A coil is connected to a constant dc supply of 100v at start, when it was at the room temperature of 25 C , it drew a current of 13 a . After sometime, its temperature was 70C and the current reduced further to 80 C . Also , find the temperature coefficient of resistance of the coil material at 25 C ??
Answers
The complete question is:
A coil is connected to a constant dc supply of 100V. At start, when it was at the room temperature of 25°C , it drew a current of 13 A . After sometime, its temperature was 70°C and the current reduced further to 8.5 A . Find the current it will draw when its temperature further increases to 80°C. Also, find the temperature coefficient of resistance of the coil material at 25°C ?
Given:
Initial voltage = 100
Initial temperature = 25°C
Current drawn = 13A
Temperature after sometime = 70°C
Final value of Current drawn = 8.5A
To find:
Temperature coefficient of resistance of the coil material at 25°C and current drawn when temperature increases to 80°C.
Solution:
By Ohm's law:
R(at 25) = V/I
Where R = resistance
V = Voltage supplied
I = current
R₂₅ = 100/ 13 ohm = 7.6923 ohm
R₇₅ = 100/ 8.5 ohm = 11.7647 ohm
R₇₅ = R₂₅(1 + αΔT)
11.7647 = 7.6923 (1 + α(70 - 25))
1.5294 = (1 + 45*α)
45α = 0.5294
α = 0.01176444
α = 0.0117644
Since I = V/R
R(at 80) = R₂₅(1 + α(80-25))
R₈₀ = 7.6923 (1 + 0.0117644*55)
R₈₀ = 7.6923 * 1.647042
R₈₀ =12.6695 ohm
I = 100/ 12.6695
I = 7.892 A
Temperature coefficient of resistance of the coil material at 25°C will be 0.0117644 and current drawn when temperature increases to 80°C will be 7.892 A.