Question

Each capacitor in figure (31-E21) has a capacitance of 10 µF. The emf of the battery is 100 V. Find the energy stored in each of the four capacitors.
Figure

Answer 1

Explanation:

Capacitors b and c are in parallel; their equivalent capacitance is 20 µF.

Thus, the net capacitance of the circuit is given by

$\text { 1 Cnet }=110+120+110 \Rightarrow 1 \text { Cnet }-2+1+220=520 \Rightarrow$ Cnet=4 μF

The total charge of the battery is given by

$Q=\text { Cnet } Y=(4, \mu E) \times(100 \vee)=4 \times 10-4 \text { CE.s } \alpha \text { and } d$$\begin{aligned}&\mathrm{q}=4 \times 10-4 \mathrm{C} \text { and } \mathrm{C}=10-5 \mathrm{F}: \therefore \mathrm{E}=\mathrm{q} 22 \mathrm{C}=4 \times 10-42 \times 10-5 \Rightarrow \mathrm{E}=8 \times 10-\\&3 \mathrm{J}=8 \mathrm{mJFor} b \text { and } c\end{aligned}$$\mathrm{q}=4 \times 10-4 \mathrm{C} \text { and } \mathrm{Ceg}=2 \mathrm{C}=2 \times 10-5 \mathrm{F}$

∴V= q Ceq4$\times$10-42$\times$10-5=20 V⇒ E=12

$\mathrm{CV} 2 \Rightarrow \mathrm{E}=12 \times 10-5 \times 400 \Rightarrow \mathrm{E}=2 \times 10-3 \mathrm{J}=2 \mathrm{mJ}$