Experiment 2 : To find resistance of a given wire / standard resistor using metre bridge.

Experiment 2 : To find resistance of a given wire / standard resistor using metre bridge.

Aim : 

To find resistance of a given wire using Whetstone’s bridge (meter bridge) & hence determine the specific resistance of the material.

Apparatus :

A meter bridge (slide Wire Bridge), a galvanometer, a resistance box, a laclanche cell, a jockey, a one- way key, a resistance wire, a screw gauge, meter scale, set square, connecting wires and sandpaper.

Theory : 

Metre bridge apparatus is also known as a slide wire bridge. It is fixed on the wooden block and consists of a long wire with a uniform cross-sectional area. It has two gaps formed using thick metal strips to make the Wheatstone's bridge.

Then according to Wheatstone's principle, we have:

$$\frac{X}{R} = \frac{l}{(100-l)}$$

The unknown resistance can be calculated as:

$$X = R \frac{l}{(100-l)}$$

Then the specific resistance of the material of the is calculated as:

$$\rho = \frac{X.\pi.D^{2}}{4L}$$

Where,

 * L is the length of the wire

 * D is the Diameter of the wire

Observations :

• Least count of screw gauge: 0.001 cms
• Pitch of screw gauge: 0.1 cm
• Total no. of divisions on circular scale: 100
• Least Count = Pitch/No. of divisions on circular scale; LC = 0.001 cm
• Length of given wire, L = 25cm

Table for length (l) & unknown resistance, X:

S.No.	Resistance from resistance box R (ohm)	Length AB = l cm	Length BC = (100-l) cm	Unknown Resistance X=R. (100-l)/l Ω
1	2	41	59	2.87
2	4	60	40	2.66
3	6	69	31	2.69

Table for diameter (D) of the wire:

Circular Scale Reading

S.No.	Linear Scale Reading (N) mm	No. of circular scale divisions coinciding (n)	Value n x (L.C.) mm	Observed diameter D = N + n x L.C. mm
1	0	34	0.34	0.34
2	0	35	0.35	0.35
3	0	36	0.36	0.36
4	0	35	0.35	0.35

Calculation:

For unknown resistance, Mean X:

$$X = \frac{X_{1}+X_{2} + X_{3}}{3}$$

$$X = \frac{2.87+2.66 +2.69}{3}$$

$$X = \frac{8.22}{3}$$

$$X= 2.74 \Omega$$

Mean diameter, D : 

$$D= \frac{D_{1}+D_{2}+ D_{3}+ D_{4}}{4}$$

$$D= \frac{0.34+0.35+ 0.36+ 0.35}{4}$$

$$D= \frac{1.4 }{4}$$

$$D=0.35mm = 0.035cm$$

Specific Resistance, ρ

$$\rho = \frac{X.\pi.D^{2}}{4L}$$

$$\rho = \frac{2.74 \times 3.14 \times 0.035^{2}}{4 \times 25}$$

$$\rho = \frac{0.01053}{100}$$

$$\rho =  0.0001053$$

$$\rho = 1.053 \times 10^{-4} \Omega cm$$

Result : 

• The value of unknown resistance X = $2.74 \Omega$
• The specific resistance of the material of the given wire = $1.053 \times 10^{-4} \Omega cm$

Precautions : 

• The connections should be neat, tight and clean.
• Plugs should be tightly connected in the resistance box.
• The movement of the jockey should be gentle and it shouldn’t be rubbed.
• The key K should be inserted only when the observations are to be taken.
• The null point should be between 45cm and 55cm.
• To avoid the error of parallax, the set square should be used to note the null point.
• There shouldn’t be any loops in the wire.
• The diameter of the wire should be measured in two perpendicular directions that are mutual.

Sources Of Error : 

• The screws of the instrument might be loose.
• The wire might be of non-uniform diameter.
• There might be a backlash error in the screw gauge.