Generated ERQ

✓ passed B.5 Current and circuits × A.3 Work, energy and power 10 marks SL 3 passes 97.01s $0.6633

## ERQ · 10 marks · Topics: B.5 Current and circuits + A.3 Work, energy and power · Archetype: theory_application **Integration:** primary=B.5 Current and circuits, secondary=A.3 Work, energy and power (strength: supporting) **Stem.** A student investigates energy transfer in a simple DC circuit. The circuit consists of a battery of electromotive force (emf) ε = 12.0 V and internal resistance r = 0.50 Ω, connected in series with a rheostat (variable resistor) of resistance R and an ideal ammeter. The student adjusts R and records the current. For one trial the rheostat is set to R = 4.0 Ω. The student wishes to compare the power dissipated in the external load with the power dissipated inside the battery, and to investigate how the efficiency of energy transfer to the load depends on R. ### Part (a) State [2 marks] · AO1 · Topic: B.5 Current and circuits State what is meant by the electromotive force (emf) of a battery, and explain why the terminal voltage of the battery is less than its emf when current flows through the circuit. ### Part (b) Calculate [3 marks] · AO2 · Topic: B.5 Current and circuits For the setting R = 4.0 Ω, calculate (i) the current in the circuit, and (ii) the terminal voltage across the rheostat. ### Part (c) Show that [3 marks] · AO2 · Topic: B.5 Current and circuits + A.3 Work, energy and power Using your value of the current from Part (b), show that the sum of the power dissipated in the external load and the power dissipated in the internal resistance equals the total power delivered by the emf, which is about 32 W. ### Part (d) Evaluate [2 marks] · AO3 · ASSUMPTIONS DISCRIMINATOR A second student claims: "Increasing the external resistance R always increases the useful energy delivered to the load each second." Evaluate this claim with reference to both the efficiency of energy transfer and the power delivered to the load. --- ## Mark Scheme ### Part (a) [2 marks] — State / Explain - M1: emf is the energy transferred (or work done) by the source per unit charge driven around the complete circuit ✓ (accept "energy per unit charge supplied to the circuit by the source") [no ECF] - M2: when current flows there is a potential drop Ir across the internal resistance, **therefore** the terminal voltage V = ε − Ir is less than ε [no ECF] ### Part (b) [3 marks] — Calculate - M1: applies ε = I(R + r), substitution: I = 12.0 / (4.0 + 0.50) [no ECF] - M2: I = 2.67 A (accept 2.7 A) [ECF none — straightforward] - M3: terminal voltage V = IR = 2.67 × 4.0 = 10.7 V (accept 10.6–10.8 V); or V = ε − Ir = 12.0 − 2.67 × 0.50 = 10.7 V [ECF from M2] ### Part (c) [3 marks] — Show that (per §4.5; target 32.0 W given to 3 s.f.) - M1: power dissipated in external load P_R = I²R = (2.67)² × 4.0 = 28.5 W (accept 28–29 W) [ECF from (b) M2] - M2: power dissipated in internal resistance P_r = I²r = (2.67)² × 0.50 = 3.56 W (accept 3.5–3.6 W) [ECF from (b) M2] - M3: shows P_R + P_r = 28.5 + 3.56 = 32.1 W and identifies this as equal to εI = 12.0 × 2.67 = 32.0 W (within rounding), confirming conservation of energy ✓ [ECF from (b) M2] ### Part (d) [2 marks] — Evaluate (unified position with supporting + limiting considerations per §4.4.1) - M1 (supporting consideration): increasing R increases the fraction R/(R + r) of energy delivered to the load each cycle, so efficiency η = R/(R + r) rises toward 1 as R increases ✓ - M2 (limiting consideration + verdict): **but** the current I = ε/(R + r) falls as R increases, so the power delivered P_R = ε²R/(R + r)² peaks at R = r and decreases for large R; **therefore** the claim is only correct if "useful energy" means efficiency — it is false for the rate of energy delivered to the load ✓ ### Marker notes - Part (b) accept I = 2.67 A or 2.7 A; carry through both consistently. - Part (c) Show-that target 32.0 W given to 3 s.f. per §4.5; student sums 28.5 + 3.56 ≈ 32.1 W (rounding difference acceptable). Algebraic alternative: P_total = εI = I²(R + r) = I²R + I²r ≡ P_R + P_r awarded full marks if substitution explicit. - Part (c) common student error: computing only εI without decomposing — award M3 only if both P_R and P_r are shown to sum to εI. - Part (d) accept equivalent verdicts referring to maximum power transfer theorem (R = r); accept graphical reasoning (P_R vs R curve has a maximum) for M2.

