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The resistance of iron wire is 10Omega a...

The resistance of iron wire is `10Omega` and `alpha=5xx(10^(-3))/( .^(@)C)`. If a current of 30A is flowing in it at `20^(@)C` keeping the potential difference across its length constant, if the temperature is incresed to `120^(@)C` what is the current flowing through that wire?

Text Solution

Verified by Experts

`=(R_(120) -R_(20))/(R_(20)(120-20)), 5 xx 10^(-3) =(R_(120) -10)/(10 xx 100)`
`therefore R_(120) = 15 Omega`, But V=IR
Here V is constant, Hence.
`(I_(120))/I_(20) =(R_(20))/(R_(120)), I_(120)/30 = 10/15, therefore I_(120) = 20A`
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