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6 identical capacitors are joined in parallel and are charged with a battery of 10 V. Now the battery is removed and they are joined in series with each other. In this condition, what would be the potential difference between the free plates of the combination?

Asked in GSEB Board July 2016 · Redistribution of charge

• (1) 30 V
• (2) 10 V
• (3) 60 V
• (4) (10)/6 V

Step-by-step solution

Why the other options are wrong

• (1) No single slip gives 30 V. Sanity check: each capacitor still holds the charge it got at 10 V, and six such voltages add in series to 6×10 = 60 V.
• (2) Assumes the combination still sits at the battery voltage. That is true only in parallel; in series (the positive plate of one joined to the negative plate of the next) the six 10 V drops add up.
• (4) Divides where it should multiply: treats the 10 V as if it were shared among six capacitors in series, as when a battery drives a series chain. Here each capacitor keeps its own 10 V, so the total is 6×10 V.

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