HSC Exam Practice

Sample response. Back emf is the electromotive force induced in the rotating coil of a DC motor that opposes the applied voltage. It arises because the rotating coil cuts through the external magnetic field lines, inducing an emf by electromagnetic induction (Faraday’s Law). Its direction is determined by Lenz’s Law, which requires the induced emf to oppose the cause (the applied current driving the rotation). The formula relating back emf to motor speed is: ε_back = NBAω, where N is the number of turns, B the magnetic field strength, A the coil area, and ω the angular speed.

Marking notes. 1 mark for defining back emf as an induced emf opposing the applied voltage; 1 mark for naming Lenz’s Law (or Faraday’s Law) as the physical basis; 1 mark for the correct formula ε_back = NBAω with at least partial identification of symbols.

Sample response. At startup, the motor coil is stationary so the angular speed ω = 0. Since ε_back = NBAω = 0, there is no back emf opposing the applied voltage. The full supply voltage drives current through the coil resistance alone: I_stall = V/R. This is the maximum possible current. As the motor accelerates, ω increases, so ε_back increases. The net voltage driving current decreases: V_net = V − ε_back, and the current I = (V − ε_back)/R falls steadily until at full speed only enough current flows to overcome friction and the load.

Marking notes. 1 mark for explaining that at startup ω = 0 so ε_back = 0 and I = V/R (maximum); 1 mark for explaining that as speed increases ε_back increases reducing net voltage; 1 mark for concluding that current decreases to a small running value at full speed.

Sample response. (1) The large current causes excessive resistive heating (P = I²R) in the coil windings, potentially melting insulation and burning out the motor. (2) High current can damage the commutator and brushes through arcing and erosion. (3) The current surge can cause a voltage dip in the power supply, disrupting other devices connected to the same circuit.

Marking notes. 1 mark each for three correctly identified damage modes. Accept: coil overheating / commutator-brush damage / voltage dip in supply. Do not accept vague answers such as “it can break the motor.”

Sample response. By conservation of energy, the electrical power supplied to a motor must equal the mechanical power output plus the heat losses in the coil: P_in = P_mech + P_heat, i.e. VI = P_mech + I²R. The back emf is the mechanism that converts electrical energy to mechanical work: as the current I flows through a coil experiencing a back emf ε_back, the work done against the back emf per unit time equals the mechanical power output: P_mech = ε_back × I. Without back emf, P_mech = 0; the motor would produce no mechanical energy while consuming all input as heat — consistent with energy conservation but useless as a machine. Back emf is the energy-conversion agent: it ensures that energy is removed from the electrical circuit and deposited as mechanical work.

Marking notes. 1 mark for stating conservation of energy: P_in = P_mech + P_heat; 1 mark for writing P_mech = ε_back × I (formula); 1 mark for explaining that back emf is the agent of energy conversion — work done against it equals mechanical output.

Sample response. When the motor is loaded, the external torque slows its rotation, reducing the angular speed ω. Since ε_back = NBAω, the back emf decreases. This increases the net voltage (V − ε_back), which drives a larger current: I = (V − ε_back)/R. The increased current produces a larger force on the conductors (F = BIL), generating more torque to meet the load demand. This is an automatic self-regulating response: a DC motor inherently adjusts its current to match the mechanical load.

Marking notes. 1 mark for stating that speed drop reduces back emf (ε_back = NBAω); 1 mark for correctly stating that the net voltage increases and therefore current increases; 1 mark for explaining the physical mechanism (increased current = increased F = BIL = increased torque to match load).

Sample response. (a) ε_back = V − IR = 30 − (1.2 × 5.0) = 30 − 6.0 = 24 V. (b) I_stall = V/R = 30/5.0 = 6.0 A. (c) P_mech = ε_back × I = 24 × 1.2 = 28.8 W.

Marking notes. 1 mark each for: (a) correct back emf = 24 V; (b) correct stall current = 6.0 A; (c) correct P_mech = 28.8 W. Accept equivalent correct methods.

Sample response (a) — Completed table.

Formulas: ε_back = V − IR = 24 − 2I; P_mech = ε_back × I; P_in = VI = 24I; η = (P_mech/P_in) × 100.

