HSC Exam Practice

Sample response. Power loss is the energy dissipated as heat per second in the resistance of the transmission line conductors, due to the current flowing through them. It is calculated using P_loss = I²R, where P_loss = power lost as heat (W), I = current in the line (A), and R = total resistance of the transmission lines (Ω).

Marking notes. 1 mark for defining power loss as heat energy dissipated per unit time in the resistance of the lines; 1 mark for stating the formula P_loss = I²R; 1 mark for correctly identifying all three variables with units.

Sample response. The national electricity grid uses AC because transformers require a continuously changing magnetic flux to induce a voltage in the secondary coil; only AC provides this changing flux. Transformers allow voltage to be stepped up (at the power station) to reduce transmission current and losses, and stepped down (at substations and homes) to safe domestic levels. DC cannot be stepped to different voltages using simple transformers, making efficient long-distance DC transmission impractical without expensive converter electronics.

Marking notes. 1 mark for explaining that transformers require changing magnetic flux (AC provides this, DC does not); 1 mark for stating that transformers allow voltage to be stepped up/down; 1 mark for linking this to efficient transmission and safe distribution at 240 V.

Sample response. (a) I = P/V = 250×10&sup6; / 500×10³ = 500 A. (b) P_loss = I²R = (500)² × 6.0 = 1.5×10&sup6; W = 1.5 MW. (c) % loss = (1.5/250)×100 = 0.60%.

Marking notes. 1 mark for correct I = 500 A; 1 mark for correct P_loss = 1.5 MW; 1 mark for % loss = 0.60%; 1 mark for correct unit conversion throughout (250 MW to W and 500 kV to V). Accept answers expressed in consistent units; penalise unit errors once only.

Sample response. At 1 000 kV: I = 250×10&sup6;/1 000×10³ = 250 A; P_loss = (250)²×6.0 = 375 kW = 0.375 MW. This is one-quarter of the original 1.5 MW. The mathematical basis is that P_loss = I²R ∝ 1/V²: doubling V halves I, so I² and therefore P_loss decrease by a factor of 4 (quarters).

Marking notes. 1 mark for correctly calculating the new power loss (0.375 MW or 375 kW); 1 mark for stating the new loss is one-quarter of the original; 1 mark for the physical/mathematical explanation invoking P_loss ∝ 1/V² or equivalent argument about I halving and I² quartering.

Sample response. (1) Generation: the power station produces AC at approximately 10–25 kV. (2) Step-up: a step-up transformer at the power station increases voltage to ~330 kV for long-distance transmission. (3) Transmission: electricity travels along high-voltage lines across hundreds of kilometres. (4) Regional step-down: a step-down transformer at a regional substation reduces voltage to ~66 kV or 11 kV for distribution across Sydney suburbs. (5) Local step-down: a pole-top or pad-mounted step-down transformer reduces voltage to 240 V for delivery to individual homes.

Marking notes. 1 mark for each of four key stages correctly named with transformer type and typical voltage: generation (~20 kV); step-up to ~330 kV; regional step-down (~66/11 kV); local step-down to 240 V. Accept a range of ±20% on voltages. Award marks for describing the correct direction of transformation at each stage.

Sample response. The student’s claim is based on a misunderstanding. High voltages are used in transmission but are reduced to safe 240 V at homes by step-down transformers, so consumers are never exposed to transmission-level voltages. The towers carrying high-voltage lines are insulated and placed well above the ground, with strict safety corridors and automatic circuit breakers that de-energise lines if a fault occurs. The economic benefit is substantial: at 330 kV, power losses are dramatically lower than at generation voltage (e.g. transmitting at 20 kV instead of 330 kV would increase resistive losses by a factor of ~270), wasting enormous amounts of energy and requiring much higher generation capacity.

Marking notes. 1 mark for explaining that high voltage is reduced to safe levels at homes by step-down transformers (consumers are not exposed to transmission voltages); 1 mark for identifying at least one safety system in place (insulation, safety corridors, circuit breakers); 1 mark for explaining the economic benefit of high-voltage transmission with reference to reduced power loss.

Sample response (a). The graph shows an inverse-square relationship: % loss decreases steeply at low voltages and flattens at higher voltages. Mathematically, % loss = (I²R/P)×100 = (P·R/V²)×100 — loss is proportional to 1/V². Doubling V reduces loss by a factor of 4. For example, doubling from 100 kV (25.6%) to 200 kV (6.4%) reduces loss by a factor of exactly 4, consistent with 1/V².

Marking notes (a). 1 mark for describing the shape as inverse-square / hyperbolic with decreasing gradient; 1 mark for stating the formula % loss ∝ 1/V² or equivalent; 1 mark for a numerical verification (e.g. 100 kV to 200 kV: factor of 4, or any other valid pair).

