Q: Two capacitors with capacitance values C_{1} = 2000 ± 10 pF and C_{2} = 3000 ± 15 pF are connected in series . The voltage applied across this combination is C = 5.00 ±0.02 V . The percentage error in the calculation of the energy stored in this combination is ………….

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$\displaystyle C = \frac{C_1 C_2}{C_1 + C_2} $

$\displaystyle -\frac{dC}{C^2} = -\frac{dC_1}{C_1^2} – \frac{dC_2}{C_2^2} $

dC = 6 pF

$\displaystyle U = \frac{1}{2}CV^2 $

$\displaystyle \frac{dU}{U} \times 100 = (\frac{dC}{C} + \frac{2dV}{V})\times 100 $

$\displaystyle \frac{dU}{U} \times 100 = (\frac{6}{1200} + 2\times \frac{0.02}{5})\times 100 $

= 13 %