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Which electrical and mechanical calculations does a product designer need, and how are they applied?

Applied electrical and mechanical calculations: electrical power and energy, Ohm's law in context, mechanical advantage, velocity ratio, efficiency and the moment of a force, and selecting and applying the right formula to a design problem.

A focused answer to OCR A-Level Product Design on applied electrical and mechanical calculations: electrical power and energy, Ohm's law in context, mechanical advantage, velocity ratio, efficiency and moments, and how to select and apply the right formula to a design problem.

Generated by Claude Opus 4.812 min answer

Reviewed by: AI editorial process; not yet individually human-reviewed

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  1. What this dot point is asking
  2. Electrical power and energy
  3. Ohm's law in context
  4. Mechanical advantage, velocity ratio and efficiency
  5. Moments and choosing the right formula

What this dot point is asking

OCR wants you to select and apply the right electrical or mechanical formula to a design problem: electrical power and energy, Ohm's law, mechanical advantage, velocity ratio, efficiency and moments. This dot point pulls the quantitative skills together and tests whether you can choose the correct formula, not just substitute into a given one.

Electrical power and energy

Ohm's law in context

Mechanical advantage, velocity ratio and efficiency

Moments and choosing the right formula

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 heater operates at 230 V and draws a current of 8 A. Calculate the electrical power it uses, and the energy used in 30 minutes in kilowatt-hours.
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A Component 01 power and energy calculation. Marks for the power and for the energy with units.

Electrical power is voltage times current: P=V×I=230×8=1840P = V \times I = 230 \times 8 = 1840 W, which is 1.841.84 kW. Energy is power times time: in 3030 minutes (0.50.5 hours), energy =1.84×0.5=0.92= 1.84 \times 0.5 = 0.92 kWh.

A common dropped mark is leaving the power in watts when the question asks for kilowatt-hours (convert 1840 W to 1.84 kW first), or using 30 instead of 0.5 hours for the time.

OCR 20226 marksA pulley system has a velocity ratio of 4. It lifts a load of 200 N with an effort of 60 N. Calculate the mechanical advantage and the efficiency, and explain why the efficiency is less than 100 percent.
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A Component 01 mechanical-advantage and efficiency calculation. Marks for the mechanical advantage, the efficiency and the explanation.

Mechanical advantage = load divided by effort: MA=20060=3.33\text{MA} = \frac{200}{60} = 3.33. Efficiency = MAVR×100=3.334×100=83%\frac{\text{MA}}{\text{VR}} \times 100 = \frac{3.33}{4} \times 100 = 83\% (since for an ideal machine MA would equal VR). The efficiency is less than 100 percent because energy is lost to friction in the pulley bearings and the rope, and to the work done lifting the moving parts of the system itself, so the effort must do more work than just lifting the load.

A common dropped mark is computing MA or efficiency wrongly (efficiency is MA over VR), or not explaining the loss in terms of friction and the weight of moving parts.

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