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/(6 xx 10^(-6)) + 1/(6 xx 10^(-6)) + 1/(6 xx 10^(-6))`

`1/C_s = 3/(6 xx 10^(-6))`

∴ C_s = 2 × 10⁻⁶ F

C_s and C₄ are parallel

∴ equivalent resistance

C = C_p = C_s + C₄

= 2 × 10⁻⁶ + 6 × 10⁻⁶

C = 8 × 10⁻⁶

∴ v = v₁ + v₂ + v₃

240 = 3v₁     ...[v₁ = v₂ = v₃]

∴ v₁ = 80 volt

Also, the charge on C₁, C₂, C₃ are the same.

Q₁ = Q₂ = Q₃ = Q

Q = C₁v₁

= 6 × 10⁻⁶ × 80

= 480 × 10⁻⁶ C and 

Q₄ = 6 × 10⁻⁶ × 240

= 144 × 10⁻⁵

Q₄ = 1.44 × 10⁻³ C