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When 2 capacitors (lets say, of same capacitence 1F) are connected to a battery of 1V (a source of charges), then the capacitors take some energy from the battery and put some charge inside them (Q=CV=1Coulomb for both capacitors). Which means the battery now has less energy.
Hence I conclude, the battery with 2 capacitors in parallel will drain out faster than a battery with individual capacitors (considering we charge the capacitors many many times, causing the battery to loose the energy). Now does this all make sense or its just baloney? Now does this all make sense or it's just baloney? It's just baloney.
Now, if those 2 capacitors were a part of seperate circuits with same volatages, they would still aquire same charge (Q=CV=1 Coulomb) as they did when they were connected in parallel. But in this case the individual batteries ( which are part of 2 seperate circuits with a capacitor each) will only loose lower energy each.
Even "directly in parallel with the batteries" isn't really directly in parallel with the batteries, thanks to wiring resistances. The capacitor should have the closest and most direct connection to the load, then this pair should be connected to the battery via wiring which gives you some control of the current drawn from the battery.
1. Suppose a parallel plate capacitor (with capacitance C1) is fully charged (to a value Qo) by a battery. The battery (which supplies a potential difference of Vo) is then disconnected.
The net charge on the combination of the two plates of the capacitor is the same (zero) before and after charging so no charge has been "supplied" by the battery. The positive terminal of the battery pulls electrons off of the capacitor plate connected to it, making that plate positively charged.
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When 2 capacitors (lets say, of same capacitence 1F) are connected to a battery of 1V(a source of charges), then the capacitors take some energy from the battery and …
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When capacitors are connected together in parallel the total or equivalent capacitance, C T in the circuit is equal to the sum of all the individual capacitors added together. This is because the top plate of capacitor, C 1 is …
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