Question

The resistance of a wire at room temperature 20°C is found to be 20W. Now to increase the resistance by 20%, the temperature of the wire must be [The temperature coefficient of resistance of the material of the wire is 0.002 per °C].

Answer 1

Answer:

Explanation:

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Answer 2

Answer:

120°C

Explanation:

R=20ohm

$R'=20+(\frac{20}{100} *20)$ =24 ohm

α=0.002°$C^{-1}$

R'=R(1+αΔT)

⇒24 = 20( 1 + 0.002ΔT)

⇒1.2 = 1  + 0.002ΔT

⇒0.2 = 0.002ΔT

⇒200 =2ΔT

⇒ΔT =100°C

temp =T +ΔT = 20+100=120°C