Question

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

Answer 1

Energy stored is given by,
$U=\frac{1}{2}\frac{C_1{C}_2}{C_1+C_2}{V}^2$
Let $C_{\text{eq}}=\frac{C_1{C}_2}{C_1+C_2}$
$\frac{1}{C_{eq}\pm \Delta C_{eq}}=\frac{1}{C_1\pm \Delta C_1}+\frac{1}{C_2\pm \Delta C_2}$
$\Rightarrow C_{eq}\pm \Delta C_{eq}\simeq \frac{C_1{C}_2+C_1\Delta C_2+C_2\Delta C_1}{C_1+C_2+\Delta C_1+\Delta C_2}$
$\Rightarrow C_{eq}\pm \Delta C_{eq}=\frac{1200(1\pm \frac{12}{1200})}{(1\pm \frac{25}{5000})}$

$C_{eq}\pm \Delta C_{eq}=1200[1\pm (\frac{1}{100}-\frac{1}{200})]$

$C_{eq}\pm \Delta C_{eq}=1200\pm 6$ pF

$U=\frac{1}{2}\Delta C_{eq}{V}^2$

$\frac{\Delta U}{U}\times 100=\frac{\Delta C_{eq}}{C_{eq}}\times 100+\frac{2\Delta V}{V}\times 100$

$\Rightarrow \frac{\Delta U}{U}\times 100=\frac{1}{200}\times 100+2\times \frac{0.02}{5}\times 100$

$\Rightarrow \frac{\Delta U}{U}\times 100=1.3\%$