Question

the resistance of resistor varies linearly with temperature.If resistance at 30° is 60ohm and at 70° is 80ohm then resistance at 60° will be​

Answer 1

resistance at 60° is 75 Ω

resistance varies as R = R_0[1 + α(T - T_0)]

where R is resistance at T, R_0 is resistance at T_0 and α is coefficient of resistance.

let T_0 is 60°C then, R_0 is resistance at 60°C

given,

at 30°, resistance = 60 Ω

so, 60 = R_0[1 + α(30 - 60)]

⇒60 = R_0[1 - 30α] .......(1)

at 70°, resistance = 80 Ω

so, 80 = R_0[1 + α(70 - 60)]

⇒80 = R_0[1 + 10α] ........(2)

from equations (1) and (2),

60/80 = (1 - 30α)/(1 + 10α)

⇒3/4 = (1 - 30α)/(1 + 10α)

⇒3 + 30α = 4 - 120α

⇒150α = 1

⇒α = 1/150

now, R_0 = 60/(1 - 30/150) = 60/(1 - 1/5)

= 75Ω

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Answer 2

$\huge\bold\purple{Answer:-}$

resistance varies as R = R_0[1 + α(T - T_0)]

where R is resistance at T, R_0 is resistance at T_0 and α is coefficient of resistance.

let T_0 is 60°C then, R_0 is resistance at 60°C

given,

at 30°, resistance = 60 Ω

so, 60 = R_0[1 + α(30 - 60)]

⇒60 = R_0[1 - 30α] .......(1)

at 70°, resistance = 80 Ω

so, 80 = R_0[1 + α(70 - 60)]

⇒80 = R_0[1 + 10α] ........(2)

from equations (1) and (2),

60/80 = (1 - 30α)/(1 + 10α)

⇒3/4 = (1 - 30α)/(1 + 10α)

⇒3 + 30α = 4 - 120α

⇒150α = 1

⇒α = 1/150

now, R_0 = 60/(1 - 30/150) = 60/(1 - 1/5)

= 75Ω