Do Capacitors Drain the Source?

[seanhbailey]

I hooked up four capacitors in parallel to a solar charger and found that they all acquired the same voltage. Why is this? I thought that capacitors took energy from the source, which would imply that the second, third, and fourth capacitor had less voltage than the previous one.

 

[Integral]

It's kind'a part of the definition of components in parallel, they all have the same voltage.


Just look at the leads, are not the sides of each of the caps connected together? You should be aware that all points connected by a wire at the same voltage with respect to any reference point.

 

[seanhbailey]

I am aware of this fact, but wouldn't charging the capacitors affect the source, i.e. drain it?

 

[rcgldr]

I am aware of this fact, but wouldn't charging the capacitors affect the source, i.e. drain it?

While the capacitors are charging from an initial state of near zero voltage, they will drain the source, but as they approach the no load voltage of the source, their load on the source also approaches some small amount, based on how "leaky" they are.

 

[S_Happens]

If they're in parallel, which one is first?

 

[rcgldr]

If they're in parallel, which one is first?

If the wiring involved has zero resistance, zero impedeance, and if speed of electrical propagation was infinite, they all charge at the same time and at the same rate. In the real world, the one closest to the voltage source is going to charge up a tiny bit sooner than the others.

 

[seanhbailey]

Ok, so if I had a solar panel source rated at 3V, all the capacitors in parallel would acquire a voltage of 3V?

 

[rcgldr]

Ok, so if I had a solar panel source rated at 3V, all the capacitors in parallel would acquire a voltage of 3V?

Yes, as long as the solar panel could produce 3 volts at the leakage rate current of the capacitors in parallel. 4 capacitors should not be an issue here.

 

[seanhbailey]

Thank you.

 

[russ_watters]

...and as long as the sun is shining on the panel at it's rated input.

 

[mrspeedybob]

Do this. Charge 1 capacitor and disconnect it from the power source, for the sake of argument let's say you charged it to 12v. Now this capacitor becomes your power source, use it to charge the second capacitor and you will find that they are now both at 6 volts. In other words charging the second capacitor drained the power source (the 1'st capacitor) from 12v down to 6v. The reason you don't see this when charging from a battery is that the capacity of the battery is huge compared to that of the capacitor. In the case of your solar cell there is no real "capacity" because it will continue to generate current as long as light is falling on it.

I should state that the above assumes that the capacitors are identical and that leakage will be negligible in the time frame of the experiment.

 

[Naty1]

They'd all have the same voltage if wired in series too..assuming ideal capacitors...but not three volts each...1/4 of 3 volts...

 

[mrspeedybob]

They'd all have the same voltage if wired in series too..assuming ideal capacitors...but not three volts each...1/4 of 3 volts...

That also assumes that they are all the same capacity.

 

[pallidin]

...and as long as the sun is shining on the panel at it's rated input.

I agree. A solar panel does not reduce it's charge potential due to load.
For a solar panel the charge potential is reduced by reduced solar illumination.

 

[S_Happens]

If the wiring involved has zero resistance, zero impedeance, and if speed of electrical propagation was infinite, they all charge at the same time and at the same rate. In the real world, the one closest to the voltage source is going to charge up a tiny bit sooner than the others.

Just for clarity, that was not a serious question for myself. It was a question meant to get the OP to think about what they were saying. Do you yourself expect that because the real world setup can allow one capacitor to start charging first that it makes any difference?

 

[pallidin]

Closer to what source? Positive or negative?
Seems to me that any parallel capacitor will be in equal range of the the TOTAL charging source; some will be closer to the positive, others closer to the negative, but since charging a capacitor requires both positive and negative it doesn't matter.

 

[rcgldr]

Closer to what source? Positive or negative?

Both, it's a parallel circuit, the capacitor array is like the steps of a ladder, and the power source is at one end of the ladder attached to the two legs.