# Numerical problems on Resistivity with solution

Last updated on August 25th, 2023 at 03:07 pm

This post presents a few important numerical problems in physics that you can solve using the concepts of resistivity. If you want a quick revision to memorize the resistivity formula then you can check this post on *resistivity formula derivation*.

Note: Remember that: (1) resistivity is a property of the material (2) the unit of resistivity is ohms times metres and not ohms per metre. (3) In calculations, *l *must be in metres and *A *must be in m^{2}.

**Resistivity – Numerical problems with solution**

**1 ] Question1**

Use the data for the **32 swg** wire given in Figure 1 to calculate a value for the resistivity of nichrome.

**Solution: Answer**

*R *= ρ*l/A*

where *A *= π (*d/*2)^{2}

Rearranging:

**ρ = RA/l**

Fetched data from Figure 1 for the **32 swg** wire.

Substituting *R *= 18.3 Ω when *l *= 1.000 m and *d *= 0.2743 mm:

ρ = (18.3 Ω) π(0.5 × 0.2743 × 10^{–3} m)^{2} / 1 m = 1.08 × 10^{–6} Ω m

**2 ] Question 2**

A carbon chip of resistivity 3.0x 10^{–5} Ω m has the dimensions shown in Figure 2.

What resistance does the chip have for a current in the direction shown?

**Solution: Answer**

Using the formula:

*R *= ρ*l/A*

Referring figure 2 we get, *l *= 10 mm *= *10 × 10^{–3 }m

and *A *= 5 mm × 1 mm *= *5 × 10^{–3} m × 1 × 10^{–3 }m

*= *5 × 10^{–6} m^{2}

Resistance of the chip = R = ρ*l/A= ( 3.0x 10*^{–5}*Ωm)(10 × 10*^{–3 }*m) /5 × 10*^{–6}*m*^{2}

*=*0.060 Ω

**3 ] Question 3**

A conductor wire of length 2.0 m and diameter 4 mm has a resistance of 16 Ω at room temperature. Find its resistivity. In your opinion, is the wire is made of pure metal or alloy?

Solution:

Length of wire, l = 2.0 m, diameter, d = 4 mm = 4 × 10^{-3}m.

So, the area of cross-section A = π d^{2}/ 4.

The resistance, R = 16 Ω.

If ρ be the resistivity of the material of the wire, we have

R = ρ l / A,

which gives

ρ = R A / l = R π d^{2}/ (4 l)

Or ρ = (16 Ω) (3.14) (4 × 10^{-3}m)2 / (4 × 2.0 m) = 100.5 × 10^{-6}Ω m.

The above value of ρ tells us that the wire is made of alloy.

**4 ] Question 4**

A metallic wire has a radius of 0.25 mm and a resistance of 5 Ω. What will be the length of this wire if the resistivity of the material of the wire is 3.2 × 10^{-8} Ω m?

Solution

Radius r = 0.25 mm = 2.5 × 10^{-4}m.

Required resistance R = 5 Ω

Resistivity of the material of the wire ρ = 3.2 × 10^{-8}Ω m

Length of wire L =?

Now, the area of the cross-section of the wire A =π r^{2}.

We have, R = ρL / A, which gives

L= R A / ρ = Rπ r^{2}/ρ

or, L = (5 Ω) × (3.14) × (2.5 × 10^{-4}m)^{2}/ (3.2 × 10^{-8}Ω m)

=> L= 30.7 m.

So, the required length of wire L = 30.7 m.