V capacitor. You cannot put 10V on the 2.7V capacitor just because there is a 10V capacitor in parallel.

You can make a 10V 100F capacitor by putting 4 400F 2.7V capacitors in series, but you need extra circuitry to ensure that they are each charged to less than 2.7V.

Bob

Thanks bob... I needed to know this just to prove that my teacher is wrong and thanks to you (and Harald kapp) we just did.

 

[hevans1944]

500 µF parallel to 100F is 100.5 F

100 F = 100,000,000 μF. So, add another 500 μF in parallel with that and you have 100,000,500 μF or 100.0005 F NOT 100.5 F.

 

[nati levia]

100 F = 100,000,000 μF. So, add another 500 μF in parallel with that and you have 100,000,500 μF or 100.0005 F NOT 100.5 F.

I know i messed up with that...

 

[Harald Kapp]

Ahh µ and m. I messed up, too
Makes the 2*100µF even more meaningless. The capacity of these is less than what the chance in capacity by the tolerance of a 100 F supercapacitor.

 

[nati levia]

Ahh µ and m. I messed up, too
Makes the 2*100µF even more meaningless. The capacity of these is less than what the chance in capacity by the tolerance of a 100 F supercapacitor.

Thanks.

 

[hevans1944]

Makes the 2*100µF even more meaningless.

Ummm. The original question involved two 1000 μF, not 100 μF, capacitors. But your are right: it is still meaningless. Based on this and similar threads started by the OP, I think they should probably lower their sights quite a bit as to which college they should seek admission after graduating form high school.

 

[KD6DYR]

1,000,000uf to the Farad. 1uf is one millionth of a farad.

 

[73's de Edd]

One undesirable effect . . . is that the super cap has very low leakage . . .the conventional electrolytics shunting across it will "bleed" down the charge on the super cap much faster than the normal charge decay time of the super cap alone..
Should you be using this as a memory backup supply application / function.