wheatstone bridge principle

wheatstone bridge principle

Principle

According to this principle,

If four resistances are connected in series and represented by the four sides of a parallelogram and one diagonal of the quadrilateral is connected to the galvanometer and the other diagonal to the electric cell, if there is no declination in the galvanometer G, then,

P/Q=R/S

Proof —

Working

Suppose four resistances P, Q,R and S are connected as four arms of the quadrilateral ABCD. The galvanometer is connected on one side b/w points B & D and the electric cell is connected b/w A & C. There are two keys K₁ and K₂. Current i is passed through the electric cell E.

At point A, the current is divide into two parts, the one part i₁ flows in side AB and the other part i₂ flows in the side AD.

The resistance PQRS is adjusted in such a way that no the current flows in the galvanometer G on pressing the key K₂.

In this case, the current in the arm BC will be i₁, & the current in the arm DC will be i₂.

 Kirchhoff’s second law for closed loop ABDA

i₁p – i₂R = 0

i₁p = i₂R ———-(1)

For closed loop BCDB

i₁Q – i₂S =0

i₁Q = i₂S ———-(2)

Equation (½)

(i₁p/i₁Q) = (i₂R/i₂S)

P/Q = R/S

By Mr. Ali

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