1.
TEMPERATURE COEFFICIENT
OF
RESISTANCE
Akshita Chaudhary
Btech 1st year
Chemical Engineering

2.
 Aim:
To find the temperature coefficient of resistance of a
given coil.
 Apparatus:
carey foster bridge, unknown low resistor, Resistance
box, Lead accumulator, jockey, one way key,
Galvanometer, connecting wires etc. .

3.
 Theory:
A Carey foster bridge is principally same as a meter bridge, which consists of four
resistances P, Q, R and X that are connected to each other as shown in figure 1.
 that are connected to each other as shown in figure 1.
 Figure 1: Wheatstone's bridge

4.
In this circuit G is the galvanometer and E is a lead
accumulator and K1and K are the galvanometer key
and accumulator key respectively. If the values of the
resistances are adjusted so that no current flows
through the galvanometer, and if any of three
resistances P,Q,R and X are known, the unknown
resistance can be determined using the relationship
P/Q=R/X
The Carey foster bridge is used to measure the
difference between two nearly equal resistances and
knowing the value of one, the other can be
calculated.

5.
In this bridge, the end resistances are eliminated in
calculation. Which is an advantage and hence it can
conveniently used to measure a given resistance.
Let P and Q be the equal resistances connected in
the inner gaps 2 and 3, the standard resistance R is
connected in gap 1 and the unknown resistance X is
connected in the gap 4.Let l1 be the balancing length
ED measured from the end E

6.
By wheatstone’s principle,
P/Q=(R+a+l1p)/(X+b+(100-l1)p) ....(1)
Where, a and b are the end corrections at the ends E
and F, and is the resistance per unit length of the
bridge wire.
If the experiment is repeated with X and R
interchanged and if l2 is the balancing length
measured from the end E,
P/Q=(X+a+l2p)/(R+b+(100-l2)p ....(2)

7.
 From equation (1) & (2)
X=R+p(l1-l2) ......(3)
 Let l1' and l2 ' are the balancing lengths when the
above experiment is done with a standard
resistance r (say 0.1) in the place of R and a thick
copper strip of zero resistance in place of X
 From equation (3),
p=r/(l1’-l2’) ......(4)

8.
If X1 and X2 are the resistance of a coil at
temperatures t1
oc and t2
oc, the temperature
coefficient of resistances is given by the equation,
A=(X2-X1)/(X1t2-X2t1) .....(5)
Also, if X0 and X100 are the resistance of the coil at
0oc and 100oc,
A=(X100-X0)/(X0*100) ......(6)

9.
 Procedure to perform the simulation
 To calibrate the Carey Fosters Bridge
 Drag the ‘Resistor’ to the gap 2 and 3 of the Carey
fosters bridge.
 Drag ‘Fractional resistor' to gap 1.
 Drag the Battery to space in between the two
‘Resistors’.
 Drag the copper strip to gap 4.
 Press the continue button, on the top.
 Click and unmark the resistance which has to be
introduced in ‘Resistor’.

10.
 Power On button to start and power off to stop the
experiment.
 Change the position of jockey to get balancing
length l1’.
 Reverse connection button to interchange copper
strip with fractional resistor.
 Move jockey to get balancing length l2’.
 Resistance per cm of bridge wire is found using
equation
p=r/(l1’-l2’)

11.
 To find the temperature coefficient of resistance
 Repeat steps 1-3.
 Drag ‘Unknown resistance 1’ to gap 4 in bridge.
 Introduce the resistance and move jockey to get
balancing length l1.
 Reverse button to interchange unknown resistance
1 with Fractional resistor.
 Move jockey to get balancing length l2.
 Unknown resistance value is found using the
equation
X=R+p(l1-l2)

12.
 Repeat the experiment for different temperature t1,
t2….and note corresponding resistance X1, X2...
 Temperature coefficient of resistance is found using
equation
A=(X2-X1)/(X1t2-X2t1)
 Repeat the experiment for unknown resistance 2.
 A graph is plotted between values of X and t ,
which gives a straight line, whose slope is α.

13.
• Observation Table
Determination of
resistance per unit length
of bridge wire
S.No.
Resistance
connected in
decimal
resistance box
X (in ohm)
Zero deflection
Position when
resistance box is
connected
(l2-l1)
( in
cms)
ρ = X/ (l2-l1)
(in ohms))
In left gap
l1 (in cms)
In right gap
l2 (in cms)
01 0.1 48.8 51.1 2.3 0.043
02 0.2 47.6 52.4 4.8 0.041
03 0.4 45.1 54.8 9.7 0.041
04 0.5 44.0 56.0 12 0.041
Mean rho
0.0415= Ohms

14.
For the
measurement of
resistance (Y) of the
given wire.
.No.
Resistance
connected in
decimal
resistance box
X (in ohm)
Zero deflection
Position when
resistance box is
connected
(l2-l1)
( in
cms)
Y = X- ρ (l2-
l1)
(in ohms)
In left gap l1 (in
cms)
In right gap
l2 (in cms)
01 0.1 49.1 51 1.9 0.021
02 0.2 47.8 52.2 4.4 0.017
03 0.4 45.3 54.7 9.4 0.010
04 0.5 44.1 55.9 11.8 0.013
Mean 0.01525
Y = Ohms

15.
• CALCULATIONS:-
• Material :- Constantan
• Length of wire(l) = 15cm
• Diameter of wire = 0.2 cm => Area(A) = 3.14 * 0.1 *0.1 = 0.0314 cm2
• Mean Rho = 0.0415 ohm-cm
• Mean Y (Calculated value) = 0.01525
• Standard Value = 0.025189
• Percentage error = (0.025189-0.015255)*100/0.025189 = 9.30 %
• Resistivity of Material = Y * A / l = 0.015255 * 0.0314 / 15 = 3.19 x 10-5
• Result :-
• Resistivity of the given wire = 3.19 x 10-5 ohm-cm

16.
Specific resistance of given
wire is : 3.19 X 10-5 Ω/cm
RESULT

17.
PRECAUTIONS AND SOURCES OF ERROR
A key should be used and it should be closed only while observation are
being made.
While checking the null point the cell key should be closed first and then
the jockey should be made to touch the bridge wire.
The jockey should be pressed gently and momentarily . It should not be
dragged along the length of wire otherwise it will spoil the bridge wire.