Two capacitors with capacitance values C1=20±1μF and C2=30±1.5μF are connected in series. The voltage across the combination is 10±0.01V .The percentage error in the calculation of energy stored in the combination of capacitors is 26x%. The value of x is

1C=1C1+1C2=120+130⇒C=12ΔCC2=ΔC1C12+ΔC2C22ΔCC=ΔC1C12+ΔC2C22C1C2C1+C2ΔCC=1400+1.5900×12=5100=120U=12C×V2ΔUU=ΔCC+2ΔVV120+2×0.0110=1+0.0420=0.052% change =5.2%

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Two capacitors with capacitance values $C_{1} = 20 \pm 1 μF$ and $C_{2} = 30 \pm 1.5 μF$ are connected in series. The voltage across the combination is $10 \pm 0.01 V$ .The percentage error in the calculation of energy stored in the combination of capacitors is $\frac{26}{x} %$ . The value of $x$ is

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answer is 5.

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Detailed Solution

$\frac{1}{C} = \frac{1}{C_{1}} + \frac{1}{C_{2}} = \frac{1}{20} + \frac{1}{30} \Rightarrow C = 12$
$\begin{array}{l} \frac{ΔC}{C^{2}} = \frac{{ΔC}_{1}}{C_{1}^{2}} + \frac{{ΔC}_{2}}{C_{2}^{2}} \\ \frac{ΔC}{C} = (\frac{{ΔC}_{1}}{C_{1}^{2}} + \frac{{ΔC}_{2}}{C_{2}^{2}}) \frac{C_{1} C_{2}}{C_{1} + C_{2}} \\ \frac{ΔC}{C} = (\frac{1}{400} + \frac{1.5}{900}) \times 12 = \frac{5}{100} = \frac{1}{20} \\ U = \frac{1}{2} C \times V^{2} \\ \frac{ΔU}{U} = \frac{ΔC}{C} + \frac{2 ΔV}{V} \\ \frac{1}{20} + \frac{2 \times 0.01}{10} = \frac{1 + 0.04}{20} = 0.052 \end{array}$
% change =5.2%