# Capacitor voltage, charge, energy working in and out

1. Feb 4, 2014

### Mr.Bomzh

Hello , It's been a little while since I havent written anything here.A recent thread was directly on topic with what I was wondering here about a little longer ago and the answers I got were good ofcourse but I'm still not convinced about the workings of my scenario , so I will redefine my question and my idea and ask it again.
By the way the recent thread to which I'm reffering to is linked right here ...

So basically here is my rewritten question.The schematic of my question will be added.
If we have a dc power source like that from a battery or rectified mains and in series with that source is a transformer primary and a varyable capacitor whose permittivity can be altered either by mechanical or electrical means , do the resulting charge flow towards and away from the capacitor induce current in the transformer secondary and be used to transfer energy through the transformer? In the previous thread we concluded that indeed it can be so.
Then the guys from PF went on to say that , since there is no physical path for the varying charges to flow because a capacitor blocks direct charge flow from one terminal of a battery to the other , then no work could be done through the transformer primary , the only way work could be done is to supply energy through the mechanism that changes the capacitance whether it would be mechanical or electrical.

This was said by forum member Bobbywhy.
Ok so here is what my intuition tells , a capacitor whose capacitance would be varied in a circuit like the one i described should be able to produce charge flow(current) towards the cap and away from it in cycles which when passed through a trasnformer could induce current in the secondary to be used to do work.The mechanism which changes the capacitance isn't adding energy to the system , the energy that moves the charges comes from the battery.

And another question similar to his one , what would happen if we had the same transformer primary and capacitor in series but this time the power source would be say 50hz AC mains ? the capacitor would still block any real charges from passing through it , but as we know capacitor pass through AC signals, so the AC would run through the transformer and still do work just like when it would be passed through the transformer without the capacitor in series, is this correct?

So in the DC battery case the transformer should be able to induce secondary current because of varying carge flow towards and away from the capacitor assuming one wouled change the dielectric constant and in the second AC case the transformer is able to do work because AC sinewave or any varying in amplitude and/or polarity current is able to induce secondary current and also is able to " pass" through a capacitor. Is that correct?

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2. Feb 5, 2014

### Mr.Bomzh

Ok over night , I was thinking I think I kinda got it finally , in a solely dc supply the charges charge up the capacitor and once the cap is charged to a given level it stops charging and the charge just stays there , now if we would want to decrease the capacitance and make some of the charge flow back into the circuit we would need to do work on the cap because those opposite charges want to attract so getting some of them flow back would mean doing work against them, so I guess the mechanism that changes the capacitance of a capcitor in a DC circuit is using energyu to do that after all, but I'm not 100% sure , so I will say that I don't know.

Also if the varyable capacitor would be self controlled like a spark gap whose rate of discharge is controlled only on the air chemical properties between the contacts , their seperation distance and voltage level on the contacts.
Imagine in my case a capacitor whose dielectric permittivity drops when the PD across it's plates reaches a certain value , like say 100v DC, the dc power source is capable of supplying 100v DC but the permittivity drops when the cap is charged to say 98v DC , what would then happen with the charge flowing towards the cap , as the capacitor isn't fully charged to the source potential while this change in dielectric constant would happen?