a copper wire has diameter 0.5mm and resistivity of 1.6×10‐⁸ . What will the length of this wire to make its resistance 10 ohm?
Share
a copper wire has diameter 0.5mm and resistivity of 1.6×10‐⁸ . What will the length of this wire to make its resistance 10 ohm?
Sign Up to our social questions and Answers Engine to ask questions, answer people’s questions, and connect with other people.
Login to our social questions & Answers Engine to ask questions answer people’s questions & connect with other people.
Step-by-step 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
Find out
We have to find the length of the wire
Let the length of the wire be L
Formula
We know that
R = ρ (L) / (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 new resistance be Rʹ
R = ρ (L) / (A)
R’ = ρ (L) / (4A)
R’ = ρ (L) X 1/(4A)
Hence, if diameter doubles, resistance becomes 1/4 times.
Answer:
l = 1.226 × 10² m
Step-by-step explanation:
R=ρ(l/a)
R=10Ω
ρ =1.6×10^-8 Ω/m
d=0.5mm
So, r = 0.25 mm = 25 × 10^-5m
Area,a = πr² = 3.14 × (25×10^-5)²
= 1962.5×10^-10 m²
l = (Ra)/ρ
= (10×1962.5×10^-10)/1.6 × 10^-8
= 1.226 × 10² m