You connect three capacitors as shown in the diagram below. If the potential difference between A and B is 24.5-V, what is the total energy stored in this system of capacitors if C₁ = 5.0 μF, C₂ = 4.0 μF, and C₃ = 3.0 μF?

Question

Quizplus · Accepted Answer

The total energy stored in the system of capacitors can be found using the formula:
$E=\frac{1}{2}\left(\frac{Q_1^2}{C_1}+\frac{Q_2^2}{C_2}+\frac{Q_3^2}{C_3}ight)$
where $Q_1$, $Q_2$, and $Q_3$ are the charges stored on the respective capacitors, and $C_1$, $C_2$, and $C_3$ are their capacitances. 
The charges on all capacitors are equal and can be found using:
$Q=CV$
where $V$ is the potential difference between A and B. 
$Q=CV=5.0	imes10^{-6}	ext{ F}	imes24.5	ext{ V}=1.225	imes10^{-4}	ext{ C}$
Now we can calculate the energy:
$E=\frac{1}{2}\left(\frac{(1.225	imes10^{-4}	ext{ C})^2}{5.0	imes10^{-6}	ext{ F}}+\frac{(1.225	imes10^{-4}	ext{ C})^2}{4.0	imes10^{-6}	ext{ F}}+\frac{(1.225	imes10^{-4}	ext{ C})^2}{3.0	imes10^{-6}	ext{ F}}ight)$
$E=6.8	imes10^{-4}	ext{ J}=6.8	imes10^{-4}	imes10^{-10}	ext{PJ}=6.8	imes10^{-14}	ext{ PJ}$
Therefore, the correct answer is option D.

You Connect Three Capacitors as Shown in the Diagram Below