Question

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 _____.

Answer 1

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