Answer:
61.5 °C
Explanation:
The resistance of the relay coil at 15 °C is R = V/I = 6/0.12 = 50 ohms. In order for the coil current to remain above 0.10 A, the resistance must remain below R = 6/0.10 = 60 ohms.
At some temperature difference ΔT from 15 °C, the resistance of the coil will be ...
R = R0(1 +α·ΔT)
where R0 is the resistance at 15 °C, α is the temperature coefficient of resistance, and ΔT is the temperature change. We want to solve this for ΔT:
R/R0 = 1 +α·ΔT
(R/R0 -1)/α = ΔT = (60/50 -1)/0.0043 ≈ 46.5 . . . . °C
The relay may fail to operate at temperatures above (15 +46.5) °C = 61.5 °C.
