Wheatstone bridge ? Give its Principle, Theory and Proof

Ques : What is Wheatstone bridge? Give its principle, theory and proof.

Solution : Wheatstone bridge is an arrangement of four resistors in the form of a bridge used for measuring one unknown resistance in terms of other three known resistances.

Construction : Four resistors of resistances $P, Q, R$ and $S$ respectively are arranged in the form of a bridge.

A source of e.m.f. '$\epsilon$' is connected between points $A$ and $C$. A galvanometer is connected between points $B$ and $D$. Unknown resistance can be in any of the four arms of the bridge $P, Q, R$ and $S$. Usually, unknown resistance is

Put at S. Out of the four resistances, one(S) is unknown, one is variable (R) and the other two (P and Q) can be standard resistors.

Principle : When key K is closed, the galvanometer shows the presence of current $I_g$ through it. The value of a resistances say R is adjusted in such a way that the galvanometer shows no deflection. At this stage, the potential at points B and D is equal and hence no current flows through the galvanometer. The Wheatstone bridge at this stage is said to be balanced and the ratio of P and Q is equal to the ratio of R and S.

That is,

$ \frac{P}{Q} = \frac{R}{S} $

    or

 $ S = \left( \frac{Q}{P} \right) R $

Knowing P, Q and R, the value of S can be calculated.

Proof : When key K is closed, the currents flowing in different arms of Wheatstone bridge are marked according to Kirchhoff's first law as shown in Figure 39.

Applying Kirchhoff's loop law to the closed loop ABDA, we get,

$ I_2P + I_gG - I_1R = 0 \dots (i) $

where G is the resistance of the galvanometer.

In the case of balanced Wheatstone bridge, no current flows through the galvanometer i.e., $I_g = 0$

Hence, eqn. (i) becomes, 

$ -I_2P + I_1R = 0$

Or $I_1R = I_2P $

Or

$ \frac{I_1}{I_2} = \frac{P}{R} \quad \dots (ii)$

Applying Kirchhoff's loop law to the closed loop BCDB, we get,

$ (I_2 - I_g)Q - (I_1 + I_g)S - I_gG=0...(iii) $

In the case of balanced Wheatstone bridge, no current flows through the galvanometer i.e., $I_g = 0$, eqn. (iii) becomes,

$ I_2Q - I_1S = 0 \quad \text{ or } \quad I_1S = I_2Q $

 Or

$ \frac{I_1}{I_2} = \frac{Q}{S} \quad \dots (iv) $

From eqns. (iii) and (iv), we get,

$\frac{P}{R} = \frac{Q}{S}$

$\frac{P}{Q} = \frac{R}{S} $. Proved.