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### Author Topic: simple overunity?  (Read 5995 times)

#### abcdxyz

• Newbie
• Posts: 4
##### simple overunity?
« on: November 04, 2015, 01:30:23 PM »
They must be three capacitors with the same capacity

Please check the voltage on the capacitors when the system is switched

#### hartiberlin

• Administrator
• Hero Member
• Posts: 8125
##### Re: simple overunity?
« Reply #1 on: November 04, 2015, 04:22:14 PM »
What should be the value of the resistor ?
DO you think the voltage goes higher all in all ?

#### abcdxyz

• Newbie
• Posts: 4
##### Re: simple overunity?
« Reply #2 on: November 04, 2015, 04:38:39 PM »
any resistor can be tested
This is a test circuit n.1
The point is that in the circuit is only 12v4700mikroF
when you turn on the switch, the sum of the voltage on the capacitors is greater
maybe I'm doing something wrong

#### citfta

• Hero Member
• Posts: 1050
##### Re: simple overunity?
« Reply #3 on: November 04, 2015, 05:12:00 PM »
I have seen this circuit discussed before.  The mistaken idea is that the energy stored in a capacitor is linear.  In other words people think that a cap charged to 12 volts has twice the energy of the same cap if it is charged to six volts.  I don't remember the formula but according to the calculations for power in a cap the energy at 12 volts is closer to 4 times the energy at six volts.  So you can easily get three caps to show a higher voltage than a single cap of the same value.  But the power is actually less.  Maybe someone else has the formula handy so you can do the calculations and see for yourself what is actually happening.

#### abcdxyz

• Newbie
• Posts: 4
##### Re: simple overunity?
« Reply #4 on: November 04, 2015, 05:44:41 PM »
you're right
I just checked with the fourth capacitor
and it is less
thx

#### memoryman

• Hero Member
• Posts: 758
##### Re: simple overunity?
« Reply #5 on: November 04, 2015, 10:31:01 PM »
E=1/2 x C x V^2

#### abcdxyz

• Newbie
• Posts: 4
##### Re: simple overunity?
« Reply #6 on: November 20, 2015, 08:36:54 AM »
If the voltage on the capacitor equals the voltage source, the circuit stops working,
keep the voltage on the capacitor as low as possible.