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
thanks
ReplyDelete