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?
Answers
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