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Capacitor question

2010 · Q187
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Capacitor question

2010 · Q187

NEETPhysicsCapacitorMCQ+4 / −1
A series combination of n1 capacitors, each of value C1, is charged by a source of potential difference 4V. When another parallel combination of n2 capacitors, each of value C2, is charged by a source of potential difference V, it has the same (total) energy stored in it. as the first combination has. The value of C2. in terms of C1, is then
  1. A
    2C1n1n2{{2{C_1}} \over {{n_1}{n_2}}}n1​n2​2C1​​
  2. B
    16n2n1C116{{{n_2}} \over {{n_1}}}{C_1}16n1​n2​​C1​
  3. C
    2n2n1C12{{{n_2}} \over {{n_1}}}{C_1}2n1​n2​​C1​
  4. D
    16C1n1n2{{16{C_1}} \over {{n_1}{n_2}}}n1​n2​16C1​​
View written solutionFree

Correct answer: D

A series combination of n1 capacitors each of capacitance C1 are connected to 4V source.

Total capacitance of the series combination of the capacitors is

1/Cs = 1/C1 + 1/C1 + 1/C1 .... upto n1 terms = n1/C1

⇒\Rightarrow⇒ Cs = C1/n1

Total energy stored in a series combination of the capacitors is

Us = (1/2) Cs (4V)2 = (1/2) (C1/n1) (4V)2    …(1)

Now a parallel combination of n2 capacitors each of capacitance C2 are connected to V source

Total capacitance of the parallel combination of capacitors is

Cp = C1 + C2 + ... + upto n2 terms = n2C2

⇒\Rightarrow⇒ Cp = n2C2

Total energy stored in a parallel combination of capacitors is

Up = (1/2)CpV2 = (1/2)(n2C2 )(V)2    ...(ii)

According to the given problem, Us = Up

(1/2) (C1/n1)(4V)2 = (1/2)(n2C2)(V)2

C1 × 16/n1 = n2C2

C2 = 16 × C1 / (n1 × n2)

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