Question

A four Capacitor circuit

A four-capacitor circuit is chargcd by a battery, as shown in the figure. The capacitances are C_1 = 1.8 mF, C_2 = 1.4 mF, C_3 = 3.6 mF, and C_4 = 4.8 mF, and the battery potential is V_B = 5.0 V. When the circuit is at equilibrium, point D has potential V_D = 0 V. What is the potential, V_A , at point A ?

Answer 1

here,

c1 = 1.8 mF = 0.0018 F
c2 = 2.4 mF = 0.0024 F
c3 = 3.6 mF = 0.0036 F
c4 = 4.8 mF = 0.0048 F

as c1 and c2 are in parallel,
c' = c1 + c2

also c3 and c4 are in parallel,
c'' = c3+c4

now c' and c'' become in series therefore equivalent capacitane will be ,
c = (c'*c'') / (c' + c'')
c = ( (c1 + c2) * (c3+c4)) / (((c1 + c2)) + (c3+c4))
c = ( (0.0018 + 0.0024) * (0.0036+0.0048)) / (((0.0018 + 0.0024)) + (0.0036+0.0048))
c = 0.0028 F

Net charge across circuit will be,

Q = c * v
Q = 0.0028 * 5
Q = 0.014 C

as c' and c'' were in series so charge will remain across them, therefore potential drop after c' will be ,

V = Q/c'
V = 0.014/(c1+c2)
V = 0.014/(0.0018 + 0.0024)
V = 3.33 V