## ERQ · 10 marks · Topics: B.5 Current and circuits + A.3 Work, energy and power
**Stem.** A student builds a portable phone charger from a battery cell of emf ε = 12.0 V and internal resistance r = 0.50 Ω. The cell is connected through ideal wires to a USB regulator that the student models as a fixed external resistor of resistance R = 2.0 Ω. While testing the charger, the student measures the terminal potential difference across the regulator using a high-resistance voltmeter, then runs the charger continuously for 5.0 minutes. The manufacturer claims the cell stores 25 kJ of chemical energy when fully charged. The student wants to know what fraction of this stored energy is actually delivered to the phone.
### Part (a) Define [2 marks] · AO1 · Topic: B.5
Define *electromotive force (emf)* of a cell and state the difference between emf and terminal potential difference when the cell supplies current.
### Part (b)(i) Calculate [2 marks] · AO2 · Topic: B.5
Calculate the current in the circuit when the charger is operating.
### Part (b)(ii) Determine [2 marks] · AO2/AO3 · Topic: B.5
Determine the terminal potential difference across the USB regulator.
### Part (c) Show that [2 marks] · AO3 · Topic: B.5 + A.3
Show that the efficiency of energy transfer from the cell to the phone (i.e. the fraction of total electrical energy delivered by the emf that is dissipated in the external resistor R, rather than in r) is approximately 80%.
### Part (d) Suggest [2 marks] · AO3 · ASSUMPTIONS DISCRIMINATOR
The student measures the actual energy delivered to the phone over 5.0 minutes and finds it is significantly *less* than 80% of 25 kJ. Suggest **two** reasons, based on the physical model used, why the experimental energy delivered is lower than the value predicted by the calculation in (c).
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## Mark Scheme
### Part (a) [2 marks]
- M1: emf defined as the energy transferred (from chemical to electrical) per unit charge passing through the cell ✓ [ECF: no]
- M2: terminal pd < emf when current flows because of pd «Ir» dropped across internal resistance r ✓ [ECF: no]
### Part (b)(i) [2 marks]
- M1: applies I = ε / (R + r) ✓ [ECF: no]
- M2: I = 12.0 / (2.0 + 0.50) = 4.8 A ✓ (2 s.f., unit required) [ECF: yes]
### Part (b)(ii) [2 marks]
- M1: uses V = IR (or V = ε − Ir) ✓ [ECF: no]
- M2: V = 4.8 × 2.0 = 9.6 V ✓ (unit + s.f.) [ECF: yes from (b)(i)]
### Part (c) [2 marks]
- M1: identifies efficiency as η = P_R / P_total = R/(R+r) **OR** computes P_R = I²R = 46.08 W and P_total = εI = 57.6 W ✓ [ECF: yes from (b)(i)]
- M2: η = 2.0/2.5 = 0.80 = 80% (or 46.08/57.6 = 0.80) — must show the ratio explicitly, not just state 80% ✓ [ECF: yes]
### Part (d) [2 marks]
Award 1 mark each, max 2, for any TWO distinct critiques of the model:
- internal resistance r is not constant — it increases as the cell discharges / heats up, so more energy is dissipated inside the cell ✓
- the USB regulator is not a pure ohmic resistor — it contains electronics that dissipate additional energy / has a conversion efficiency < 100% ✓
- connecting wires and contacts have non-zero resistance that was neglected ✓
- the phone battery itself has internal resistance, so not all energy delivered to the USB output reaches chemical storage ✓
- emf of the cell is not constant over 5 minutes — it drops as the cell discharges, reducing current and delivered power ✓
- energy losses as heat radiated from the circuit / Joule heating in components other than R ✓
### Marker notes
- (b)(i): accept 4.80 A; do not penalise 3 s.f.
- (c): a candidate who writes η = R/(R+r) directly and substitutes scores both marks; a candidate who only writes "80%" with no working scores 0 (process over product).
- (d): do **not** accept vague answers such as "experimental error" or "the battery is old" without linking to the physical model. The reason must identify a specific *modelling assumption* that fails.
- ECF chain: an incorrect current in (b)(i) propagates through (b)(ii) and the numerical route of (c); the ratio route in (c) is independent and still earns both marks if attempted cleanly.