Physics • Year 12 • Module 6 • Lesson 17
DC Motors in Depth
Build HSC Band 5–6 extended-response technique on back emf, energy conservation, starting protection, and quantitative motor analysis in engineering and transport contexts.
1. Multi-step calculation, tram motor analysis (Band 4–6)
8 marks Band 4–6
Scenario. A DC motor in an early Melbourne tram has coil resistance 0.80 Ω and operates from a 48 V supply. At startup, a series starting resistor of 5.2 Ω is connected in series with the motor. Once at full speed, the starting resistor is bypassed. At full speed, the motor draws 3.0 A and the back emf is 45.6 V.
Q1(a) Calculate the current drawn at startup (i) without a starting resistor and (ii) with the starting resistor in circuit. Show all working. State why the starting resistor is essential for this motor. 3 marks
Q1(b) At full speed (starting resistor bypassed), calculate: (i) the electrical power supplied by the battery; (ii) the power dissipated as heat in the coil; (iii) the mechanical power output; (iv) the efficiency of the motor. 4 marks (1 each)
Q1(c) Explain, using conservation of energy, why the back emf must exist in this motor. Refer to the energy flows calculated above. 1 mark
2. Data + scenario: evaluating motor protection strategies (Band 5–6)
8 marks Band 5–6
Scenario. Two large industrial DC motors (Motor X and Motor Y) with identical specifications (R = 1.0 Ω, V = 100 V) are started differently. Motor X uses a mechanical starting resistor of 19 Ω (bypassed once running). Motor Y uses a modern electronic pulse-width modulation (PWM) controller that gradually increases the effective voltage from 5 V to 100 V over 10 seconds, eliminating the need for a starting resistor. The table below records measured data at startup and at full running speed.
| Quantity | Motor X, startup | Motor Y, startup (PWM, Veff = 5 V) | Motor X, full speed | Motor Y, full speed |
|---|---|---|---|---|
| Current (A) | 5.0 | 5.0 | 2.0 | 2.0 |
| Back emf (V) | 0 | 0 | 98 | 98 |
| Heat in coil (W) | 25 | 25 | 4.0 | 4.0 |
| Mechanical power (W) | 0 | 0 | 196 | 196 |
| Electrical power in (W) | 500 | 25 | 200 | 200 |
Note: Motor X startup uses full 100 V across (R + Rstart) = 20 Ω; current = 5.0 A. Motor Y startup uses effective voltage 5 V; current = 5 V/1.0 Ω = 5.0 A (back emf still zero).
Q2. Analyse and evaluate the data to compare the two motor starting strategies. In your response you must:
- Explain why both strategies limit the startup current to 5.0 A even though they operate differently.
- Identify the key difference in startup electrical power input between Motor X and Motor Y, using the data, and explain its significance for system design.
- Explain what happens to the “excess” power drawn by Motor X at startup that does not appear as heat in the motor coil. Use the formula for heat in the starting resistor.
- Evaluate which strategy better protects the motor coil and supply system, referring to at least two pieces of quantitative evidence.
- State one limitation of the PWM strategy that might still be relevant in practice.
Q1(a), Startup currents (3 marks)
Without starting resistor: I = V/R = 48/0.80 = 60 A [1].
With starting resistor: I = V/(R + Rstart) = 48/(0.80 + 5.2) = 48/6.0 = 8.0 A [1].
Why essential: Without the resistor, 60 A would flow, producing Pheat = (60)² × 0.80 = 2880 W in the coil, far exceeding its design rating and causing instant burn-out. The resistor limits startup current to a safe 8.0 A while still providing adequate starting torque [1].
Q1(b), Full-speed power analysis (4 marks)
(i) Pin = VI = 48 × 3.0 = 144 W [1].
(ii) Pheat = I²R = (3.0)² × 0.80 = 9.0 × 0.80 = 7.2 W [1].
(iii) Pmech = εback × I = 45.6 × 3.0 = 136.8 W. (Check: 144 − 7.2 = 136.8 W ✓) [1].
(iv) η = (Pmech / Pin) × 100 = (136.8/144) × 100 = 95.0% [1].
Q1(c), Conservation of energy and back emf (1 mark)
The motor receives 144 W from the supply. Of this, 7.2 W is dissipated as heat and 136.8 W is delivered as mechanical work. By the law of conservation of energy, this mechanical energy must be accounted for by a force (the back emf) opposing the current: Pmech = εback × I = 45.6 × 3.0 = 136.8 W. If back emf did not exist, no work would be done against any opposing force, meaning the motor would produce mechanical energy from nothing, violating conservation of energy. Back emf is therefore the mechanism that converts electrical energy to mechanical energy, required by energy conservation.
Q2, Sample Band 6 response (8 marks), annotated
Why both limit startup current to 5.0 A: Motor X inserts a 19 Ω starting resistor in series, giving a total circuit resistance of 20 Ω. The full 100 V drives I = 100/20 = 5.0 A. Motor Y reduces the effective applied voltage to 5 V via PWM; with only the 1.0 Ω coil in circuit, I = 5/1.0 = 5.0 A. Both achieve the same safe startup current by different means, one by adding resistance, one by reducing voltage [1].
Startup electrical power difference: Motor X draws 500 W from the supply at startup (P = VI = 100 × 5.0); Motor Y draws only 25 W (P = 5 × 5.0). This 20× difference in supply load is significant: in a system with many motors starting simultaneously (e.g. a tram depot), Motor X causes large voltage dips on the grid that can disrupt other loads. Motor Y’s PWM approach draws minimal power at startup, reducing grid stress [1].
Fate of Motor X’s “excess” startup power: The 500 W drawn by Motor X at startup is split: 25 W heats the motor coil (I²R = 25 × 1.0 = 25 W) and 475 W heats the starting resistor (I²Rstart = 25 × 19 = 475 W). This wasted energy in the starting resistor is a significant inefficiency that Motor Y avoids entirely [1].
Evaluation, coil and supply protection: Both strategies protect the motor coil equally: both limit coil heat to 25 W at startup [1]. However, Motor Y better protects the supply system: it draws only 25 W vs Motor X’s 500 W at startup [1]. Motor X wastes 475 W in the starting resistor, energy that serves no useful purpose and heats the resistor, which itself requires thermal management [1]. Motor Y’s gradual voltage ramp also means smoother torque buildup, reducing mechanical stress on drive train components. On balance, PWM (Motor Y) is significantly superior for supply-side protection while providing equal coil protection [1].
Limitation of PWM: PWM controllers contain sensitive power electronics that can fail in high-temperature, vibrating, or contaminated environments (e.g. dusty mines or flooded workshops). They are also more expensive to manufacture and repair than a simple resistor bank [1].
Marking criteria (8 marks): 1 = correctly explains why both limit I to 5.0 A with numerical reasoning for each; 1 = identifies and explains the supply power difference (500 W vs 25 W) with significance for system design; 1 = correctly identifies that excess Motor X power goes to Rstart with calculation (475 W); 1 = states both protect coil equally (same I²R); 1 = quantitative comparison of supply-side burden with a specific data point; 1 = evaluates PWM as superior on supply-side basis with evidence; 1 = reaches clear evidence-based comparative judgement; 1 = states one specific practical limitation of PWM.