Question

Derive the equation of the balanced state in a Wheatstone bridge using Kirchhoff’s laws.

Answer 1

In this case, Kirchhoff’s junction rule applied to junctions D and B (see the figure) immediately gives us the relations I₁ = I₃ and I₂ = I₄. Next, we apply Kirchhoff’s loop rule to closed loops ADBA and CBDC.

The first loop gives 

–I₁R₁ + 0 + I₂R₁ = 0  ...(I_g = 0)

And the second loop gives, upon using I₃ = I₁, I₄ = I₂

I₂R₄ + 0 – I₁R₃ = 0

From the equation, we obtain,

`I_1/I_2 = R_2/R_1`

Whereas from the equation, we obtain,

`I_1/I_2 = R_4/R_3`

Hence, we obtain the condition

`R_2/R_1 = R_4/R_3`

This last equation relating the four resistors is called the balance condition for the galvanometer to give zero or null deflection.

The Wheatstone bridge and its balance condition provide a practical method for the determination of unknown resistance. Let us suppose we have an unknown resistance, which we insert in the fourth arm; R₄ is thus not known. Keeping known resistances R₁ and R₂ in the first and second arm of the bridge, we go on varying R₃ till the galvanometer shows a null deflection. The bridge then is balanced, and from the balance condition the value of the unknown resistance R₄ is given by, R₄ = `R_3 R_2/R_1`