A copper wire has diameter 0.5mm and resistivity of 1.6×10 −8 Ωm. What will be the length of this wire to make its resistance 10Ω? How much does the resistance change if the diameter is doubled?
Answers
Explanation:
Resistivity (ρ) = 1.6 × 10-8 Ω m
Resistance (R) = 10 Ω
Diameter (d) = 0.5 mm
d = 5 × 10⁻⁴ m
Hence, we will get radius
Radius (r) = 0.25 mm
r = 0.25 × 10⁻³ m
r = 2.5 × 10⁻⁴ m
We need to find the area of cross-section
A = πr2
A = (22/7)(2.5 × 10⁻⁴)2
A = (22/7)(6.25×10⁻⁸)
A = 1.964 × 10-7 m2
Answer:
L = 122.72 m
R'=(1/16)R
Explanation:
Given:
Resistivity (ρ) = 1.6 × 10-8 Ω m
Resistance (R) = 10 Ω
Diameter (d) = 0.5 mm
d = 5 × 10⁻⁴ m
∴Radius (r) = 0.25 mm
r = 0.25 × 10⁻³ m
r = 2.5 × 10⁻⁴ m
area of cross-section (A) = πr²
A = (22/7)(2.5 × 10⁻⁴)²
A = (22/7)(6.25×10⁻⁸)
A = 1.964 × 10-7 m²
To Find:
Length of the wire
Formula used:
R = ρL/A
L =R×A/ρ
Substituting the values in the above equation we get,
L =10 × 1.964 × 10⁻⁷/ 1.6 × 10⁻⁸ m
L = 1.964×10⁻⁶ /1.6 × 10⁻⁸
L = 122.72 m
If the diameter of the wire is doubled, the new diameter = 2 × 0.5 = 1mm = 0.001m
Let the new resistance be Rʹ
d=2r
2d=2*2r
=4r
∴A'=π(4r)²
=16πr²
=16A
∴R'=ρl/(16A)
R'=R×1/16
So, the new resistance will be 1/16 times of the original resistance.
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