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The equivalent capacitance for the netwo...

The equivalent capacitance for the network shown in the figure is

A

`(1200)/(7)pF`

B

`(1000)/(4)pF`

C

`(1800)/(7)pF`

D

`(1300)/(3)pF`

Text Solution

Verified by Experts

The correct Answer is:
A

Capacitance of `C_(1) =C_(4) = 100 pF`
Capactiance of `C_(2) = C_(3) = 400 pF`
Supply voltage , `V = 400V` Capacitors `C_(2)` and `C_(3)` are connected is series,
Equivalent capacitance `C = (1)/(400) + (1)/(400) = (2)/(400)` or `C = 200pF`
Capacitors `C_(1)` and C are in parallel
Their equivalent capacitance.
`C = C + C_(1) = 200 + 100 = 300 pF`
Capacitors C'' and `C_(4)` are connected in series Equivalent capacitance `(1)/(C_(eq)) =(1)/(C'') +(1)/(C_(4)) = (1)/(300) +(1)/(400) +(1)/(C_(eq)) = (7)/(1200)`
`therefore " "C_(eq) = (1200)/(7)pF`
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Knowledge Check

  • The equivalent capacitance of the combination show in figure is

    A
    `C `
    B
    `2C`
    C
    `(C /2)`
    D
    none of these
  • The equivalent capacitance of the arrangement shown in figure, if A is the area of each plate is

    A
    `C=(epsilon_0A)/d[K_1/2+(K_2+K_3)/(K_2K_3)]`
    B
    `C=(epsilon_0A)/d[K_1/2+(K_2K_3)/(K_2+K_3)]`
    C
    `C=(epsilon_0A)/(2d)[K_1+(K_2+K_3)/(K_2+K_3)]`
    D
    `C=(epsilon_0A)/d[K_1+(K_2K_3)/(K_2+K_3)]`
  • The equivalent capacitance C_(AB) of the circuit shown in the figure is. .

    A
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    C
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    D
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