QJEEAdv20Ph6739: JEE Advanced 2020 Physics - Class 12 JEEAdv20Ph6739 Solution: Two capacitors with capacitance values C 1 = 2000 10 pF and C 2 …

Question

Verified Solution • 2025–26 Session moderate

Two capacitors with capacitance values C₁ = 2000

\pm

10 pF and C₂ = 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 ________.

Accepted Answer

Given that,

C_1=(2000 \pm 10) \mathrm{pF} ; \quad C_2=(3000 \pm 15) \mathrm{pF}

and

V=(5.00 \pm 8.02) \mathrm{V}

.

Combine capacitance of series combination

\begin{aligned}\frac{1}{C} &amp; =\frac{1}{C_1}+\frac{1}{C_2} ~~~~....(1)\\\\\Rightarrow &amp; \frac{1}{C}=\frac{1}{2000}+\frac{1}{3000}=\frac{(6+4)}{12000} \\\\\frac{1}{C} &amp; =\frac{10}{12000} \Rightarrow C=1200 \mathrm{pF}\end{aligned}

Differentiating Eq. (1), we get

\begin{aligned}-\frac{1}{C^2} \cdot d C &amp; =-\frac{1}{C_1^2} d C_1+\left(-\frac{1}{C_2^2}\right) \cdot d C_2 \\\\\Rightarrow \frac{d C}{C^2} &amp; =\left(\frac{d C_1}{C_1^2}+\frac{d C_2}{C_2^2}\right) \\\\\Rightarrow \frac{d C}{C} &amp; =\left[\frac{d C_1}{C_1^2}+\frac{d C_2}{C_2^2}\right] \cdot C \\\\\Rightarrow \frac{d C}{C} &amp; =\left[\frac{10}{(2000)^2}+\frac{15}{(3000)^2}\right] \times 1200 \\\\\Rightarrow \frac{d C}{C} &amp; =\left[\frac{10}{4} \times 10^{-6}+\frac{15}{9} \times 10^{-6}\right] \times 1200 \\\\&amp; =\left[\frac{10}{4}+\frac{15}{9}\right] \times 12 \times 10^{-4}\end{aligned}

\Rightarrow \frac{d C}{C}=[2.5+1.66] \times 12 \times 10^{-4}=[4.17 \times 12] \times 10^{-4}

And

\frac{d V}{V}=\frac{0.02}{5}

Energy stored in the capacitor is given by

U=\frac{1}{2} C V^2

.....(2)

Differentiating Eq. (2), we get

\begin{aligned}\frac{d U}{U} \times 100 &amp; =\frac{d C}{C} \times 100+2 \frac{d V}{V} \times 100 \\\\&amp; =\left[50.04 \times 10^{-4} \times 100+\frac{2 \times 0.02}{5} \times 100\right] \\\\\left(\frac{d U}{U}\right)^{100} &amp; =[0.5004+0.8] \% \Rightarrow 1.30 \%\end{aligned}

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