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4.4.5Electrons, Waves and Photons

Energy Transfer — OCR A Level Physics A H556 Revision Notes & Topical Questions

5 min read 5 parts Lesson 38 of 62 Unit quiz

📚 OCR A Level Physics A H556 Hub / Topical Revision
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Part 1 of 5Power in the Circuit

4.4.5 Energy Transfer

Because of internal resistance, not all the energy supplied by the chemical reaction reaches the load.

Total power
  • The source supplies electrical power εI to the whole circuit, found by multiplying e.m.f. by current.
Useful power
  • The power delivered to the load resistor is I²R, which is also the terminal p.d. multiplied by current.
Wasted power
  • The power transferred inside the cell is I²r, which heats the cell and is not available to the load.
Conservation
  • Energy is conserved, so εI = I²R + I²r for every circuit with a single source.

Part 2 of 5Efficiency

Efficiency
Definition
  • Efficiency is useful power output divided by total power input, often shown as a percentage.
In terms of p.d.
  • Efficiency equals terminal p.d. divided by e.m.f., which is V/ε.
In terms of resistance
  • Efficiency equals R / (R + r), so a large load resistance compared with r gives high efficiency.
Trade-off
  • Reducing the load resistance draws more current, but a larger share of the energy is wasted in the cell.

Part 3 of 5Steps for Energy Problems

1
Find the current using I = ε / (R + r) from the given e.m.f. and resistances.
2
Calculate the total power εI and the load power I²R from this current.
3
Subtract, or use I²r, to find the power wasted inside the cell.
4
Divide load power by total power to obtain the efficiency and express it as a fraction or percentage.
5
For energy rather than power, multiply by the time in seconds.

Part 4 of 5Worked Example

Question

  • A cell of e.m.f.
  • 6.0 V and internal resistance 1.0 Ω is connected to a 5.0 Ω resistor.
  • Find the power in the resistor, the power wasted in the cell and the efficiency.

  1. Current I = 6.0 / (5.0 + 1.0) = 1.0 A.
  2. Total power = εI = 6.0 × 1.0 = 6.0 W.
  3. Load power = I²R = 1.0² × 5.0 = 5.0 W; wasted power = I²r = 1.0 × 1.0 = 1.0 W.
  4. Efficiency = 5.0 / 6.0 = 0.83, or 83%.

Answer Load 5.0 W, wasted 1.0 W, efficiency 83%.

Part 5 of 5Exam Checklist & Tips

4.4.5 – Key Points for the Exam

  • Total power The rate at which the source supplies energy, εI.
  • Useful power The rate at which energy is transferred to the load, I²R.
  • Efficiency Useful power output divided by total power supplied.
  • Dissipation The transfer of electrical energy to heat in a resistor.

Common Mistakes

  • Using the e.m.f. instead of the terminal p.d. when calculating power in the load.
  • Forgetting the internal resistance wastes power even when the load looks fine.
  • Quoting an efficiency greater than 100% after a calculation slip.
  • Using ε²/R for load power when the internal resistance has not been included.

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