A copper wire has diameter 0.5 mm and resistivity of 1.6 × 10−8Ω m. What will be the length of this wire to make its resistance 10 Ω? How much does the resistance change if the diameter is doubled?
Area of cross-section of the wire, A =π (d/2) 2
Diameter= 0.5 mm = 0.0005 m
Resistance, R = 10 Ω
We know that
radius of wire r = 0.5/2 mm = 0.25 mm = 0.00025 m
`A=pir^2=3.14 xx (0.00025)^2=0.000000019625 m^2`
If the diameter (radius) is doubled, the new radius r = 0.5 mm = 0.0005 m
`A=pir^2=3.14 xx (0.0005)^2=0.000000785m^2`
So, the new resistance will be
`R'=(rho l)/A=(1.6 xx 10^(-8)xx 122.72)/(0.000000785)=2.5 Omega`
Hence, the new resistance will become 1⁄4 times the original resistance.
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