When subjected to a voltage of 10 V, the current through a resistor at a temperature of `40^@C` is 0.1 A. The temperature coefficient of resistance of the material of the resistor is `2 xx 10^(-4) ""^@C` . The temperature of the resistor in `""^@C` when the current drops to 0.098 A is

Options

A. 142
B. 167
C. 181
D. 206

Answer

A

Explanation

Given , Voltage , `V_1 = 10 V`
Current through resistor , `I_1 = 0.1 A ` ,
temperature , `t_1 = 40^@C`
and temperature coefficient , ` alpha = 2 xx 10^(-4) ""^@C^(-1)`
` therefore` Resistance of resistor at `t_1 = 40^@C`
` R_(t_1) = (V_1)/(I_1) = (10)/(0.1) = 100 Omega`
Since , `R_t = R_0 (I + alpha t)`
`R_(t_1) = R_0 (1 + alpha t_1)`
` 100 = R_0 (1 + 2 xx 10^(-4) xx 40 )`
` R_0 = (100)/(1.008) = 99.21 Omega`
At temperature `t_2` , current drops to 0.098 A, then resistance of resistor is given by
`R_(t_2) = V/I = (10)/(0.098) = 102.04 Omega`
Since `R_(t_2) = R_0 (1 + alpha t_2) `
` 102.04 = 99.21 ( 1+ 2 xx 10^(-4) t_2) `
`rArr 1 + 2 xx 10^(-4) t_2 = 1.0285 rArr t_2 = (0.0285)/(2 xx 10^(-4) ) `
` = 142.6 ~~ 142^@C`