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How do potential difference, resistance and resistivity determine energy transfer in a component?

Energy, power and resistance: potential difference and electromotive force, resistance and Ohm's law, the I-V characteristics of conductors, lamps, diodes and thermistors, resistivity, and electrical power and energy.

A focused answer to the OCR H556 content on energy, power and resistance, covering potential difference and electromotive force, resistance and Ohm's law, the I-V characteristics of metallic conductors, filament lamps, diodes and thermistors, resistivity and its temperature dependence, and electrical power and energy.

Generated by Claude Opus 4.812 min answer

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  1. What this dot point is asking
  2. The answer
  3. Examples in context
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What this dot point is asking

OCR wants you to define potential difference and electromotive force, use resistance and Ohm's law, interpret the I-V characteristics of a metallic conductor, filament lamp, diode and thermistor, use resistivity and its temperature dependence, and calculate electrical power and energy.

The answer

Potential difference and emf

Resistance and Ohm's law

I-V characteristics

Resistivity, power and energy

Examples in context

Thermistors are used as temperature sensors in thermostats and engine management, exploiting the fall in resistance with temperature. Diodes rectify alternating current and protect circuits from reverse polarity. Resistivity values guide the choice of conductor: copper for low-loss wiring, nichrome for heating elements because of its high resistivity. Power ratings on appliances use P=IVP = IV to set the correct fuse and cable. The OCR resistivity practical finds ρ\rho from the gradient of a resistance-against-length graph.

Try this

Q1. Define potential difference. [1 mark]

  • Cue. The energy transferred per unit charge between two points, V=WQV = \frac{W}{Q}.

Q2. A resistor carries 0.25 A0.25\ \text{A} at a p.d. of 5.0 V5.0\ \text{V}. Find its resistance and the power dissipated. [2 marks]

  • Cue. R=VI=20 ΩR = \frac{V}{I} = 20\ \Omega; P=IV=0.25×5.0=1.25 WP = IV = 0.25 \times 5.0 = 1.25\ \text{W}.

Q3. Describe how the resistance of an NTC thermistor changes as its temperature rises, and why. [2 marks]

  • Cue. Resistance decreases, because more charge carriers are released as temperature rises.

Exam-style practice questions

Practice questions written in the style of OCR exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

OCR 20194 marksA nichrome wire of length 0.80 m0.80\ \text{m} and cross-sectional area 1.2×10−7 m21.2 \times 10^{-7}\ \text{m}^2 has a resistance of 7.3 Ω7.3\ \Omega. Calculate the resistivity of nichrome.
Show worked answer →

Rearrange R=ρLAR = \frac{\rho L}{A} for resistivity: ρ=RAL\rho = \frac{RA}{L}.

Substitute: ρ=7.3×1.2×10−70.80=8.76×10−70.80=1.1×10−6 Ω m\rho = \frac{7.3 \times 1.2 \times 10^{-7}}{0.80} = \frac{8.76 \times 10^{-7}}{0.80} = 1.1 \times 10^{-6}\ \Omega\,\text{m}.

Markers reward rearranging the resistivity equation, correct substitution, and the value about 1.1×10−6 Ω m1.1 \times 10^{-6}\ \Omega\,\text{m} (a typical value for nichrome).

OCR 20223 marksA 12 V12\ \text{V} heater has a power rating of 36 W36\ \text{W}. Calculate the current through it and its resistance.
Show worked answer →

Current from P=IVP = IV: I=PV=3612=3.0 AI = \frac{P}{V} = \frac{36}{12} = 3.0\ \text{A}.

Resistance from V=IRV = IR: R=VI=123.0=4.0 ΩR = \frac{V}{I} = \frac{12}{3.0} = 4.0\ \Omega. (Equivalently R=V2P=14436=4.0 ΩR = \frac{V^2}{P} = \frac{144}{36} = 4.0\ \Omega.)

Markers reward the current 3.0 A3.0\ \text{A} and the resistance 4.0 Ω4.0\ \Omega, by either route.

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