Q51 / 54 · JEE Advanced 2020 (Paper 2) · 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 __________.

Correct answer: 1.3

Solution

For the purpose of calculation of error, fundamental formula is considered

1C=1C1+1C2⇒C=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%

Hence, the correct answer is 1.3.

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