Question

Obtain balancing condition in case of Wheatstone's network.

Answer 1

Solution-

In balanced Wheatstone's bridge, zero current flow through galvanometer. Means while applying Kirchoff's law we can neglet this path.

Look at the figure,

No current flows through galvanometer G if circuit is balanced.

Applying Kirchoff's law to

1 ) Mesh ABDA,

-I1R1 + (I-I1)R4 = 0

I1R1 = (I-I1)R4 ….(1)

2 ) Mesh BCDB,

-I1R2 + (I-I1)R3 = 0

I1R2 = (I-I1)R3 ….(2)

Dividing (1) by (2),

I1R1 / I1R2 = (I-I1)R4 / (I-I1)R3

R1/R2 = R4/R3

This is balancing condition for Wheatstone's bridge.

Best luck for exams buddy...

Answer 2

Answer:

In balanced Wheatstone's bridge, zero current flow through galvanometer. Means while applying Kirchoff's law we can neglet this path. Look at the figure, No current flows through galvanometer G if circuit is balanced