A gold wire 5.44 m long and of diameter 0.840 mm carries a current of 1.25 A .a.Find the resistance of this wire.Express your answer in ohms.b.Find the potential difference between its ends.Express your answer in volts.

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Bryanna Villarreal · Accepted Answer

Solution:Resistivity of gold is 0.022 mmlenght=5.44 mdiameter   =  0.84  ×      10          −      3        m  marea   =  π  [            0.84      ×              10                  −          3                      2        ]    2  a) Resistance   =            ρ      l        A    =            0.022      ×      5.44              (                        0.84          ×                      10                          −              3                                      2                    )        2              R  =      0.1197          0.5534      ×              10                  −          6                      =  0.216  ×      10    6    nb) Potential difference V=IR  =  1.25  ×  0.216  ×      10    6    =  0.27  ×      10    6       V  V  =  0.27  ×      10    6       V

A gold wire 5.44 m long and of diameter 0.840 mm carries a current of 1.25 A . a. Find the resistance of this wire. Express your answer in ohms. b. Find the potential difference between its ends.

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