Answer
a. 1.91 mJ.
b. 0.424 mJ.
c. 3.81 mJ.
Work Step by Step
a. The circuit consists of a $25.4\mu F$ capacitor in series with a $25.4\mu F$ capacitor, and the combination is in parallel with a $25.4\mu F$ capacitor.
$$C_{eq}= (\frac{1}{25.4\mu F }+\frac{1}{25.4\mu F })^{-1}+25.4\mu F =38.1\mu F $$
Use the equivalent capacitance to calculate the stored energy.
$$PE=\frac{1}{2}C_{eq}V^2=\frac{1}{2}(38.1\times10^{-6} F)(10.0V)^2=1.91\times10^{-3}J$$
b. Now all the identical capacitors are in series. The equivalent capacitance is 1/3 the value of one of the capacitors.
$$C_{eq}= (\frac{1}{25.4\mu F }+\frac{1}{25.4\mu F }+\frac{1}{25.4\mu F })^{-1} =8.467\mu F $$
Use the equivalent capacitance to calculate the stored energy.
$$PE=\frac{1}{2}C_{eq}V^2=\frac{1}{2}(8.467\times10^{-6} F)(10.0V)^2=4.23\times10^{-4}J$$
c. Now all the identical capacitors are in parallel. The equivalent capacitance is 3 times the value of one of the capacitors.
$$C_{eq}=25.4\mu F +25.4\mu F +25.4\mu F =76.2\mu F $$
Use the equivalent capacitance to calculate the stored energy.
$$PE=\frac{1}{2}C_{eq}V^2=\frac{1}{2}(76.2\times10^{-6} F)(10.0V)^2=3.81\times10^{-3}J$$
