Question

Four capacitors are connected as shown in the figure below. (C = 16.0

Answer 1

ans a : 5.98uF

ans b :140.18uC , 42.056uC , 5.65uC , 17.74uC

Cp = C1 + C2

Cs = C1C2/C1+C2

Cx= 16 x 3/ (16 + 3) = 2.53uF( c and 3)

Cy = 2.53 + 6 = 8.53 uF

Cab = 8.53 x 20/(20 +8.53) = 5.98uF

Vab = 10v

Qab = Vab x Cab = 10 x 5.98 = 59.8uc

let q1 be charge across c and 3

let q2 be charge across 6

V = Q/C

at a constant voltage difference q1/q2 = c1/c2

q1/q2 = cx/6 =2.53/6

q2 = 2.37q1

q1 + q2 = qab

q1 = 17.74uc

q on 6 uF = 2.37 q1 = 42.056uc

V across 6 uF = Vab - q on 6Uf/c = 10 - 7.009 = 2.99 v

q on 20 uf =2.99 x 20 = 59.8uc

q on c = q1 = 17.74uc

V across c = q on c / c = 17.74/16 = 1.10875v

Vacross 3 uF = 10 - 1.10875 - 2.099 = 6.79225 v

q on 3uF = V across 3 uF x 3 = 6.79925 x 3 = 20.38 uc