Chapter 3: Electricity

Q&A -Ask Doubts and Get Answers

Q
Electricity

Question:

A copper wire has a diameter 0.5 mm and resistivity of 1.6 × 10–8 Ω 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?

Answer:

Given:

Diameter, d = 0.5\ mm

Radius,r = \frac{d}{2} = 0.25\ mm

r = 0.25 \times 10^{-3}\ m

Resistivity, \rho = 1.6 \times 10^{-8}\ \Omega m

Resistance, R = 10\ \Omega

Length of wire,L =\ ?

We know,

R = \frac{\rho L}{A} ----------(1)

Here, the area of cross section

A = \pi{r}^{2}

A = 3.14 \times 0.25 \times (10^{-3})^{2}

A = 3.14 \times 0.0625 \times 10^{-6}

A = 0.19625 \times 10^{-6}

Putting this value in (1)

10 = \frac{1.6 \times 10^{-8} L}{0.19625 \times 10^{-6}}

L = \frac{10 \times 0.19625}{1.6 \times 10^{-2}}

L = 10 \times 0.12265625 \times 10^2

L = 0.12265625 \times 10^3

L = 122.65\ m

 

Again using,

R = \frac{\rho L}{A}

R = \frac{\rho L}{\pi r^2}

R = \frac{\rho L}{\pi (\frac{d}{2})^{2}}

R = \frac{ \rho L}{\pi \frac{d^2}{4}}

R \propto \frac{1}{d^2}----------(2)

The new resistanceR' when the diameter is double is given by,

R' \propto \frac{1}{(2d)^2}

R' \propto \frac{1}{4(d)^2}---------(3)

Dividing (3) by (2)

\frac{R'}{R} = \frac{\frac {1}{4d^2}}{\frac{1}{d^2}}

\frac{{R}'}{R} = \frac{1}{4}

R' = \frac{R}{4}

R' = \frac{10}{4}

R' = 2.5\ \Omega

So, the decrease in the resistance = R - R'

Decrease in the resistance = 10 - 2.5 = 7.5\ \Omega


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