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Friday, September 6, 2019

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.


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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