Electrical

### Wheatstone Bridge Circuit | Potentiometer and its principle

757 / May-30-2021 09:50:37 By- srijan / 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. 