Sum

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

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#### Solution

C_{1},C_{2} and C_{3} 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 C_{4} are parallel

∴ equivalent resistance

C = C_{p} = C_{s} + C_{4}

= 2x10^{-6} + 6x10^{-6}

C = 8x10^{-6}

∴v = v1+ v2+ v3

240 = 3v_{1} {v_{1}=v_{2}=v_{3}}

∴ v_{1} 80 volt

Also charge on C_{1},C_{2} ,C_{3} are same.

Q_{1} = Q_{2} = Q_{3} = Q

Q = C_{1}v_{1}

= 6x10^{-6} x 80

Q = 480x10^{-6}C and Q_{4} = 6x10^{-6}x 240= 144 × 10^{-5}

Q_{4} = 1.44 × 10^{-3} C

Concept: Condensers in Series and Parallel,

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