The equivalent capacitance for the network shown in the figure is

Options

A. `(1200)/(7)pF`
B. `(1000)/(4)pF`
C. `(1800)/(7)pF`
D. `(1300)/(3)pF`

Answer

A

Explanation

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`