Consider the network of capacitors as shown in the figure. Find equivalent capacitance between points A and C. Assume `C_(1)=1muF, C_(2)=0.5muF and C_(3)=1muF`.

Explanation

In the given network of capacitors we can see thatequivalent capacitance cannot be obtained by using the formulae for series and parallel combinations. Apart from that it is nto the case of a balanced Wheatstone bridge. We can still make use of reverse symmetric values of capacitances of the given network to cut short the calculation to a certain extent.
When we connect a battery between the points A and C of circuit, then charge flowing from A to B can be taken as same as flowing from D to C. Similarly charge flowing from A to D will be same as charge flowing form B to C. Flow of charge through the bridge capacitor form B to D can be written accordingly. See the following diagram how total charge Q is distributed in the circuit when a potential difference of V is applied between the points A and C. Here, we must note the fact that, the ratio Q/V will be the equivalent capacitance between A and C, which is required as the final answer to this quesiton.

Kirchhoff.s loop Law: Algebraic sum of all the potential differences in a loop must add to zero. This a due to conservative nature of electric field.
On applying Kirchhoff.s law for the loop EFABCE we get the following equation:
`V-(Q_(1))/(1)-(Q-Q_(1))/(0.5)=0`
`rArr V-Q_(1)-2Q+2Q_(1)=0 rArr V=2Q-Q_(1)" " ...(i) `
On applying Kirchhoff.s law for the loop ABDA we get the following equation:
`-(Q_(1))/(1)-(2Q_(1)-Q)/(1)+(Q-Q_(1))/(0.5)=0`
`rArr -Q_(1)-2Q_(1)+Q+2Q-2Q_(1)=0`
`rArr 5Q_(1)=3Q rArr Q_(1)=(3Q)/(5) " " ...(ii)`
Substituting `Q_(1)` from equation (ii) in equation (i) we get the following:
`V=2Q-(3Q)/(5)=(7Q)/(5) rArr C_(eq)=(Q)/(V)=(5)/(7)muF`