ω = 0: ε_back = 24 − 24 = 0 V; P_mech = 0 W; P_in = 288 W; η = 0%.
ω = 50: ε_back = 24 − 16 = 8 V; P_mech = 8 × 8 = 64 W; P_in = 192 W; η = 33%.
ω = 100: ε_back = 24 − 8 = 16 V; P_mech = 16 × 4 = 64 W; P_in = 96 W; η = 67%.
ω = 150: ε_back = 24 − 4 = 20 V; P_mech = 20 × 2 = 40 W; P_in = 48 W; η = 83%.
ω = 200: ε_back = 24 − 0.40 = 23.6 V; P_mech = 23.6 × 0.20 = 4.72 W; P_in = 4.8 W; η = 98%.

Marking notes (a): 1 mark for each correct column of back emf (5 rows), P_mech (5 rows), η (5 rows) — award 2 marks per column if all 5 rows are correct and 1 mark if at least 3 rows are correct. Total 6 marks: 2 + 2 + 2.

Sample response (b) — Efficiency trend. Efficiency increases steadily as angular speed increases from 0% at startup to 98% at no-load speed. Efficiency is maximised at no-load speed (200 rad s−1) because back emf nearly equals V; current is minimal, so heat loss I²R is tiny relative to input power [1]. Efficiency is zero at startup because ε_back = 0 so P_mech = 0 regardless of how large the current is; all input power goes to coil heating [1]. Although efficiency appears highest at no-load speed, useful mechanical power output (28.8 W at no-load vs 64 W at medium loads) is actually much lower there; maximum power output occurs around ω = 50–100 rad s−1 where the product ε_back × I is greatest [1]. A well-designed motor for maximum power delivery would operate in this intermediate range, not at no-load speed [1].

Marking notes (b): 1 mark for describing efficiency increasing with speed; 1 mark for explaining zero efficiency at startup (back emf = 0, P_mech = 0); 1 mark for explaining near-maximum efficiency at no-load (tiny heat loss fraction); 1 mark for distinguishing efficiency from useful power output and identifying that maximum power occurs at intermediate speed.

Sample response. At startup, a DC motor’s coil is stationary so angular speed ω = 0 and back emf ε_back = NBAω = 0. With no opposing emf, the full supply voltage drives current through the coil resistance alone: I_stall = V/R. This is the maximum possible current and can be 5–10 times the running current, generating dangerous heat in the coil (P_heat = I_stall²R) and potentially damaging the commutator or causing supply voltage dips. A starting resistor R_start in series limits the initial current to I = V/(R + R_start), protecting the motor during this vulnerable phase when back emf has not yet built up. As the motor accelerates, ω increases, so ε_back = NBAω grows. The net voltage V − ε_back and thus the current I = (V − ε_back)/R both decrease. The starting resistor is no longer needed and is bypassed. At full running speed (no external load), ε_back approaches V, leaving only a tiny net voltage to drive the small friction-overcoming current. The motor reaches a self-regulated equilibrium speed. The role of back emf in energy conversion is fundamental to understanding motor performance: by conservation of energy, P_in = P_mech + P_heat, i.e. VI = ε_backI + I²R. The mechanical power output P_mech = ε_backI is the work done per unit time against the back emf — it is the mechanism by which electrical energy is converted to mechanical energy. Without back emf, no mechanical work would be produced, contradicting conservation of energy (useful motion cannot come from nothing). Efficiency η = (P_mech/P_in) × 100 = (ε_backI)/(VI) × 100 = (ε_back/V) × 100. As speed increases, ε_back/V approaches 1, so efficiency approaches 100%; the motor becomes nearly lossless at no-load speed, though at that point it delivers negligible useful mechanical power. In summary, back emf plays three roles: (1) it limits the running current, keeping the motor safe and the coil cool; (2) it is the thermodynamic agent that converts electrical energy to mechanical work; and (3) its relationship to speed governs the torque–speed and efficiency–speed characteristics that determine how a motor must be matched to its mechanical load.

Marking criteria (7 marks). 1 = correctly explains startup: ω = 0, ε_back = 0, I_stall = V/R; identifies danger (excessive heat / damage). 1 = correctly explains role of starting resistor with formula I = V/(R + R_start). 1 = explains how back emf grows with speed, reducing current, using ε_back = NBAω and I = (V − ε_back)/R. 1 = states conservation of energy: P_in = P_mech + P_heat and links P_mech = ε_backI. 1 = explains why back emf is required by conservation of energy (energy conversion mechanism). 1 = correctly derives or states efficiency formula η = ε_back/V × 100 and explains trend. 1 = reaches a clear, integrated evaluative judgement connecting all three roles of back emf (current limiting, energy conversion, efficiency determination) across the operating cycle.