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JEE Advanced · Physics · 1. Math in Physics

Two capacitors with capacitance values C1=2000±10 pF  and C2=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 __________.

  1. A 1.4
  2. B 1.3
  3. C 1.2
  4. D 1.1
Verified Solution

Answer & Solution

Correct Answer

(B) 1.3

Step-by-step Solution

Detailed explanation

For the purpose of calculation of error, fundamental formula is considered
1C=1C1+1C2C=1200 pF
-dCC12=-dC1C12-dC2C22
dC=6 pF
Equivalent capacitance =1200±6 pF
E=1/2 CV2
(dE/E=dC/C+2dV/V)×100
=1.3%
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