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The coil of a relay takes a current of 0.12A when at room temperature of 15 ° C and connected to a 6V supply. If the minimum operating current of relay is 0.1A, calculte the temperature above which the relay will fail to operate when connected to the supply. Take temperature resistance of material as 0.0043/°C

Sagot :

Sqdancefanidn

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.

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