JEE Advanced · 2020 · 27 Sep 2020 · Paper 2 · Units & Measurement

Two capacitors with capacitance values

C_{1} = 2000 \pm 10

pF and

C_{2} = 3000 \pm 15

pF are connected in series. The voltage applied across this combination is

V = 5.00 \pm 0.02

V. The percentage error in the calculation of the energy stored in this combination of capacitors is ____.

Answer: 1.3

Solution

The stored energy is

U = \frac{1}{2} C_{e q} V^{2}

with

C_{e q} = \frac{C _{1} C _{2}}{C _{1} + C _{2}} = 1200

pF. Because

C_{e q}

depends on both capacitances, its absolute error is

Δ C_{e q} = \frac{C _{2}^{2} Δ C _{1} + C _{1}^{2} Δ C _{2}}{( C _{1} + C _{2} ) ^{2}} = \frac{300 0 ^{2} \times 10 + 200 0 ^{2} \times 15}{500 0 ^{2}} = 6

pF, so

\frac{Δ C _{e q}}{C _{e q}} = \frac{6}{1200} = 0.5%

. Adding the voltage term,

\frac{Δ U}{U} = \frac{Δ C _{e q}}{C _{e q}} + 2 \frac{Δ V}{V} = 0.5% + 2 \times \frac{0.02}{5.00} = 0.5% + 0.8% = 1.3%

JEE Advanced · 2020 · 27 Sep 2020 · Paper 2 · Numerical answer