Percentage Error in Stored Energy of Series Connected Capacitors

To find the percentage error in the total energy stored in the series combination of capacitors, we analyze the energy equation $U = \frac{1}{2} C_{\text{eq}} V^2$.

The equivalent capacitance $C_{\text{eq}}$ for two capacitors connected in series is given by: $\frac{1}{C_{\text{eq}}} = \frac{1}{C_1} + \frac{1}{C_2}$

Differentiating to find the maximum error in $C_{\text{eq}}$ gives: $\frac{\Delta C_{\text{eq}}}{C_{\text{eq}}^2} = \frac{\Delta C_1}{C_1^2} + \frac{\Delta C_2}{C_2^2} \implies \frac{\Delta C_{\text{eq}}}{C_{\text{eq}}} = \left(\frac{C_{\text{eq}}}{C_1}\right) \frac{\Delta C_1}{C_1} + \left(\frac{C_{\text{eq}}}{C_2}\right) \frac{\Delta C_2}{C_2}$

Given:

• $C_1 = 2000\text{ pF}$, $\Delta C_1 = 10\text{ pF} \implies \frac{\Delta C_1}{C_1} = 0.5\%$
• $C_2 = 3000\text{ pF}$, $\Delta C_2 = 15\text{ pF} \implies \frac{\Delta C_2}{C_2} = 0.5\%$
• $C_{\text{eq}} = \frac{2000 \times 3000}{2000 + 3000} = 1200\text{ pF}$
• $V = 5.00\text{ V}$, $\Delta V = 0.02\text{ V} \implies \frac{\Delta V}{V} = 0.4\%$

Calculating the fractional error in $C_{\text{eq}}$: $\frac{\Delta C_{\text{eq}}}{C_{\text{eq}}} = \left(\frac{1200}{2000}\right)(0.5\%) + \left(\frac{1200}{3000}\right)(0.5\%) = 0.3\% + 0.2\% = 0.5\%$

The total percentage error in the energy $U$ is: $\frac{\Delta U}{U} \times 100 = \frac{\Delta C_{\text{eq}}}{C_{\text{eq}}} \times 100 + 2 \left(\frac{\Delta V}{V} \times 100\right) = 0.5\% + 2(0.4\%) = 1.3\%$

The percentage error in the calculation of the stored energy is 1.3 (or 1.30%).