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4.29 Using these three formulas V = IR R = (pL)/A A = r(0.5d)2 find the difference in current (I) that a copper wire (p = 1.72 × 10-8 Q · m) can carry over an aluminum wire (p = 2.75 × 10−8 Q ·m) with equal diameters (d) of 0.5 cm and a length (L) of 10 000 m carrying 110 V (volts).

4.29 Using these three formulas V = IR R = (pL)/A A = r(0.5d)2 find the difference in current (I) that a copper wire (p = 1.72 × 10-8 Q · m) can carry over an aluminum wire (p = 2.75 × 10−8 Q ·m) with equal diameters (d) of 0.5 cm and a length (L) of 10 000 m carrying 110 V (volts).

Introductory Circuit Analysis (13th Edition)
Introductory Circuit Analysis (13th Edition)
13th Edition
ISBN:9780133923605
Author:Robert L. Boylestad
Publisher:Robert L. Boylestad
Chapter1: Introduction
Section: Chapter Questions
Problem 1P: Visit your local library (at school or home) and describe the extent to which it provides literature...
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4.29 Using these three formulas V = IR R = (pL) /A A = n(0.5d)2 find the difference in current (I) that a copper wire (p = 1.72 x 10-82 · m) can carry over an aluminum wire (p = 2.75 × 10−8 № • m) with equal diameters (d) of 0.5 cm and a length (L) of 10 000 m carrying 110 V (volts).
Transcribed Image Text:4.29 Using these three formulas V = IR R = (pL) /A A = n(0.5d)2 find the difference in current (I) that a copper wire (p = 1.72 x 10-82 · m) can carry over an aluminum wire (p = 2.75 × 10−8 № • m) with equal diameters (d) of 0.5 cm and a length (L) of 10 000 m carrying 110 V (volts).
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