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# Physics/How can I oscillate the charges between ground and the top of the secondary windings of a transformer ?

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QUESTION: Hello Steve,

Here I am again with a new question. It is like a puzzle.......Lets assume we have a signal generator and send a sinus wave with an energy of 3 volts and 1 ampere to a transformer with a relation  of 10/1, the frequency is 60 Hz, so, if we assume we have not any losses, we would get 0,30 volts and 10 ampere at the secondary windings of the transformer. Hmmmm, I want to take one end of my secondary winding into the earth (grounding) and the other end is free. So, can these 10 amperes flow between the ground and top of the secondary winding ?

b-  I made a simple calculation, 60 Hz, would mean, I would have the maximum charge which is transported between ground and the top at the time intervals between 0-Pi/2 (positive) and Pi and 3Pi/2 (negative), this would be equal to
10 A/60x4 1/s = 0,0416 coulombs, am I right ?

c- If I am wrong, can you show me my mistake and what can I do, If I want this 0,0416 coulombs to flow between ground and top of the secondary winding ?

d- What about connecting a metallic ball (aerial capacitor)to the top of the secondary windings, with a certain capacity, how would this change the result ?

Many thanks,
Birol

ANSWER: a) Yes, but you'll have about 5% loss or so, depending on the design of your transformer.

b,c) No, not even close.  This calculation requires an integral.  There's no simple way to explain that, since you didn't try one it appears that you need more calculus before you can attempt it.

d) You would need an enormous sphere, bigger than the Earth itself, since the capacitor could never exceed 0,30 Volts and you're attempting to store charge in the milliCoulomb range on it.  The capacitance of a sphere is easily calculated, if you know enough calculus to do it, and it should be equal to 4*pi*epsilon_0*r, where epsilon_0 is the electric constant for free space (8.85*10^-12) and r is the radius of the sphere.

---------- FOLLOW-UP ----------

QUESTION: Hello Steve,

You upset me badly, smile......Assume the capacity is 50 pF, if the potential is maximum 0,30 Volts, this would mean we have ultra less current as the peak value (at Pi/2 and 3Pi/2), this would be equal to Q = 1,50 x10 exp (-11) Coulombs (CxV), this would give 3,60 nano Amperes, but we have 10 ampere output, why is there such an enormous difference ?
Is there any equation to calculate the oscillating i value between the aerial capacitor and ground ?

b- Is this a RLC Circuit, you see that it is a single wire construct.

c- What happens when we increase the frequency with time, lets say 1 MHz, this time the peak charge we transport seems very less, at 10 ampere, Q would be equal to = 10/4x10 exp (+6) = 2,50x10 exp (-7) Coulombs, anyway, if the capacity is 50 pF, we would create a potential of 5000 Volts, much higher than 0,30 Volts, am I right ?

Thank you for your help,
Birol

Answer
I don't know why you're upset, but that's not for such a forum as this.  You really need more calculus-based physics that involves phasor diagrams in order to really understand  your own question.  One of my freshman or sophomore students could answer this, given time.

You keep oversimplifying the problem.  Currents are integrated on capacitors to get charges, not set to maximum values.  And your load that you specify can't handle 10 Amperes of output (yes it's capitalized, give Mr. Ampere his due credit), so that's why the huge difference exists. The load you specified will never draw the full current because it will over-volt and you'll cut off the current.  This is not a simple algebraic problem, or a static problem.  It is an RLC circuit, but you have in no way involved the inductance of the transformer or the power supply it is attached to.  Go take the second semester of physics and study.

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