Sample response (b). From the graph, 2% loss occurs at approximately 360–380 kV. Verification: % = (800×10&sup6;×8/V²)×100 = 2; V² = 800×10&sup6;×8×100/2 = 3.2×10¹1;; V = 566 kV. Accept graph reading of 350–400 kV for 1 mark and correct calculation for the remaining 2 marks. Note the graph shows % at 600 kV ≈ 0.71%, so 2% is reached at approximately 360 kV from the graph; calculation gives 566 kV. Discrepancy arises because the 2% gridline is below the 200 kV data point. Award marks for method even if graph reading differs from calculation.

Marking notes (b). 1 mark for reading approximately 350–380 kV from the graph with explicit reference to the 2% gridline; 1 mark for setting up the formula correctly; 1 mark for correct calculation (V ≈ 566 kV) and acknowledging any discrepancy between graph and calculation.

Sample response (c). From the graph: at 200 kV, % loss ≈ 6.4%; at 600 kV, % loss ≈ 0.71%. For P = 800 MW: at 200 kV, P_loss = 6.4% × 800 = 51.2 MW. At 600 kV, P_loss = 0.71% × 800 = 5.68 MW. Power saving = 51.2 − 5.68 ≈ 45.5 MW. [2 marks: 1 for correctly reading both graph values, 1 for calculating the power difference] One practical reason engineers do not always use the highest possible voltage is that higher voltages require more expensive and heavier insulation, taller pylons with larger electrical clearances, and more complex switchgear, all of which greatly increase the capital cost of the transmission line. The marginal energy saving above a certain voltage may not justify the additional infrastructure expense. [2 marks: 1 for correct reason, 1 for linking to cost or practicality from the lesson context]

Marking notes (c). 1 mark for correctly reading % loss at 200 kV (~6.4%) and 600 kV (~0.71%) from the graph; 1 mark for calculating the absolute power saving in MW (accept 44–47 MW); 1 mark for stating a valid practical reason against maximum voltage (cost, insulation complexity, pylon size, safety — any reason supported by the lesson); 1 mark for correctly explaining the link between high voltage and that practical constraint.

Sample response. Transformers are the critical enabling technology of the modern electricity grid because they allow voltage to be changed efficiently using electromagnetic induction — but only when operated with alternating current. A transformer uses a changing AC primary current to create a continuously changing magnetic flux in an iron core. By Faraday’s law, this changing flux induces an EMF in the secondary coil proportional to its number of turns: V_s/V_p = n_s/n_p. DC provides a constant (non-changing) flux, induces no secondary EMF, and cannot be transformed. This is why the entire Australian National Electricity Market runs on AC.

At the power station (e.g. Liddell coal in the Hunter Valley), electricity is generated at approximately 11–23 kV. A step-up transformer (n_s > n_p) increases this to 330 kV before it enters the transmission network. Using P = VI, at 330 kV, the current for 1 000 MW is I = 1 000×10&sup6;/330×10³ ≈ 3 030 A. If instead the same power were transmitted at 20 kV (generation voltage), the current would be 50 000 A — approximately 16.5 times larger. Since P_loss = I²R, this would increase power loss by a factor of (50 000/3 030)² ≈ 272 — an enormous waste of energy. High-voltage transmission is therefore essential for efficiency over the hundreds of kilometres between Hunter Valley generators and Sydney load centres.

At the other end of the system, safety is ensured by a cascade of step-down transformers (n_s < n_p). Regional substations reduce 330 kV to 66 kV or 11 kV for distribution through suburbs. Finally, pole-top transformers outside each street reduce 11 kV to 240 V — the standard household voltage. Consumers are therefore never exposed to transmission-level voltages. The safe 240 V level is compatible with standard appliance insulation and fuse ratings.

In summary, step-up transformers enable efficient long-distance transmission by reducing current (and therefore I²R losses), while step-down transformers restore electricity to safe voltages for consumers. AC is the essential common thread: it is the only form of electricity that can be transformed by simple electromagnetic induction, making the entire grid architecture possible.

Marking criteria (8 marks). 1 = correctly explains that transformers require AC / changing magnetic flux (Faraday’s law); DC cannot be transformed. 1 = states the turns ratio formula V_s/V_p = n_s/n_p and identifies step-up (n_s > n_p) vs step-down (n_s < n_p). 1 = explains quantitatively how a step-up transformer reduces current for fixed power (I = P/V, higher V ⇒ lower I). 1 = uses a specific numerical example to demonstrate the reduction in power loss (e.g. at 330 kV vs 20 kV, loss factor of ~272, or any valid calculation). 1 = explains the cascade of step-down transformers (330 kV → 66 kV → 11 kV → 240 V) in the correct sequence. 1 = links the final 240 V to consumer safety (not exposed to transmission voltages; compatible with appliance insulation). 1 = makes an explicit evaluative judgement that integrates efficiency (step-up) and safety (step-down) as complementary roles of transformers. 1 = uses precise physics terminology throughout: electromagnetic induction, Faraday’s law, turns ratio, P_loss = I²R, AC/DC, step-up/step-down, 240 V / 330 kV.