Answer to Question #216991 in Electricity and Magnetism for david chulapani

Question

For the system of capacitors shown in Figure, find :

a. The equivalent capacitance of the system

b. The potential across each capacitor

c. The charge on each capacitor

d. The total energy stored by the group

picrure reference : https://ibb.co/Sn6fwRv

Answer 1

$$C_1=3\mu F,C_2=6\mu F,C_3=2\mu F,C_4=4\mu F$$

$$C'=\frac{C_1C_2}{C_1+C_2}=\frac{3\times6}{3+6} \times10^{-6}=2\times10^{-6}F$$

$$C''=\frac{C_3C_4}{C_3+C_4}=\frac{2\times4}{2+4} \times10^{-6}=\frac{4}{3}\times10^{-6}F=1.33\mu F$$

$$C_{net}=C'+C''={3}=2+1.33=3.33\mu F$$

Series combination charge is same

$$C=\frac{Q}{∆V}\\Q'={90}\times{2}=180\mu C$$

$Q''=1.33\times90=120\mu C$

$C_1,C_2$ Each capacitor charge 180$\mu C$

$C_3,C_4$ each capacitor charge 120$\mu C$

C_1,C₂ ,C₃,C₄net potential

$V_{c_1}=\frac{Q'}{C_1}=\frac{180}{3}=60V\\V_{c_2}=\frac{Q'}{C_2}=\frac{180}{6}=30V$$V_{c_3}=\frac{Q''}{C_3}=\frac{120}{2}=60V\\V_{c_2}=\frac{Q''}{C_2}=\frac{120}{4}=30V$

Q=CV

$U=\frac{1}{2}C_{net}V^2$

$$U=\frac{1}{2}\times 3.33\times10^{-6}\times90^2=13.4mJ$$