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An electrical cable consists of 200 strands of fine wire, each having 2.26 µΩ resistance. The same potential difference is applied between the ends of all the strands and results in a total current of 0.876 A. (a) What is the current in each strand? (b) What is the applied potential difference? (c) What is the resistance of the cable?

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Answer:

4.38 × 10⁻³ A

0.395mΩ

452 µΩ

Explanation:

(a)the current in each strand is = the total current/ number of strand

0.876/ 200 = 4.38 × 10⁻³ A

(b) potential difference = IR

the total resistance = 2.26 µΩ × 0.876

= 0.395mΩ

c) resistance of the cable =  2.26 µΩ  × 200

=452 µΩ

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