A copper conductor was measured to have a resistance of 6.5 ohms at 20°C. When the temperature increased to 42°C, the resistance increased to 7.135 ohms due to copper's positive temperature coefficient. Similarly, a motor winding measured 12 ohms at 20°C and 26 ohms after running, calculating the temperature increase to be 259°C, making the total temperature 279°C.

1. A copper conductor is measured at room temperature(20 °C) to be 6.5 Ω, I f the temperature of the room increased to 4 2 °C Calculate the new resistance of the conductor R = R 0 (1+ α 0 Δ t) copper α 0 =0.0045 at 20 °C

2. A copper conductor is measured at room temperature(20 °C) to be 6.5 Ω, I f the temperature of the room increased to 4 2 °C Calculate the new resistance of the conductor R = R 0 (1+ α 0 Δ t) copper α 0 =0.0045 at 20 °C Put values into formulae

3. A copper conductor is measured at room temperature(20 °C) to be 6.5 Ω , If the temperature of the room increased to 4 2 °C Calculate the new resistance of the conductor R = R 0 (1+ α 0 Δ t) copper α 0 = 0.0045 at 20 °C R = 6.5Ω (1+ 0.0045 ( 4 2 °C - 20 °C ))

4. A copper conductor is measured at room temperature(20 °C) to be 6.5 Ω , If the temperature of the room increased to 4 2 °C Calculate the new resistance of the conductor R = R 0 (1+ α 0 Δ t) copper α 0 = 0.0045 at 20 °C R = 6.5Ω (1+ 0.0045 ( 4 2 °C - 20 °C )) R = 6.5Ω (1+ 0.0045 x 22 )

5. A copper conductor is measured at room temperature(20 °C) to be 6.5 Ω , If the temperature of the room increased to 4 2 °C Calculate the new resistance of the conductor R = R 0 (1+ α 0 Δ t) copper α 0 = 0.0045 at 20 °C R = 6.5Ω (1+ 0.0045 ( 4 2 °C - 20 °C )) R = 6.5Ω (1+ 0.0045 x 22 ) R = 6.5Ω (1+ 0.099)

6. A copper conductor is measured at room temperature(20 °C) to be 6.5 Ω , If the temperature of the room increased to 4 2 °C Calculate the new resistance of the conductor R = R 0 (1+ α 0 Δ t) copper α 0 = 0.0045 at 20 °C R = 6.5Ω (1+ 0.0045 ( 4 2 °C - 20 °C )) R = 6.5Ω (1+ 0.0045 x 22 ) R = 6.5Ω (1+ 0.099) R = 6.5Ω x 1.099

7. A copper conductor is measured at room temperature(20 °C) to be 6.5 Ω , If the temperature of the room increased to 4 2 °C Calculate the new resistance of the conductor R = R 0 (1+ α 0 Δ t) copper α 0 = 0.0045 at 20 °C R = 6.5Ω (1+ 0.0045 ( 4 2 °C - 20 °C )) R = 6.5Ω (1+ 0.0045 x 22 ) R = 6.5Ω (1+ 0.099) R = 6.5Ω x 1.099 R = 7.135 Ω

8. A motor winding is measured at room temperature(20 °C) to be 12 Ω, after running the motor the resistance is measured to be 26 Ω. Calculate the new temperature of the motor R = R 0 (1+ α 0 Δ t) copper α 0 =0.0045 at 20 °C

9. A motor winding is measured at room temperature(20 °C) to be 12 Ω , after running the motor the resistance is measured to be 26 Ω . Calculate the new temperature of the motor R = R 0 (1+ α 0 Δ t) copper α 0 = 0.0045 at 20 °C 26Ω = 12Ω (1+ 0.0045 Δ t)

10. A motor winding is measured at room temperature(20 °C) to be 12 Ω , after running the motor the resistance is measured to be 26 Ω . Calculate the new temperature of the motor R = R 0 (1+ α 0 Δ t) copper α 0 = 0.0045 at 20 °C 26Ω = 12Ω (1+ 0.0045 Δ t) 26Ω/ 12Ω = (1+ 0.0045 Δ t)

11. A motor winding is measured at room temperature(20 °C) to be 12 Ω , after running the motor the resistance is measured to be 26 Ω . Calculate the new temperature of the motor R = R 0 (1+ α 0 Δ t) copper α 0 = 0.0045 at 20 °C 26Ω = 12Ω (1+ 0.0045 Δ t) 26Ω/ 12Ω = (1+ 0.0045 Δ t) 2.166 = (1+ 0.0045 Δ t)

12. A motor winding is measured at room temperature(20 °C) to be 12 Ω , after running the motor the resistance is measured to be 26 Ω . Calculate the new temperature of the motor R = R 0 (1+ α 0 Δ t) copper α 0 = 0.0045 at 20 °C 26Ω = 12Ω (1+ 0.0045 Δ t) 26Ω/ 12Ω = (1+ 0.0045 Δ t) 2.166 = (1+ 0.0045 Δ t) 2.166 -1 = 0.0045 Δ t

13. A motor winding is measured at room temperature(20 °C) to be 12 Ω , after running the motor the resistance is measured to be 26 Ω . Calculate the new temperature of the motor R = R 0 (1+ α 0 Δ t) copper α 0 = 0.0045 at 20 °C 26Ω = 12Ω (1+ 0.0045 Δ t) 26Ω/ 12Ω = (1+ 0.0045 Δ t) 2.166 = (1+ 0.0045 Δ t) 2.166 -1 = 0.0045 Δ t 1.166 = 0.0045 Δ t

14. R = R 0 (1+ α 0 Δ t) copper α 0 = 0.0045 at 20 °C 26Ω = 12Ω (1+ 0.0045 Δ t) 26Ω/ 12Ω = (1+ 0.0045 Δ t) 2.166 = (1+ 0.0045 Δ t) 2.166 -1 = 0.0045 Δ t 1.166 = 0.0045 Δ t Δ t = 1.166/ 0.0045

15. R = R 0 (1+ α 0 Δ t) copper α 0 = 0.0045 at 20 °C 26Ω = 12Ω (1+ 0.0045 Δ t) 26Ω/ 12Ω = (1+ 0.0045 Δ t) 2.166 = (1+ 0.0045 Δ t) 2.166 -1 = 0.0045 Δ t 1.166 = 0.0045 Δ t Δ t = 1.166/ 0.0045 Δ t = 259 °C

16. Δ t = 259 °C This is the change in the temperature

17. Δ t = 259 °C This is the change in the temperature The room temperature was 20 °C and we have changed by 259 °C, so therefore our new temperature will be,

18. Δ t = 259 °C This is the change in the temperature The room temperature was 20 °C and we have changed by 259 °C, so therefore our new temperature will be, t = Δ t + 20 °C

19. Δ t = 259 °C This is the change in the temperature The room temperature was 20 °C and we have changed by 259 °C, so therefore our new temperature will be, t = Δ t + 20 °C t = 279°C