Question

A carbon rod with a radius of 1.9 mm is used to make a resistor. What length of the carbon rod should be used to make a 3.7 Ω resistor? The resistivity of this material is 1.8 × 10−5 Ω · m .

Answer 1

Answer:

Length = 2.32 m

Explanation:

Let the length required be 'L'.

Given:

Resistance of the resistor (R) = 3.7 Ω

Radius of the rod (r) = 1.9 mm = 0.0019 m [1 mm = 0.001 m]

Resistivity of the material of rod (ρ) = $1.8\times 10^{-5}\ \Omega\cdot m$

First, let us find the area of the circular rod.

Area is given as:

$A=\pi r^2=3.14\times (0.0019)^2=1.13\times 10^{-5}\ m^2$

Now, the resistance of the material is given by the formula:

$R=\rho( \frac{L}{A})$

Express this in terms of 'L'. This gives,

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

Now, plug in the given values and solve for length 'L'. This gives,

$L=\frac{3.7\ \Omega\times 1.13\times 10^{-5}\ m^2}{1.8\times 10^{-5}\ \Omega\cdot m}\\\\L=\frac{4.181}{1.8}=2.32\ m$

Therefore, the length of the material required to make a resistor of 3.7 Ω is 2.32 m.

Answer 2

The required length of material will be "2.32 m".

Resistance

According to the question,

Resistance, R = 3.7 Ω

Rod's radius, r = 1.9 mm or,

                        = 0.0019 m

Rod's resistivity, ρ = 1.8 × 10⁻5 Ω

We know the area or circular rod,

→ A = πr²

By substituting the values, we get

     = 3.14 × (0.0019)²

     = 1.13 × 10⁻⁵ m²

We know the relation,

→ R = ρ ($\frac{L}{A}$)

or,

ρ × L = R × A

Now, the length will be:

     L = $\frac{3.7\times 1.13\times 10^{-5}}{1.8\times 10^{-5}}$

        = $\frac{4.181}{1.8}$

        = 2.32 m

Thus the above answer is appropriate.

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