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,
Volatage `V _(1) = 10V`
Current through resistor, `I _(1) = 0.1A`
Temperature, `T _(1) = 40^(@)C`
Temperature coefficient of resistance, `alpha =2 xx10 ^(-14) ""^(@)C ^(-1)`
Current drops to `0.098 A`
`I _(2) = 0.098A`
At `T_(1),`
`R _(T1) = (V_(1))/(I _(1)) = (10)/(0.1)= 100 Omega`
`R _(T) =R_(0) (1 + alpha T)" "...(1)`
`R_(T1) =R_(0) (1+ alpha T_(1))`
`100 = R_(0) (1+2xx 10 ^(-14) xx 40)`
`100 = R_(0) (1+ 8xx 10 ^(-13))`
`100 = R_(0) (1.008)`
`implies R_(0) = 99.21 Omega`
`At T _(2),`
`R _(T2) = (V_(1))/(I _(2)) `
`= (10)/( 0.098) = 102.04Omega`
Then from equation (1)
`RT _(2) = (1+ alpha T_(2))`
`102. 02 = (1+ 2xx 10 ^(-14) T_(2))`
` 1 + 2xx 10 ^(-14) T _(2) = 102. 04`
` 1+ 2 xx 10 ^(-14) T_(2) = 1. 0285`
`2xx 10 ^(-14) T_(2) = 0.0285`
`T_(2)=142.6 ~~ 142 ^(@) C`