Question 20
A student applies a potential difference V
of (4.0 ± 0.1) V across a resistor of resistance
R of
(10.0 ± 0.3) Ω for a time t
of (50 ± 1) s.
The student calculates the energy E
dissipated using the
equation below.
E = V2t
/ R = (4.02 × 50)
/ 10.0 = 80 J
What is the absolute uncertainty in the calculated energy
value?
A 1.5 J B 3 J C 6 J D 8 J
Reference: Past Exam Paper – June 2018 Paper 12 Q4
Solution:
Answer: D.
E = V2t / R
ΔE / E = 2(ΔV/V) + Δt/t + ΔR/R
ΔE = [2(ΔV/V) + Δt/t + ΔR/R] × E
ΔE = [2(0.1/4.0)
+ (1/50) + (0.3/10.0)] × 80
ΔE = 0.1 × 80
ΔE = 8 J
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