Potentiometer — A/L Physics
Aligned with the NIE syllabus
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1 question with a worked solution
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A potentiometer wire of length 100 cm and resistance 5.0 Ω is connected in series with a 4.0 V driver battery of negligible internal resistance and a series limiting resistor Rs = 15.0 Ω. What is the potential gradient along the potentiometer wire?
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Correct answer: D) 1.0 V m⁻¹ (0.01 V cm⁻¹)
Total circuit resistance Rtotal = Rwire + Rs = 5.0 Ω + 15.0 Ω = 20.0 Ω. Current in potentiometer wire I = Vdriver / Rtotal = 4.0 V / 20.0 Ω = 0.20 A. Potential drop across the wire Vwire = I × Rwire = 0.20 A × 5.0 Ω = 1.0 V. Potential gradient k = Vwire / L = 1.0 V / 1.0 m = 1.0 V m⁻¹ (0.01 V cm⁻¹).
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Step 1: I = 4.0 V / (5.0 + 15.0) Ω = 0.20 A. Step 2: Vwire = I · Rwire = 0.20 A × 5.0 Ω = 1.0 V. Step 3: Potential gradient k = Vwire / L = 1.0 V / 1.0 m = 1.0 V m⁻¹. Answer: 1.0 V m⁻¹ (0.01 V cm⁻¹).Modelled on: the A/L Physics examination pattern
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Where does Potentiometer fit in the syllabus?
Unit 7.8 · Year 2 · Term 1
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