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.