Finding equivalent capacitance between A and B with isolated component

[cnh1995]

But then I get confused which one will be denominator and which one will be numerator.

It doesn't matter which should be numerator or denominator. Just pick any ratio on one side, use the same on the other side.

 

[gracy]

Let say if I would have taken
C1/C2=C4/C3
After putting the values
I could have thought the bridge is not balanced.

 

[gneill]

But then I get confused which one will be denominator and which one will be numerator.

You want to see if the two voltage dividers in the bridge circuit will produce the same voltage division. So a simple procedure is to make ratios of the components that make up each voltage divider. Compare the ratios. You can't go wrong if you select corresponding pairs (use symmetry and common sense).

 

[gracy]

You can't go wrong if you select corresponding pairs (

You mean if I have selected C1 as numerator for one potential divider I shall select the one which is adjacent to it but at the other side of bridge as numerator of second potential divider,Right?

 

[cnh1995]

Let say if I would have taken
C1/C2=C4/C3
After putting the values
I could have thought the bridge is not balanced.

You can call the two resistors on one side as 'upper' and 'lower'. So, upper/loewer on one side=upper/lower on the other side.

 

[cnh1995]

You can call the two resistors on one side as 'upper' and 'lower'. So, upper/loewer on one side=upper/lower on the other side.

Its just a memory trick..To understand the bridge balance conceptually, you should analyse it using voltage divider.

 

[gracy]

upper/lower on one side=upper/lower on the other side.And if bridge is of this sort

upper left/upper right on one side=lower left/lower right on the other side

 

[cnh1995]

upper/lower on one side=upper/lower on the other side.And if bridge is of this sort

upper left/upper right on one side=lower left/lower right on the other side

Right. Once you're fluent, you won't need this trick.