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PhysicsNumerical+4Question 15 of 54

JEE Advanced 2020 · Paper 2, Physics question 15

Two capacitors with capacitance values C1=2000±10C_1 = 2000 \pm 10 pF and C2=3000±15C_2 = 3000 \pm 15 pF are connected in series. The voltage applied across this combination is V=5.00±0.02V = 5.00 \pm 0.02 V. The percentage error in the calculation of the energy stored in this combination of capacitors is ______.

Answer and worked solution

Official answer: 1.3

Series capacitance.
C=C1C2C1+C2=1200 pF.C=\frac{C_1C_2}{C_1+C_2}=1200\text{ pF}.

Its error. From 1C=1C1+1C2\frac1C=\frac1{C_1}+\frac1{C_2}:
ΔCC2=ΔC1C12+ΔC2C22  ⟹  ΔCC=1200(1020002+1530002)=0.005=0.5%.\frac{\Delta C}{C^2}=\frac{\Delta C_1}{C_1^2}+\frac{\Delta C_2}{C_2^2}\implies\frac{\Delta C}C=1200\left(\frac{10}{2000^2}+\frac{15}{3000^2}\right)=0.005=0.5\%.

Energy. U=12CV2U=\frac12CV^2, so
ΔUU=0.5%+2×0.025.00×100%=0.5%+0.8%=1.3%.\frac{\Delta U}U=0.5\%+2\times\frac{0.02}{5.00}\times100\%=0.5\%+0.8\%=1.3\%.
Answer: 1.30

Topics:Errors in measurementSeries and parallel combinationsEnergy stored in a capacitor
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