Ce diaporama a bien été signalé.
Le téléchargement de votre SlideShare est en cours. ×

# TEMP COEFF OF RESISTANCE 001.pptx

Publicité
Publicité
Publicité
Publicité
Publicité
Publicité
Publicité
Publicité
Publicité
Publicité
Publicité
Chargement dans…3
×

## Consultez-les par la suite

1 sur 17 Publicité

# TEMP COEFF OF RESISTANCE 001.pptx

Temp coeff

Temp coeff

Publicité
Publicité

## Plus De Contenu Connexe

Publicité

### TEMP COEFF OF RESISTANCE 001.pptx

1. 1. TEMPERATURE COEFFICIENT OF RESISTANCE Akshita Chaudhary Btech 1st year Chemical Engineering
2. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 16. Specific resistance of given wire is : 3.19 X 10-5 Ω/cm RESULT
17. 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.