Question

Two capacitors with capacitance values $C_1=2000\pm 10pF$ and $C_2=3000\pm 15pF$ are connected in series. The voltage applied across this combination is $V=5.00\pm 0.02V$. The percentage error in the calculation of the energy stored in this combination of capacitors is _____.

Answer 1

Answer: 1.30

$\frac{1}{C}=\frac{1}{C_1}+\frac{1}{C_2}$
$\Rightarrow \frac{1}{C^2}\Delta C=\frac{1}{C_1^2}\Delta C_1+\frac{1}{C_2^2}\Delta C_2$
$\Rightarrow \frac{\Delta C}{C}=C\left(\frac{\Delta C_1}{C_1^2}+\frac{\Delta C_2}{C_2^2}\right)$
$=1200\left(\frac{10}{4\times 10^6}+\frac{15}{9\times 10^6}\right)=5\times 10^{-3}$
Energy stored
$U=\frac{1}{2}C{V}^2\Rightarrow \frac{\Delta U}{U}=\frac{\Delta C}{C}+2\frac{\Delta V}{V}$
Percentage error in energy stored in the combination of capacitors.
$\frac{\Delta U}{U}\times 100=\left(5\times 10^{-3}+2\frac{0.02}{5}\right)\times 100=1.30\%$