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Q.

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 _______.

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answer is 1.30.

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Detailed Solution

U=12C1C2C1+C2V2 Let Ceq=C1C2C1+C2=200030002000+3000=1200pFSince, 1Ceq=1C1+1C2⇒−dCeqCeq2=−dC1C12−dC2C22⇒dCeq12002=1020002+1530002⇒dCeq=6pFU=12CeqV2⇒ΔUU×100=ΔCeqCeq×100+2ΔVV×100⇒ΔUU×100=61200×100+2×0.025×100⇒ΔUU×100=1.3%
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