Physics • Year 12 • Module 6: Electromagnetism • Lesson 21
Module 6 Consolidation
Build HSC Band 5–6 technique across extended-response reasoning, multi-concept integration, and critical evaluation of electromagnetic systems.
1. Multi-concept scenario: a DC motor under load
8 marks Band 5–6
Scenario. A DC motor is connected to a 240 V supply. The armature resistance is 2.0 Ω. The table below shows measurements taken at three different operating conditions.
| Condition | Armature current (A) | Back emf (V) | Input power (W) | Mechanical power output (W) |
|---|---|---|---|---|
| Stall (rotor not moving) | 120 | 0 | 0 | |
| Running at half load | 20 | |||
| Running at full speed | 8 |
Motor supply voltage = 240 V. Armature resistance R = 2.0 Ω.
Q1. Analyse the motor data above. In your response you must:
- Complete the missing values in the table (show working for each calculated value).
- Explain why the armature current is highest at stall and decreases as the motor accelerates, with reference to back emf.
- Calculate the efficiency of the motor at full speed (efficiency = mechanical power out ÷ electrical power in, expressed as a percentage).
- Explain why high stall current can damage a motor’s armature windings, and describe one design strategy used to limit start-up current.
2. Evaluate and explain — designing a power transmission system
7 marks Band 5–6
Context. An engineering team is designing a system to transmit 200 MW from a remote wind farm to a city 300 km away. The transmission lines have a total resistance of 10 Ω. The team is considering two options: Option A: Transmit at 100 kV. Option B: Transmit at 500 kV. Both options require the same final delivery voltage of 11 kV to local distribution substations.
Q2. Evaluate the two transmission options. In your response you must:
- Calculate the line current and power loss for both Option A and Option B, showing all working.
- Express each power loss as a percentage of the transmitted power (200 MW).
- Evaluate which option is more economically and practically preferable, referring to your calculations.
- Describe the role of transformers at each end of the transmission line in Option B (power station end and city substation end), including whether each transformer is a step-up or step-down type.
- State one additional factor an engineer would consider when deciding between options (beyond power loss alone).
Q1 — Sample Band 6 response (8 marks), annotated
Completed table:
Stall: I = 120 A; εback = 0 V (given); Pin = VI = 240 × 120 = 28 800 W; Pmech = 0 (rotor stationary).
Half load: εback = V − IR = 240 − (20)(2.0) = 200 V; Pin = 240 × 20 = 4 800 W; Pmech = I × εback = 20 × 200 = 4 000 W.
Full speed: εback = 240 − (8)(2.0) = 224 V; Pin = 240 × 8 = 1 920 W; Pmech = 8 × 224 = 1 792 W.
[2 marks — all five missing cells correctly calculated, showing working. Deduct 0.5 per error.]
Why current decreases as motor speeds up: When the rotor is stationary (stall), back emf = 0. The only resistance limiting current is the armature resistance R, so I = V/R = 240/2.0 = 120 A. As the motor accelerates, the rotating coil generates a growing back emf (εback = V − IR). Since I = (V − εback)/R, increasing back emf reduces the net voltage driving current through the armature, so the current decreases. At full speed, the back emf (224 V) is close to the supply voltage, leaving only 16 V across R, giving I = 8 A. [2 marks]
Efficiency at full speed: η = Pmech/Pin × 100% = 1792/1920 × 100% ≈ 93.3%. The remaining 6.7% is dissipated as heat in the armature resistance: Pheat = I2R = 64 × 2 = 128 W. [1 mark]
Why high stall current damages windings: At stall, current is 120 A — 15 times the full-speed current of 8 A. Heat dissipation in the armature is P = I2R = (120)2 × 2.0 = 28 800 W at stall compared with 128 W at full speed. This extreme heating can melt insulation on the copper windings, causing short circuits and motor failure. [1 mark] Design strategy: a starting resistor is connected in series with the armature at start-up, increasing total resistance and limiting stall current to a safe value. As the motor accelerates and back emf grows, the starting resistor is progressively switched out (manually or automatically). [1 mark]
Marking criteria summary (8 marks): 2 = all five missing values calculated correctly; 2 = clear explanation of back emf increasing with speed and reducing armature current, with formula reference; 1 = efficiency calculated correctly as a percentage; 1 = correct explanation of why high current causes heating (I2R); 1 = names a valid start-up current limiting strategy (starting resistor, series resistance, variable voltage supply, or soft-starter); 1 = uses precise terminology throughout (back emf, armature resistance, torque, efficiency).
Q2 — Sample Band 6 response (7 marks), annotated
Option A (100 kV): I = P/V = 200 × 106 ÷ 100 × 103 = 2 000 A. Ploss = I2R = (2000)2 × 10 = 40 MW. As percentage: 40/200 × 100% = 20%. [1 mark]
Option B (500 kV): I = 200 × 106 ÷ 500 × 103 = 400 A. Ploss = (400)2 × 10 = 1.6 MW. As percentage: 1.6/200 × 100% = 0.8%. [1 mark]
Evaluation: Option B (500 kV) is strongly preferable. It loses only 0.8% of transmitted power compared to 20% for Option A — a factor of 25 less loss for the same delivered power. Economically, 40 MW of lost power at ~$0.10/kWh represents approximately $35 million/year wasted on line heating alone (Option A); Option B wastes only $1.4 million/year. The capital cost of higher-voltage insulators and switch gear for Option B is justified over the lifetime of the line. [2 marks — comparative evaluation with reference to calculated percentages and economic reasoning]
Role of transformers in Option B: At the wind farm: a step-up transformer raises the generator output (typically 11–33 kV) to 500 kV for transmission. With Np < Ns, the turns ratio boosts voltage and reduces current. At the city substation: a step-down transformer reduces 500 kV to 11 kV for local distribution. With Np > Ns, the turns ratio reduces voltage and increases current for local loads. Both transformers operate on Faraday’s Law (alternating flux in the shared core induces emf in the secondary coil) and conserve power if ideal. [2 marks]
Additional factor: Any one valid engineering consideration, for example: (1) cost of insulation and tower infrastructure needed to safely handle 500 kV vs 100 kV — higher voltage requires larger clearances and more expensive hardware; (2) safety risks of corona discharge and electromagnetic interference at ultra-high voltages; (3) reactance losses in AC lines over 300 km (reactive power) which are not captured by the simple I2R model; (4) maintenance and reliability considerations. [1 mark]
Marking criteria summary (7 marks): 1 = correct line current and power loss for Option A with percentage; 1 = correct line current and power loss for Option B with percentage; 2 = well-reasoned evaluation citing specific calculated values and economic/practical justification; 2 = correct description of step-up transformer at power station end and step-down at city end, with turns ratio direction and Faraday’s Law reference; 1 = one valid additional engineering factor named and briefly justified.