The resistance of a wire of iron is `10 ohm` and temperature coefficient of resistivity is `5 xx 10^-3//.^@C`, At `20^@C` it carries `30 mA` of current. Keeping constant potential difference between its ends. The temperature of the wire is raised to `120^@C`. The current in `mA` that flows in the wire now is.

Options

A. 10 mA
B. 20 mA
C. 5 mA
D. 15 mA

Answer

B

Explanation

Potential difference applied `= V = IR`
`= 30 xx 10^(-3)xx10`
`=0.3V`
Resistance at `120^(@)C` is given by,
`R_(120)=R_(20)(1+alpha xx 100)=15 Omega`.
`:.` Current at `120^(@)C` is given by,
`I=(V)/(R_(120))=(0.3)/(15)=0.02 A = 20 mA`