The current in a block of semiconductor is 30.0 mA when there is a potential difference (p.d.) of 10.0 V across it.

Question 16

The current in a block of semiconductor is 30.0 mA when there is a potential difference (p.d.) of 10.0 V across it. The dimensions of the block and the direction of the current in it are as shown.

The electrical meters used are accurate to ± 0.1 mA and ± 0.1 V. The dimensions of the block are accurate to ± 0.2 mm.

What is the resistivity of the semiconductor?

A 10.0 ± 0.2 Ω m

B 10.0 ± 0.3 Ω m

C 10.0 ± 0.5 Ω m

D 10.0 ± 0.8 Ω m

Reference: Past Exam Paper – June 2017 Paper 12 Q4

Solution:

Answer: C.

We need to tackle this question step-by-step.

Resistance R = ρL / A

We need to make the resistivity the subject of the formula.

The best approach is to re-arrange the formula and only replace the numerical values as the last step.

Resistivity ρ = RA / L

From Ohm’s law, R = V / I

So, replacing R = V / I in to the equation for resistivity,

ρ = VA / IL

Cross-sectional area, A = l × w = (15×10^-3) × (30×10^-3)

Resistivity ρ = Vlw / IL

V = 10.0 V

l = 15×10^-3 m               w = 30×10^-3 m             L = 15×10^-3 m

I = 30×10^-3 A

Resistivity ρ = [10.0 × (15×10^-3) × (30×10^-3)] / [(30×10^-3) × (15×10^-3)]

Resistivity ρ = 10 Ωm

 

Since all the values given in the choices are 10, we can understand that we are eon the right track.

Considering the uncertainties,

Δρ / ρ = ΔV/V + Δl/l + Δw/w + ΔI/I + ΔL/L

Δρ = (ΔV/V + Δl/l + Δw/w + ΔI/I + ΔL/L) × ρ

Δρ = (0.1/10 + 0.2/15 + 0.2/30 + 0.1/30 + 0.2/15) × 10

Δρ = 0.467 = 0.5 Ωm

Uncertainties should be given to only 1 s.f.

 
Resistivity ρ = 10 ± 0.5 Ωm