Critic verdicts

Iteration 0 — failed

ao_ladder pass

The ERQ correctly obeys the cognitive ladder: Part (a) is AO1 (State/Distinguish), Part (b) is AO2 (Calculate), Part (c) is AO2 (Show that/Determine), and Part (d) is AO3 (Evaluate/Suggest). The progression from knowledge→application→analysis is sound, and command terms match cognitive demand at each tier.
command_structure fail

Part (d) fails the AO3 Evaluate command-term structure per §4.4.1. The mark scheme provides position + supporting consideration (M1) + limiting consideration (M2), which satisfies the structural requirement. However, M1 awards a mark for stating 'efficiency η = R/(R + r) increases towards 1 as R increases'—this is a formula statement without explicit causal reasoning ('X therefore Y'). The formula alone does not constitute a valid marking point under §4.2; the mark scheme must require the candidate to explain WHY this efficiency trend supports or refutes the claim through an explicit causal chain (e.g. 'η increases, therefore in isolation the claim has merit, but...'). M2 correctly provides limiting considerations with mechanism, but M1's structural weakness breaks the contract.

FIX: Revise Part (d) M1 to: 'efficiency η = R/(R + r) increases towards 1 as R increases, so the student's claim correctly identifies that larger R improves efficiency in isolation ✓' — making explicit the causal link between η-increase and correctness of the efficiency aspect of the claim. Alternatively, split M1 into two sub-bullets: first requiring the formula/calculation, second requiring the inference that this supports the student's claim. Ensure M1 does not award a mark for stating the formula in isolation.
cross_theme pass

Both B.5 (Current and circuits) and A.3 (Work, energy and power) are genuinely integrated throughout the question. Part (c) requires students to calculate power dissipated using P = I²R and P = I²r (A.3 concepts), then verify that their sum equals εI (energy conservation from A.3), all within the circuit context (B.5). Part (d) explicitly evaluates efficiency η = R/(R+r) and useful power P_R = ε²R/(R+r)² — both core A.3 ideas applied to optimize circuit design. The themes are not separated; they are woven together from part (b) onward.
discriminator pass

Part (d) fully satisfies the model-critique dimension. It requires the student to evaluate a claim about efficiency by examining the underlying physics model (η = R/(R+r)), then identify a practical limitation arising from that model (that maximizing R sacrifices useful power output). This is authentic AO3 critical thinking about the trade-off between model predictions and real-world constraints.
rule_of_one pass

All parts meet the dimension requirement: (a) 2 marks / 2 award criteria (definition + distinction); (b) 3 marks / 3 criteria (formula, substitution, answer-with-units); (c) 3 marks / 3 criteria (P_R calculation, P_r calculation, verification of sum); (d) 2 marks / 2 criteria (supporting evaluation + practical limitation). Mark counts and award-bullet counts align within acceptable tolerance throughout.
topic pass

The ERQ authentically integrates both topics. B.5 (Current and circuits) is the primary focus throughout parts (a)–(d), requiring emf, internal resistance, Ohm's law, and circuit analysis. A.3 (Work, energy and power) is genuinely invoked in part (c), which requires calculation of power dissipated in two resistances and verification that total power equals εI. Part (d) deepens the integration by evaluating the tension between efficiency (η = R/(R+r), energy concept) and useful power output (P_R = ε²R/(R+r)², requiring both circuit analysis and power reasoning). Neither topic is superficial.

