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Given, diameter,d= 0.5mm
resistivity,p =1.6×10-8m
Resistance,R =10m
let the length of wire be 1.
A=πd^2/4
R= P1 /A
= P1 /πd^2/4
= 1 = Rπd^2/4p
1 = 10× 3.14 ×[0.5×10^-3]2/4×1.6×10^-8
1= 122m
R prposnal 1/d^2
If the diameter is doubled,
resistance will be one_ fourth .
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2
Answer:
d= 0.5mm
rho= 1.6 × 10^-8
R = 10mm
A= pid^2/4
R= rho l/A
After solving......
l= 122m
Dis doubled so R = 1/4
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