Physics • Year 12 • Module 6 • Lesson 17
DC Motors in Depth
Apply your understanding of back emf, motor current, power, and efficiency to real experimental data, quantitative scenarios, and motor performance comparisons.
1. Interpret experimental data — DC motor at different loads
A student tested a DC motor connected to a 20 V supply with a coil resistance of 2.0 Ω. The motor was run at three different loads. The table below records the current drawn at each load. 8 marks
| Load condition | Current (A) | Back emf (V) | Mechanical power (W) | Efficiency (%) |
|---|---|---|---|---|
| No load (full speed) | 0.50 | |||
| Medium load | 2.0 | |||
| Heavy load (near stall) | 8.0 |
1.1 Use the formula εback = V − IR to complete the “Back emf” column. Show one calculation in full. 3 marks (1 per row)
1.2 Complete the “Mechanical power” column using Pmech = εback × I. Show one calculation. 3 marks
1.3 Complete the “Efficiency” column. Identify the load condition at which the motor is most efficient and explain why. 2 marks
2. Interpret graph — back emf vs motor speed
The graph below shows how the back emf of a DC motor varies with angular speed (ω) for two different magnetic field strengths (strong field B1 and weak field B2). The motor is connected to a 24 V supply. 7 marks
Figure 2.1. Back emf vs angular speed for a DC motor under two field conditions. Red dashed line shows the applied voltage (24 V). Illustrative data based on εback = NBAω.
2.1 Describe the relationship between back emf and angular speed shown for each field strength. 2 marks
2.2 The motor with strong field B1 reaches its no-load speed at ω = 200 rad s−1. Explain what happens to the motor current and torque at this point. 2 marks
2.3 A student claims: “The motor with B2 reaches higher speeds, so it is always more powerful.” Evaluate this claim using the graph data and the formula Pmech = εback × I. 3 marks
3. Compare DC motor behaviour at startup and full speed
Complete the two-column table below. For each feature, write a concise description contrasting motor behaviour at startup and at full running speed. 10 marks (1 per cell)
| Feature | At startup (omega = 0) | At full running speed |
|---|---|---|
| Back emf | ||
| Current drawn | ||
| Torque produced | ||
| Mechanical power output | ||
| Efficiency |
4. Predict and justify — motor seizure scenario
A DC motor with coil resistance 0.50 Ω is connected to a 12 V supply and running at full speed with a back emf of 10 V. The motor suddenly seizes (the rotor locks, unable to rotate) due to a mechanical fault.
5 marks
4.1 Calculate the current flowing just before the seizure and the current flowing immediately after the seizure. Show all working. 3 marks
4.2 Explain, using energy and power reasoning, why the seizure could permanently damage the motor. Use a calculation to support your answer. 2 marks
Q1.1 — Back emf column
Formula: εback = V − IR = 20 − I × 2.0
No load: εback = 20 − (0.50 × 2.0) = 20 − 1.0 = 19 V.
Medium load: εback = 20 − (2.0 × 2.0) = 20 − 4.0 = 16 V.
Heavy load: εback = 20 − (8.0 × 2.0) = 20 − 16 = 4.0 V.
Q1.2 — Mechanical power column
Pmech = εback × I
No load: Pmech = 19 × 0.50 = 9.5 W.
Medium load: Pmech = 16 × 2.0 = 32 W.
Heavy load: Pmech = 4.0 × 8.0 = 32 W.
Q1.3 — Efficiency column and most efficient load
Pin = VI = 20I
No load: Pin = 20 × 0.50 = 10 W; η = (9.5/10) × 100 = 95%.
Medium load: Pin = 20 × 2.0 = 40 W; η = (32/40) × 100 = 80%.
Heavy load: Pin = 20 × 8.0 = 160 W; η = (32/160) × 100 = 20%.
Most efficient at no load (95%). At no load, back emf is nearly equal to V; most input power is converted to mechanical output. At heavy load, large I means large I²R heat loss, making efficiency low.
Q2.1 — Back emf vs angular speed relationship
For both field strengths, back emf increases linearly with angular speed (a straight line through the origin), consistent with εback = NBAω. The strong-field motor (B1) has a steeper gradient (higher NBA constant), so it reaches a given back emf at a lower angular speed than the weak-field motor (B2).
Q2.2 — No-load speed for B1
At ω = 200 rad s−1 for B1, the back emf equals the applied voltage (24 V). The net voltage (V − εback) approaches zero, so the current drops to near zero (just enough to overcome friction). With negligible current, the torque produced also approaches zero. The motor spins at its maximum unloaded speed.
Q2.3 — Evaluate “higher speed = more powerful”
The claim is not always valid. At its no-load speed, both motors have back emf ≈ 24 V, so current ≈ 0 and Pmech = εback × I ≈ 0 W. A motor at very high speed with minimal load produces minimal mechanical power because the current (and thus torque) is negligible [1]. Maximum mechanical power for a motor typically occurs at an intermediate speed, not the maximum speed [1]. The B2 motor reaches higher no-load speed, but under the same load it runs at lower speed with a larger torque deficit than B1; the claim fails to account for the current and torque drop at high speed [1].
Q3 — Compare and contrast table
Back emf: Startup: zero (omega = 0) | Full speed: near-maximum (close to V applied).
Current drawn: Startup: maximum (stall current = V/R) | Full speed: minimum (only enough to overcome friction and load).
Torque produced: Startup: maximum (torque is proportional to current via F = BIL) | Full speed: minimum (just enough to maintain speed against friction).
Mechanical power output: Startup: zero (Pmech = εback × I = 0 × Istall = 0) | Full speed: maximum useful output.
Efficiency: Startup: zero (all input goes to heat; Pheat = Istall²R, Pmech = 0) | Full speed: maximum (most input goes to mechanical output, minimal heat loss at low current).
Q4.1 — Currents before and after seizure
Before seizure (running at full speed with εback = 10 V):
I = (V − εback) / R = (12 − 10) / 0.50 = 2.0 / 0.50 = 4.0 A.
Immediately after seizure (omega = 0, so εback = 0):
I = V / R = 12 / 0.50 = 24 A.
The current increases from 4.0 A to 24 A — a factor of 6 increase.
Q4.2 — Why seizure damages the motor
After seizure, all electrical power is dissipated as heat in the coil resistance (no mechanical work is being done). Heat loss before seizure: Pheat = I²R = (4.0)² × 0.50 = 8.0 W. After seizure: Pheat = (24)² × 0.50 = 288 W — a 36-fold increase [1]. This extreme heat can melt the wire insulation, warp the commutator, or burn out the coil windings, permanently destroying the motor [1].