# Capacitor loop charging

During my journey to electronics I came across several questions about capacitors.

1. If we connect an empty capacitor to a fully charged re-chargeable battery, after the battery charged the capacitor, it then became empty (wires still connected and never touched), what will happen in this situation? Will the capacitor re-charge the battery? Or will its charge drop due to the battery’s internal resistance without charging it?

2. If we replaced the battery with a charged capacitor (empty capacitor and charged capacitor with same values) what will happen in this situation?

3. If we connect a charged capacitor and an empty capacitor and a charged battery all in parallel, what will happen?

Any help to clear my doubts is preciously appreciated.

• It looks like a homework assignment from a Community College class. Aug 19, 2016 at 16:44

Any help to clear my doubts is preciously appreciated

Consider two tanks of water: -

Tank 1 is your fully charged battery and tank 2 is your capacitor. After a short while the levels will equalize. Tank 1 will have dropped its height a little bit in order to fill up tank 2 to a pretty high level. If you had a really massively big capacitor (tank 2) you might deplete tank 1 quite a lot.

Voltage = water level in each tank and tank cross sectional area (plan view) is equivalent to capacitance.

This pretty much applies to all mentioned scenarios.

Take any big cap and look up the leakage R and thus the RC decay time constant can be seconds , minutes or sometimes longer.

Consider a battery as a massive cap with 10k Farad capacitance depending on A-HR capacity.

This is mainly the difference and the effective RC time constant.

Batteries even they have a similar leakage resistance will have a much longer decay time constant. Alkaline and Lithium can last years.

Sometimes it helps to think of things in a different mindset. I may suggest thinking of capacitors as tanks of water with the flow inlet/outlet at the bottom. And think of a voltage source as a pump. This may help you intuitively answer these questions.

However, it is important to understand the fundamental equation that describes current flow into a capacitor:

I(t) = C dV(t)/dt

and that current is just the rate of charge:

I=Q/t

Watching this video may help.

1. I do not think the battery will be able to completely charge the capacitor. Initially the due to a difference in potential between the battery and capacitor, the latter will start getting charged as a result its potential will increase causing decrease in the potential difference between battery and capacitor. This the after some time the battery will be able to charge the capacitor by half of its value.

2. Same concept applies to a charged capacitor. Both them will end up having same potential as a result zero potential difference.

3. Same potential in this as well. Think in terms of potential and potential difference you will be able get the answer.