The resistance of a wire of iron is 10 ohms and temp. coefficient of resistivity is 5 times 10^{−3}/^{circ} C. At 20^{circ} C it carries 30 milliamperes of current. Keeping constant potential difference between its ends, the temperature of the wire is raised to 120^{circ} C. The current in milliamperes that flows in the wire is

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The resistance of a wire of iron is  10 ohms and temp. coefficient of resistivity is 5 times 10^{−3}/^{circ} C. At 20^{circ} C it carries 30 milliamperes of current. Keeping constant potential difference between its ends, the temperature of the wire is raised to 120^{circ} C. The current in milliamperes that flows in the wire is

Veronella6nk · Accepted Answer

Step 1: Calculation of resistance at       120          ∘           CLet variables at       20          ∘           C be marked by subscript 1 and at       120          ∘           C be marked by subscript 2.Given that,            R      1              R      2        =            1      +      α              t        1                    1      +      α              t        2            Where       t    1    =      20          ∘        C and       R    1     is resistance at        t    1     and       t    2    =      120          ∘        C and       R    2   is resistance at       t    2  .We have,       R    1    =  10  Ω  ,  α  =  5  ×      10          −      3                  /              ∘        C    ∴      10          R      2        =            1      +      (      5      ×              10                  −          3                    ×      20      )              1      +      (      5      ×              10                  −          3                    ×      120      )          ⇒      R    2    ≈  15  ΩStep 2: Calculation of current at       120          ∘        Cwhen the potential difference is constant,  i  ∝      1    R      ∴            i      1              i      2        =            R      2              R      1        =      15    10    =  1.5It is also provided,   i  1  =  30  m  A    ∴  i  2  =  301.5    ⇒  i  2  =  20  m  ATherefore, current flowing through the circuit at       120          ∘       C is 20mA.

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