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Question

Six identical capacitors each of 2$$\mu$$F are joined in parallel and each is charged to 10V. They are then disconnected and joined in series so that positive plate of one is joined to the negative plate of the adjacent capacitor. What is the potential difference of the combination?


A
10 V
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B
30 V
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C
60 V
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D
120 V
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Solution

The correct option is C 60 V
When they are then disconnected and joined in series so that positive plate of one is joined to the negative plate of the adjacent capacitor, each capacitor has potential $$10 V$$.
The potential difference of the combination $$=10+10+10+10+10+10=60 V$$
234977_142490_ans.png

Physics
NCERT
Standard XII

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