Question

Drive the Principle of Wheatstone bridge using 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

Q

P

​  

=  

S

R

​  

 

This is the principle of Wheatstone bridge.

Let the current i is divided into two parts i  

1

​  

 and i  

2

​  

 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

i  

1

​  

P−i  

2

​  

R=0

or i  

1

​  

p=i  

2

​  

R               ........(i)

Similarly, in the closed mesh BCDB, we have

i  

1

​  

Q−i  

2

​  

S=0