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 . . . . . .
IIT 2020, Medium
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$U =\frac{1}{2} \frac{ C _1 C _2}{ C _1+ C _2} V ^2$
Let $C_{\text {oq }}=\frac{C_1 C_2}{ C _1+C_2}$
$\frac{1}{ C _{ eq } \pm \Delta C _{ oq }}=\frac{1}{ C _1 \pm \Delta C _1}+\frac{1}{ C _2 \pm \Delta C _2}$
$\Rightarrow C _{ eq } \pm \Delta C _{ oq } \simeq \frac{ C _1 C _2+ C _1 \Delta C _2+ C _2 \Delta C _1}{ C _1+ C _2+\Delta C _1+\Delta C _2}$
$=\frac{1200\left(1 \pm \frac{12}{1200}\right)}{\left(1 \pm \frac{25}{5000}\right)}$
$=1200\left[1 \pm\left(\frac{1}{100}-\frac{1}{200}\right)\right]$
$\frac{\Delta U }{ U } \times 100=\frac{\Delta C _{ oq }}{ C _{ eq }} \times 100+\frac{2 \Delta V }{ V } \times 100$
$=\frac{1}{200} \times 100+2 \times \frac{0.02}{5} \times 100$
$=1.3 \%$
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