Question

Two capacitors with capacitance values ${\mathrm{C}}_1=2000\pm 10\mathrm{pF}$ and ${\mathrm{C}}_2=3000\pm 15\mathrm{pF}$ are connected in series. The voltage applied across this combination is 𝑉 = 5.00 ± 0.02 V. The percentage error in the calculation of the energy stored in this combination of capacitors is _______.

Answer 1

Step 1: Given data

It has been given that, the capacitance of capacitors ${\mathrm{C}}_1$and ${\mathrm{C}}_2$ are $(2000\pm 10)\mathrm{pF}$ and $(3000\pm 15)\mathrm{pF}$ connected in series. The voltage applied across this combination is $\mathrm{V}=(5\pm 0.02)\mathrm{V}$.

We have to find the percentage error in the calculation of the energy stored in this combination of capacitors.

Step 2: Formula to be used

The energy stored in a capacitor or capacitors are,
$\mathrm{E}=\frac{1}{2}\mathrm{C}{\mathrm{V}}^2$
Here, $\mathrm{E}$ is the energy stored, $\mathrm{C}$ is the capacitance and $\mathrm{V}$ is the potential difference.

The charge is,

$\mathrm{Q}=\mathrm{CV}$$$

Where $\mathrm{Q}$ is the charge on capacitance.

For series combination, the resultant capacitance is,

$\frac{1}{{\mathrm{C}}_{\mathrm{s}}}=\frac{1}{{\mathrm{C}}_1}+\frac{1}{{\mathrm{C}}_2}$

Here, ${\mathrm{C}}_{\mathrm{s}}$ is the resultant capacitance.

Step 3: Find the percentage error in the calculation of the energy stored in this combination of capacitors

Let,

$\frac{1}{{\mathrm{C}}_{\mathrm{eq}}}=\frac{1}{{\mathrm{C}}_1}+\frac{1}{{\mathrm{C}}_2}$ -------- $\left(1\right)$

$=\frac{2000\times 3000}{5000}$

$=1200\mathrm{pF}$

Step 4: Differentiate the equation

Now, differentiate equation $\left(1\right)$. We get,

$\frac{-{\mathrm{dC}}_{\mathrm{eq}}}{{{\mathrm{C}}_{\mathrm{eq}}}^2}=\frac{-{\mathrm{dC}}_1}{{\mathrm{C}}_1^2}-\frac{{\mathrm{dC}}_2}{{\mathrm{C}}_2^2}$

$\frac{{\mathrm{dC}}_{\mathrm{eq}}}{{{\mathrm{C}}_{\mathrm{eq}}}^2}=\frac{{\mathrm{dC}}_1}{{\mathrm{C}}_1^2}+\frac{{\mathrm{dC}}_2}{{\mathrm{C}}_2^2}$

$\frac{{\mathrm{dC}}_{\mathrm{eq}}}{{\left(1200\times {10}^{-12}\right)}^2}=\frac{10\times {10}^{-12}}{{\left(2000\times {10}^{-12}\right)}^2}+\frac{15\times {10}^{-12}}{{\left(3000\times {10}^{-12}\right)}^2}$

${\mathrm{dC}}_{\mathrm{eq}}=6\mathrm{pF}$

We know that,

$\mathrm{U}=\frac{1}{2}{\mathrm{C}}_{\mathrm{eq}}{\mathrm{V}}^2$

Here, $\mathrm{U}$ is energy stored and $\mathrm{V}$ is applied voltage.

So,

$\frac{\mathrm{dU}}{\mathrm{U}}\times 100=\left(\frac{{\mathrm{dC}}_{\mathrm{eq}}}{{\mathrm{C}}_{\mathrm{eq}}}+\frac{2\mathrm{dV}}{\mathrm{V}}\right)\times 100$

$=\left(\frac{6}{1200}+\frac{2\times 0.02}{5}\right)\times 100$

$=\left(5\times {10}^{-3}+8\times {10}^{-3}\right)\times 100$

$=0.013\times 100$

$=1.3\%$

Hence, two capacitors with capacitance values ${\mathrm{C}}_1=2000\pm 10\mathrm{pF}$ and ${\mathrm{C}}_2=3000\pm 15\mathrm{pF}$ are connected in series. The voltage applied across this combination is 𝑉 = 5.00 ± 0.02 V. The percentage error in the calculation of the energy stored in this combination of capacitors is $1.3\%$.