Six capacitors of capacities `10muF, 10 muF, 10 muF, 10 mu F, 20 mu F` and `X mu F` are connected in a network as shown below. If the network is balanced, find X and the resultant capacitance between A and C.

Explanation

(a) The effective capacitance of the series combination in the arm DC is `(20X)/(20 + X) muF`
Using the balancing condition for the network,
`10/10 = 10/((20X)/(20 + X))`
`therefore (20X)/(20 + X) = 10`
`therefore 20 X = 200 + 10X`
`therefore 10 X = 200 therefore X=20`

(b) As the network is balanced, no charge flows through the arm BD.
Resultant capacitance
`C_(ABC) = (10 xx 10)/(10 + 10) = 100/20 = 5 muF` (series combination)
Similarly,
`C_(ADC) = (10 xx 10)/(10 + 10) = 100/20 = 5 muF` (Series combination)
The resultant capacitance between A and C.
`=C_(ABC) + C_(ADC)` (Parallel combination)
`=5+5= 10 muF`