Question

A network of four capacitors of 6 μF each is connected to a 240 V supply. Determine the charge on each capacitor.

Answer 1

C₁,C₂ and C₃ are in series

`:.1/C_s=1/C_1+1/C_2+1/C_3`

`=1/(6xx10^(-6))+1/(6xx10^(-6))+1/(6xx10^(-6))`

`1/C_s=3/(6xx10^(-6))`

∴ Cs = 2x10^-6 F

C_s and C₄ are parallel

∴ equivalent resistance

C = C_p = C_s + C₄

= 2x10^-6 + 6x10^-6

C = 8x10^-6

∴v = v1+ v2+ v3

240 = 3v₁  {v₁=v₂=v₃}

∴ v₁  80 volt

Also charge on C₁,C₂ ,C₃ are same.

Q₁ = Q₂ = Q₃ = Q

Q = C₁v₁

= 6x10^-6 x 80

Q = 480x10^-6C and Q₄ = 6x10^-6x 240= 144 × 10^-5

Q₄ = 1.44 × 10^-3 C