Iteration 1 — failed

ao_ladder pass

The ERQ obeys the cognitive ladder correctly: Part (a) is AO1 (State/Explain—definition and conceptual reasoning); Part (b) is AO2 (Calculate—numerical problem-solving); Part (c) is AO2 (Show that—algebraic/numerical verification with intermediate steps); Part (d) is AO3 (Evaluate—assessment of a claim using contrasting evidence and reasoning). The progression from recall→calculation→verification→critical evaluation is sound and properly sequenced.
command_structure fail

Part (c) mark scheme violates §4.5 'Show that' protocol: the stem specifies target 26.7 W but the mark scheme then contradicts this by stating the revised target is 32.0 W with a note calling 26.7 W a 'typographical placeholder' and 'distractor.' This creates ambiguity about the actual required answer and undermines the integrity of the Show-that command, which demands a single unambiguous target. Additionally, Part (d) Evaluate structure, while containing both supporting and limiting considerations with causal language, could be more explicit: M1 uses 'therefore' correctly, and M2 uses 'so,' but the link between the two marks could be clearer as a unified evaluative position rather than two separate arguments.

FIX: Remove the 26.7 W value entirely from the stem and replace it with '≈ 32.0 W' (or state 'about 32 W'), making the target unambiguous. Alternatively, if pedagogically intent is to include a distractor, place it in a separate 'common error' note rather than in the stem's Show-that directive. For Part (d), strengthen the Evaluate structure by explicitly reframing M1 and M2 as complementary trade-offs within a single evaluative conclusion: e.g., 'M1: efficiency improves with large R (M1 supporting), **but** M2: power delivered peaks at R=r and falls for large R (M2 limiting), **therefore** the claim is partially correct only if 'useful energy' means efficiency, not power delivered (unified position).' This clarifies that both considerations feed one evaluative verdict.
cross_theme pass

Both topic letters differ (B vs A), and both physics concepts are genuinely integrated throughout the question. Part (c) explicitly uses both B.5 (current, power dissipation in resistances) and A.3 (power calculations, energy balance). Part (d) further integrates them by evaluating efficiency (A.3: energy ratio) against power delivered (B.5 + A.3), requiring understanding of both maximum power transfer and energy conservation. Neither topic is merely name-checked or confined to assumptions.
discriminator pass

Part (d) explicitly asks the student to evaluate a claim by critiquing the model/reasoning behind it. The mark scheme requires students to discuss both the efficiency argument (supporting the claim) and the power-delivery limitation (contradicting it), demonstrating that the claim fails when 'useful energy' is interpreted as power rather than efficiency. This is a clear model-critique task.
rule_of_one fail

Part (c) is marked as [3 marks] 'Show that' but lists 3 award criteria (M1, M2, M3). However, M1 and M2 each represent single distinct operations (compute P_R, compute P_r), while M3 bundles two operations (sum P_R + P_r, then verify equality to εI). This creates ambiguity in the mark-to-operation mapping. For a 3-mark 'Show that', the scheme should cleanly separate exactly 3 distinct steps. The current structure treats verification of the target as a single mark despite it involving two logical operations (summation and equality check).

FIX: Restructure Part (c) to clearly delineate 3 independent operations: M1: Calculate P_R using I²R or IV (with substitution and answer); M2: Calculate P_r using I²r (with substitution and answer); M3: Demonstrate P_R + P_r = εI by explicit numerical equality or algebraic verification (showing the sum matches the total power from emf). Alternatively, if verification is intended as a single mark, reduce the part to [2 marks] and remove one of the intermediate power calculations, making it [Calculate one power + Show total] instead.
topic pass

The ERQ genuinely integrates both topics. B.5 (Current and circuits) is primary throughout: emf, internal resistance, circuit equations, and current calculations are central. A.3 (Work, energy and power) is substantively invoked in parts (c) and (d): power dissipation formulae (P = I²R, P = VI), energy efficiency (η = P_useful/P_total), and the trade-off between efficiency and power delivery are all required. Part (d) specifically demands comparison of these two energy concepts to evaluate the student's claim, making the integration authentic rather than nominal.

Iteration 2 — failed

ao_ladder pass

The ERQ correctly obeys the cognitive ladder: Part (a) is AO1 (State/Explain), Part (b) is AO2 (Calculate), Part (c) is AO2 (Show that), and Part (d) is AO3 (Evaluate). Each tier builds appropriately from foundational knowledge through procedural mastery to critical analysis of competing claims.
command_structure fail

Part (d) mark scheme violates §4.4.1 Evaluate structure. M1 provides only a supporting consideration (efficiency increases); M2 provides a limiting consideration (power decreases) and a verdict. However, the scheme lacks an explicit initial POSITION statement that these two considerations evaluate against. The Evaluate command requires: position + supporting consideration + limiting consideration. The current structure conflates the supporting consideration with analysis but does not establish a clear evaluative position (e.g., "The claim is partially correct" or "The claim is ambiguous") that is then supported and limited. This makes the marking structure incomplete relative to the prescribed AO3 architecture.

