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A student needs to make a 0.12 Ω resistor. She has some copper wire of 0.80 mm diameter. Resistivity of copper is 1.8 × 10–8 Ωm

(a) Determine the cross-sectional area of the wire.

(b) Calculate the length of wire required for the 0.12 Ω resistor

Current Electricity

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Answer

Given,

  • Resistance (R\text R) = 0.12 Ω
  • Diameter of the wire (d\text d) = 0.80 mm = 0.80 x 10–3 m
  • Resistivity of copper (ρ\text ρ) = 1.8 × 10–8 Ωm

(a) Area of cross-sectional (A\text A) is given by,

A=π(d2)2≈3.14×(0.80×10−32)2≈3.14×(0.40×10−3)2≈3.14×0.16×10−6⇒A≈0.5024×10−6⇒A≈5.024×10−7 m2\text A = \pi \left(\dfrac{\text {d}}{2}\right)^2 \\[1em] \approx 3.14 \times \left(\dfrac {0.80 \times 10^{-3}}{2}\right)^2 \\[1em] \approx 3.14 \times (0.40 \times 10^{-3})^2 \\[1em] \approx 3.14 \times 0.16 \times 10^{-6} \\[1em] \Rightarrow \text A \approx 0.5024 \times 10^{-6} \\[1em] \Rightarrow \text A \approx 5.024 \times 10^{-7}\ \text m^2

Thus, the cross-sectional area is approximately 5.02 x 10-7 m2.

(b) Let length of the conductor be 'l\text l'.

Now, resistance (R) of the wire is given by,

R=ρlA⇒l=RAρ\text R = \text ρ \dfrac{\text l}{\text A} \\[1em] \Rightarrow \text l = \dfrac {\text {RA}}{\text ρ} \\[1em]

On putting values,

⇒l=0.12×5.02×10−71.8×10−8=0.12×5.02×10−7+81.8=0.12×5.02×101.8=1.2×5.021.8=12×5.0218=2×5.023≈2×1.67⇒l=3.34 m\Rightarrow \text l = \dfrac {0.12 \times 5.02 \times 10^{-7}}{1.8 \times 10^{-8}} \\[1em] = \dfrac {0.12 \times 5.02 \times 10^{-7+8}}{1.8} \\[1em] = \dfrac {0.12 \times 5.02 \times 10}{1.8} \\[1em] = \dfrac {1.2 \times 5.02}{1.8} \\[1em] = \dfrac {12 \times 5.02}{18} \\[1em] = \dfrac {2 \times 5.02}{3} \\[1em] \approx 2 \times 1.67 \\[1em] \Rightarrow \text l = 3.34\ \text m

The student needs a length of approximately 3.34 m of given copper wire to make a 0.12 Ω resistor.

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