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JEE Advanced Physics Mathematics in Physics 2020 JEE Advanced 2020 (Paper 2)

JEE Advanced Physics Question (2020) — Solution

Question

Two capacitors with capacitance values C 1 = 2000 ± 10   pF   and  C 2 = 3000 ± 15   pF  are connected in series. The voltage applied across this combination is  V = 5 . 00 ± 0 . 02   V . The percentage error in the calculation of the energy stored in this combination of capacitors is __________.

Options

  1. A. A
  2. B. B
  3. C. C
  4. D. D

Answer

A. A

Step-by-step solution

For the purpose of calculation of error, fundamental formula is considered 1 C = 1 C 1 + 1 C 2 ⇒ C = 1200   p F - d C C 1 2 = - d C 1 C 1 2 - d C 2 C 2 2 d C = 6   p F Equivalent capacitance = 1200 ± 6   pF E = 1 / 2   CV 2 ( d E / E = d C / C + 2 d V / V ) × 100 = 1 . 3 %

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