Example 1:
A coil consists of 2,000 turns of copper wire having a cross-sectional area of 0.8 mm2. The mean length per turn is 80 cm, and the resistivity of copper is 0.02 µ Ω-m. Find the resistance of the coil.
Solution:
Length of the coil, L= Number of turns × mean length per turn
= 2,000 × 0.8 = 1,600 m
Cross-sectional area of wire, a = 0.8 mm2 = 0.8 × 10-6 m2
Resistivity of copper, ρ = 0.02 µΩ-m = 2 × 10-2 × 10-6 Ω-m = 2 × 10-8Ω-m
Current = V/R
Resistance of the coil, R = l/a = 2 × 10-8 × 1,600 / 0.8 × 10-6 = 40 Ω Ans.
Power = voltage × current
= V × I
Current = V/R
= 110/40
= 2.75A
P= 2.75×110
= 302.5W
Check site
https://www.engineeringenotes.com/electrical-engineering/current/resistance-meaning-and-laws-current-electrical-engineering/27352
life VIDEO for this ASSIGNMENTS (the video consists of how we arrive at our answer).
Answer for resistance is 40 ohms
Power = 302.5W
AREA in mm squares = 10^-6. And not 10^-3
Relax and watch please like and subscribe for more videos that will help you
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