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?

The resistance of iron wire is `10Omega` and `alpha=5xx(10^(-3))/( .^(

1.	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?
Answer» `alpha=(R_(120)-R_(20))/(R_(20)(120-20)),5xx10^(-3)=(R_(120)-10)/(10xx100)` `thereforeR_(120)=15Omega,but V=IR` Here V is constant Hence, `(I_(120))/(I_(20))=(R_(20))/(R_(120)),(I_(120))/(30)=(10)/(15),becauseI_(120)=20A`

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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?

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