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Problem 19.21 - Enhanced -with Solution 3 of 16 Constants Part A A gold wire 6.32 m long and of diameter 0.810 mm carries a current of 1.27 A You may want to review (Pages 602-606) For related problem-solving tips and strategies, you may want to Find the resistance of this wire. view a Video Tutor Solution of Electrical hazards in heart surgery Submit Request Answer Part B Find the potential difference between its ends. Submit Request Answer Provide Feedback Next>

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Answer #1

A) Resistance = pL/A

the resistivity of gold p = 2.44*10^-8 ohm -m

R = 2.44*10^-8*6.32 / pi*(0.405*10^-3)^2

= 0.3 ohm

B) V = IR

= 1.27*0.3 = 0.38 V

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