Find the equivalent capacitances of the combinations shown in figure between the indicated points.

Explanation

(a) In loop 1256 , applying charge distribution
` q_1 /1 + q_3 / 4 - q_2/3 = 0 `
`rArr 12 q_1 + 3 q_3 - 4 q_2 = 0 `
rArr 12 q_1 - 4q_2 = 0`
In loop 2345 `(q_1 - q_3 / 3) - (q_2 - q_3 / 1 ) - (q_3/ 4 )= 0 `
`rArr 4q_1 - 4q_3 - 12q_2 - 12q_2 - 12q_3 - 3q_3 = 0 `
`rArr 4q_1 - 12q_2 - 19q_3 =0 `
`rArr 4q-1 - 12q_2 = 19 q_3`
From (i) and (ii)
` q_1 = - 7 /8 q_3`
` q_2 = - 15 /8 q_3`
Since P. d. `= q_1 / 1 + q_1 - q_3 / 3`
`= 4q_1 - q_3 / 3
So, Equivalent capacitance
` = Toatal charge / P.d. `
`= q_1 + q_2 / (4q_1 - q_3) / 3 `
`= 11 / 6 mu F `
By star Delta conversion (fig). ` 1 + 1 = 2 mu F`
(c) It is balanced diagram
So Equivalent capacitance `= 4 /3 + 8/ 3 + 4 `
`= 8 mu F`
(d) It is also a balanced diagram
So , equivalent capacitance
`= 4 / 3 + 8 / 3 + 8 / 3 + 4 /3 `
`= 8 mu F`.
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