TEMP COEFF OF RESISTANCE 01.pptx

Temperature coefficient

1. TEMPERATURE COEFFICIENT OF RESISTANCE

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. I n 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 l 1  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 l 2  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 l 1 ' and l 2  ' 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 t 1 o c and t 2 o c, the temperature coefficient of resistances is given by the equation, A=(X2-X1)/(X1t2-X2t1) .....(5) Also, if X  and X 100  are the resistance of the coil at 0 o c and 100 o c, 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 l 1 ’. Reverse connection button to interchange copper strip with fractional resistor. Move jockey to get balancing length l 2 ’. 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 l 1 . Reverse button to interchange unknown resistance 1 with Fractional resistor. Move jockey to get balancing length l 2 . Unknown resistance value is found using the equation X=R+p(l1-l2)

12. Repeat the experiment for different temperature t 1 , t 2 ….and note corresponding resistance X 1 , X 2 ... 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 (l 2 -l 1 ) ( in cms ) ρ = X/ (l 2 -l 1 ) (in ohms) ) In left gap l 1  (in cms ) In right gap l 2  (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 (l 2 -l 1 ) ( in cms) Y = X- ρ (l 2 -l 1 ) (in ohms) In left gap l 1  (in cms) In right gap l 2  (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 cm 2 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