Question

Use Kirchhoff's rules to obtain conditions for the balance condition in a Wheatstone bridge.

Answer 1

Let us consider a Wheatstone bridge arrangement as shown below.

Applying Kirchhoff's loop law to the closed loop ADBA, we get 

I₁R−I_gG−I₂P=0 .....(1)

Here, G is the resistance of the galvanometer.

Applying Kirchhoff's loop law in the closed loop BDCB, we get

I_gG+(I₁+I_g)S−(I₂−I_g)Q=0.....(2)

When the Wheatstone bridge is balanced, no current flows through the galvanometer, i.e. I_g=0

∴ From (1), we get

I₁R−I₂P=0

⇒I₁R=I₂P

`=>I_1/I_2=P/R" ......(3)"`

Similarly, from (2), we get

I₁S−I₂Q=0

⇒I₁S=I₂Q

`=>I_1/I_2=Q/S" .....(4)"`

From (3) and (4), we get

`P/R=Q/S`

or `P/Q=R/S`

This is the required balance condition in a Wheatstone bridge arrangement.