It is the combination of four resistors in which three of them is known and one is unknown. The combination of resistor are as shown in figure. In the figure P,Q and R are known resistance and X is unknown.

At the balance condition, that is for null deflection of galvanometer, we can write,

P/Q = X/R

This condition is called **Wheatstone bridge** condition.

### Wheatstone Bridge Condition:

The Wheatstone bridge circuit consist of four resistance P,Q,X and R as shown in figure. Let I 1,I1,I1,I1 are the current through resistor P,X,Q,R respectively. A galvanometer of resistance Rg is connected between the terminal B and D through which a current I g flow.

Applying the Kirchhoff's current law at a junction point B and D. The value of current through a resistance Q and R is:

I 3= I1- Ig

I 4= I2 + Ig

and taking the loop ABDA and applying Kirchhoff's voltage law:

ΣE + ΣIR = 0

or, 0 + I1 x P + Ig x Rg - I 2 x X = 0

at a balance condition (null deflection condition)

I g = 0, we get,

I 1P - I2 x X = 0

I 1P = I2X <-------- (i)

Taking the loop BCDB and applying Kirchhoff's voltage law,

ΣE + ΣIR = 0

0 + I3 x Q - I4 x R - I x Rg = 0

at balance condition (I g=0) we get,

I 3 x Q = I4 x R

I 1Q = I2R <------ (ii)

dividing equation (i) and equation (ii) we get,

P/Q = X/R

**which is the required Wheatstone bridge condition.**

**Potentiometer**

**It is a device which is used to measure the emf of the cell.** Compare the EMF of the two cell and determine the internal resistance of a cell.

It consist of 10 m long uniform cross section wire made up of constant or Eureka which is divided to equal segment. Each segment is 1 m long is placed over a wooden board by the help of copper stripes. A meter scale is placed in the lower copper's part of the wooden board provided two way scale.

**Principle of Potentiometer:**

On sliding the jockey over a potentiometer wire AB a null point is obtained at a point 'C' of length 'lac' .

from the ohm's law,

V= IRac <-------- (i)

I= current through the potentiometer wire.

R ac= (ρ x lac)/A

where, ρ= resistivity of the material of the potentiometer wire.

A= cross sectional area.

from equation (i) and (ii),

V= I. (ρ x l ac)/A

V= k. lac

where, k= Iρ/A = constant

V ∝ l ac

V ∝ l

When a constant current is passed through the potentiometer then the potential dropped across any portion is directly proportional to the length of that portion.

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I found one wrong