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

Question #216991

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



1
Expert's answer
2021-07-14T13:39:23-0400
C1=3μF,C2=6μF,C3=2μF,C4=4μFC_1=3\mu F,C_2=6\mu F,C_3=2\mu F,C_4=4\mu F

C=C1C2C1+C2=3×63+6×106=2×106FC'=\frac{C_1C_2}{C_1+C_2}=\frac{3\times6}{3+6} \times10^{-6}=2\times10^{-6}F

C=C3C4C3+C4=2×42+4×106=43×106F=1.33μFC''=\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

Cnet=C+C=3=2+1.33=3.33μFC_{net}=C'+C''={3}=2+1.33=3.33\mu F

Series combination charge is same


C=QVQ=90×2=180μCC=\frac{Q}{∆V}\\Q'={90}\times{2}=180\mu C

Q=1.33×90=120μCQ''=1.33\times90=120\mu C

C1,C2C_1,C_2 Each capacitor charge 180μC\mu C

C3,C4C_3,C_4 each capacitor charge 120μC\mu C

C1,C2 ,C3,C4net potential

Vc1=QC1=1803=60VVc2=QC2=1806=30VV_{c_1}=\frac{Q'}{C_1}=\frac{180}{3}=60V\\V_{c_2}=\frac{Q'}{C_2}=\frac{180}{6}=30VVc3=QC3=1202=60VVc2=QC2=1204=30VV_{c_3}=\frac{Q''}{C_3}=\frac{120}{2}=60V\\V_{c_2}=\frac{Q''}{C_2}=\frac{120}{4}=30V

Q=CV




U=12CnetV2U=\frac{1}{2}C_{net}V^2

U=12×3.33×106×902=13.4mJU=\frac{1}{2}\times 3.33\times10^{-6}\times90^2=13.4mJ


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