JEE Advance - Physics (2020 - Paper 2 Offline - No. 15)
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 ________.
Answer
1.3
Explanation
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
$$ \begin{aligned} \frac{1}{C} & =\frac{1}{C_1}+\frac{1}{C_2} ~~~~....(1)\\\\ \Rightarrow & \frac{1}{C}=\frac{1}{2000}+\frac{1}{3000}=\frac{(6+4)}{12000} \\\\ \frac{1}{C} & =\frac{10}{12000} \Rightarrow C=1200 \mathrm{pF} \end{aligned} $$
Differentiating Eq. (1), we get
$$ \begin{aligned} -\frac{1}{C^2} \cdot d C & =-\frac{1}{C_1^2} d C_1+\left(-\frac{1}{C_2^2}\right) \cdot d C_2 \\\\ \Rightarrow \frac{d C}{C^2} & =\left(\frac{d C_1}{C_1^2}+\frac{d C_2}{C_2^2}\right) \\\\ \Rightarrow \frac{d C}{C} & =\left[\frac{d C_1}{C_1^2}+\frac{d C_2}{C_2^2}\right] \cdot C \\\\ \Rightarrow \frac{d C}{C} & =\left[\frac{10}{(2000)^2}+\frac{15}{(3000)^2}\right] \times 1200 \\\\ \Rightarrow \frac{d C}{C} & =\left[\frac{10}{4} \times 10^{-6}+\frac{15}{9} \times 10^{-6}\right] \times 1200 \\\\ & =\left[\frac{10}{4}+\frac{15}{9}\right] \times 12 \times 10^{-4} \end{aligned} $$
$$ \Rightarrow \frac{d C}{C}=[2.5+1.66] \times 12 \times 10^{-4}=[4.17 \times 12] \times 10^{-4} $$
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
$$ \begin{aligned} \frac{d U}{U} \times 100 & =\frac{d C}{C} \times 100+2 \frac{d V}{V} \times 100 \\\\ & =\left[50.04 \times 10^{-4} \times 100+\frac{2 \times 0.02}{5} \times 100\right] \\\\ \left(\frac{d U}{U}\right)^{100} & =[0.5004+0.8] \% \Rightarrow 1.30 \% \end{aligned} $$
Combine capacitance of series combination
$$ \begin{aligned} \frac{1}{C} & =\frac{1}{C_1}+\frac{1}{C_2} ~~~~....(1)\\\\ \Rightarrow & \frac{1}{C}=\frac{1}{2000}+\frac{1}{3000}=\frac{(6+4)}{12000} \\\\ \frac{1}{C} & =\frac{10}{12000} \Rightarrow C=1200 \mathrm{pF} \end{aligned} $$
Differentiating Eq. (1), we get
$$ \begin{aligned} -\frac{1}{C^2} \cdot d C & =-\frac{1}{C_1^2} d C_1+\left(-\frac{1}{C_2^2}\right) \cdot d C_2 \\\\ \Rightarrow \frac{d C}{C^2} & =\left(\frac{d C_1}{C_1^2}+\frac{d C_2}{C_2^2}\right) \\\\ \Rightarrow \frac{d C}{C} & =\left[\frac{d C_1}{C_1^2}+\frac{d C_2}{C_2^2}\right] \cdot C \\\\ \Rightarrow \frac{d C}{C} & =\left[\frac{10}{(2000)^2}+\frac{15}{(3000)^2}\right] \times 1200 \\\\ \Rightarrow \frac{d C}{C} & =\left[\frac{10}{4} \times 10^{-6}+\frac{15}{9} \times 10^{-6}\right] \times 1200 \\\\ & =\left[\frac{10}{4}+\frac{15}{9}\right] \times 12 \times 10^{-4} \end{aligned} $$
$$ \Rightarrow \frac{d C}{C}=[2.5+1.66] \times 12 \times 10^{-4}=[4.17 \times 12] \times 10^{-4} $$
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
$$ \begin{aligned} \frac{d U}{U} \times 100 & =\frac{d C}{C} \times 100+2 \frac{d V}{V} \times 100 \\\\ & =\left[50.04 \times 10^{-4} \times 100+\frac{2 \times 0.02}{5} \times 100\right] \\\\ \left(\frac{d U}{U}\right)^{100} & =[0.5004+0.8] \% \Rightarrow 1.30 \% \end{aligned} $$
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