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If n capacitors each of capacitance C are connected in series with a battery of V volt, then electrostatic energy stored in all the capacitors will be
  • a)
    nCV2
  • b)
    1 4 nCV2
  • c)
    1 2n CV2
  • d)
    None of these
Correct answer is option 'C'. Can you explain this answer?
Verified Answer
If n capacitors each of capacitance C are connected in series with a b...
The equivalent capacitance Cs of these series-connected capacitors is
1/C s = 1/C + 1/C + 1/C ..... n times = n/C
∴ Cs = C/ n
Energy stored in all capacitors = 1 2 CsV2 = 1/2 C/n V2 = 1/2n CV2
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Most Upvoted Answer
If n capacitors each of capacitance C are connected in series with a b...
Explanation:

When n capacitors each of capacitance C are connected in series with a battery of V volt, then the effective capacitance of the combination is given by:

1/Ceq = 1/C1 + 1/C2 + … + 1/Cn

Ceq = C/n

The potential difference across each capacitor will be V as they are connected in series.

The electrostatic energy stored in each capacitor is given by:

U = 1/2 * C * V^2

Therefore, the total electrostatic energy stored in n capacitors connected in series is given by:

UTotal = n * U

UTotal = n * (1/2 * C * V^2)

UTotal = 1/2 * n * C * V^2

Substituting Ceq = C/n

UTotal = 1/2 * n * Ceq * V^2

UTotal = 1/2 * n * (C/n) * V^2

UTotal = 1/2 * V^2 * C

Hence, the correct option is (c) 1/2nCV^2.
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