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Thursday, February 14, 2019

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

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