Question

Prove Principle of Wheatstone Bridge with Kirchhoff's law.

Answer 1

Kirchhoff's law: 

To study complex circuits, kirchhoff's gave following too laws: 

(i) If a network of conductors the algebraic sum of all currents meeting at any junction of my circuit is always zero.

i.e. ∑ I = 0

(ii) In any closed mesh (or loop) of an electrical circuit the algebraic sum of the product of the currents and resistance is equal to the total e.m.f. of the mesh.

i.e. ∑ IR =  ∑ E

Formula Derivation: 

Referring fig. Four resistances P, Q, R and S are connected to form a quadrilateral ABCD. A cell E is connected across the diagonal AC and a galvanometer across BD. When the current is flown through the circuit and galvanometer does not give any deflection, then the bridge is balanced and when the bridge is balanced, then

\(\frac{P}{Q}\) = \(\frac{R}{S}\)

This is the principle of Wheatstone bridge. Let the current i is divided into two parts i₁ ​ and i₂ ​ flowing through P, Q and R, S respectively. In the position of equilibrium, the galvanometer shows zero deflection, i.e. the potential of B and D will be equal. 

In the closed mesh ABDA, by Kirchhoff's second law, we get i₁P − i₂R = 0

or i₁P =i₂R ...(i)

Similarly, in the closed mesh BCDB, we have

i₁Q = i₂S = 0

or i₁θ = i₂ S .....(ii)

Dividing equation (i) by equation (ii) we get

\(\frac{i_1P}{i_1Q}\) = \(\frac{i_2R}{i_2S}\)

 \(\frac{P}{Q}\) = \(\frac{R}{S}\)

This is the condition for balance of wheat stone bridge.