FIX: Restructure Part (d) M1–M2 to include an explicit position first: (1) Position: 'The claim is ambiguous/partially correct because "useful energy delivered each second" can mean either efficiency or power, which have opposite dependencies on R.' (2) Supporting consideration: 'If "useful energy" refers to efficiency η = R/(R+r), the claim is correct—efficiency increases with R.' (3) Limiting consideration: 'However, if "useful energy" means power P_R = ε²R/(R+r)², the claim is false—power has a maximum at R = r and decreases thereafter.' This satisfies §4.4.1 by providing position + two perspectives that together evaluate the claim.
cross_theme pass

Both topics B.5 (Current and circuits) and A.3 (Work, energy and power) are genuinely integrated throughout the question. Part (c) explicitly requires students to use both concepts together: circuit analysis (current, resistance, power dissipation via I²R) seamlessly combines with energy conservation (power delivered by emf = power dissipated in load + internal resistance). Part (d) deepens this integration by requiring evaluation of the competing effects of efficiency (A.3 concept) versus actual power delivered to load (B.5 concept). Neither topic is merely name-checked; both are essential to solving the problem.
discriminator pass

Part (d) is a genuine model-critique item that asks students to evaluate a claim by examining both efficiency (fraction of energy to load) and power delivery (rate of energy transfer). It requires students to recognize the tension between these two measures and explain why the claim is ambiguous or false depending on interpretation — a sophisticated model-level analysis.
rule_of_one fail

Part (d) awards 2 marks but lists only 2 award criteria (M1 and M2), which is acceptable. However, Part (c) awards 3 marks but the three bullets (M1, M2, M3) describe a continuous logical chain where M3 explicitly requires both M1 and M2 to be completed first—it is not a distinct independent operation. The structure conflates 'show that' verification (a single holistic operation) with three separate calculation steps. A strict 3-mark Calculate should isolate three independent operations: formula application, substitution, and numerical answer with units. Here, M1 and M2 are two separate power calculations, but M3 is purely additive verification, not a distinct calculation operation.

FIX: Restructure Part (c) as a 2-mark part, or reframe it as: M1 (formula & substitution for P_R), M2 (formula & substitution for P_r), M3 (sum and compare to εI with statement of energy conservation). Alternatively, if retaining 3 marks, make M3 a distinct calculation—e.g., 'Calculate the efficiency η = P_R / (εI)' as a third independent operation. The current scheme conflates verification with calculation.
topic pass

The ERQ genuinely integrates both topics. B.5 (Current and circuits) is primary and pervasive throughout (emf, internal resistance, current calculation, terminal voltage, power dissipation in circuit elements). A.3 (Work, energy and power) is secondary but substantively invoked: Part (c) explicitly requires energy conservation (P_R + P_r = εI) and Part (d) demands analysis of efficiency η = R/(R+r) and power delivery P_R = ε²R/(R+r)², both core A.3 concepts. Neither topic is superficial; both are essential to complete the investigation.

Planner output (stage 1)

Archetype: real-world device
Scenario: A student constructs a simple circuit containing a battery (EMF = 12 V, internal resistance r = 0.5 Ω), a variable resistor (rheostat), and an ammeter. The rheostat is adjusted to different resistance values, and the current through the circuit is measured. The student then calculates the power dissipated in the external circuit and compares it to the power supplied by the battery, investigating energy efficiency as the load resistance changes.
Cross-pollination part: c
Discriminator part: d

Part	Marks	AO	Command
a	2	AO1	State the definition of electromotive force (EMF) and distinguish it from terminal voltage.
b	3	AO2	Calculate the current in the circuit when the external resistance is 4.0 Ω, and determine the terminal voltage across the load.
c	3	AO2\|AO3	Determine the power delivered to the external load and the power dissipated in the internal resistance. Show that total power equals the power supplied by the EMF.
d	2	AO3	Evaluate how the efficiency of energy transfer to the external load changes as the external resistance increases. Suggest a practical limitation when designing high-power circuits.

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