Question

Derive condition of a balanced Wheatstone’s bridge.

Or

Draw the diagram of Wheatstone bridge. Why does no current flow through the galvanometer when the bridge is balanced?

Or

What is Wheatstone bridge? Deduce the condition for which Wheatstone bridge is balanced.

Answer 1

Wheatstone Bridge: Wheatstone bridge is an electrical arrangement which forms the basis of the instruments used to determine an unknown resistance.

Principle of Wheatstone bridge: If four resistances P, Q, R and S are arranged as shown in figure, if the galvanometer does not show any defection, then

\(\frac{P}{Q} = \frac{R}{S}\)

Derivation

Wheatstone bridge consists of four resistances P, Q, R and S arranged as shown in Fig. with a cell E between the points A and C and a galvanometer G between B and D.

Let I be the total current given out by the cell which on reaching the point A divides itself into two parts : (a) I₁ flowing through P and (b) (I - I₁) through R. The current I₁, on reaching the point B, divides itself into two parts : (a) Ig through the galvanometer G and (b) (I₁ - I_g) through Q. The current Ig through the arm BD and (I - I₁) through AD combine at D to send a current (I - I₁ + Ig) through S. On reaching the point C, the current (I₁ - I_g) through BC and (I - I₁ + I_g) through DC combine to give the total current I, thus completing the circuit. (The values of the currents at a junction can be verified by applying Kirchhoffs first law at the junction).

Applying Kirchhoffs second law to the closed mesh ABDA, we get

I₁P + I_gG - (I - I₁) R = 0. ...................(1)

where G is the resistance of the galvanometer.

Again applying Kirchhoffs second law to the closed mesh BCDB we get,

(I - I_g) Q - (I - I₁ + I_g) S - I_gG = 0 ...........................(2)

The value of R is so adjusted that the points B and D are at the same potential and as such no current flows in the arm BD i.e. I_g = 0. This is indicated by zero deflection in the galvanometer G. In this position, the bridge is said to be balanced. Putting I_g = 0 in equations (1) and (2), we get 

I₁P  - (I - I₁) R = 0

or  I₁P = (I - I₁)R .......................(3)

and I₁ Q - (I - I₁) S = 0

i.e. I₁ Q = (I - I₁) S ........................(4) 

Dividing (3) by (4), we get

\(\frac{I_1P}{I_1Q} = \frac{(I-I_1)R}{(I-I_1)S}\)

or \(\frac{P}{Q} = \frac{R}{S}\)