Question

State Kirchhoff's rules for an electric network. Using Kirchhoff's rules, obtain the balance condition in terms of the resistances of four arms of Wheatstone bridge.

Answer 1

Kirchhoff’s First Law − Junction Rule

The algebraic sum of the currents meeting at a point in an electrical circuit is always zero.

Let the currents be I₁, I₂I₃, and I₄

Convention:

Current towards the junction − positive

Current away from the junction − negative

I₃ + (− I₁) + (− I₂) + (− I₄) = 0

Kirchhoff’s Second Law − Loop Rule

In a closed loop, the algebraic sum of the emfs is equal to the algebraic sum of the products of the resistances and current flowing through them.

For closed part BACB, E₁ − E₂ = I₁R₁ + I₂ R₂ − I₃R₃

For closed part CADC, E₂ = I₃R₃ + I₄R₄ + I₅R₅

Wheatstone Bridge:

The Wheatstone Bridge is an arrangement of four resistances as shown in the following figure.

R₁, R₂, R₃,and R₄ are the four resistances.

Galvanometer (G) has a current I_g flowing through it at balanced condition, I_g = 0

Applying junction rule at B,

∴ I₂ = I₄

Applying junction rule at D,

∴ I₁ = I₃

Applying loop rule to closed loop ADBA,

`-I_1R_1+0+I_2R_2 = 0`

`∴ I_1/I_2  = R_2/R_1  ...... (1)`

Applying loop rule to closed loop CBDC,

`I_2R_4+0-I_1R_3 = 0`      ∵`I_3 =I_1,I_4 =I_2`

`∴ I_1/I_2  = R_4/R_3  ...... (2)`

From equations (1) and (2),

`R_2/R_1 = R_4/R_3`

This is the required balanced condition of Wheatstone Bridge.