AQA AS Level Physics Paper 1, November 2021: Question 6

✅ Correct Answer

The amount of energy transferred from chemical energy to electrical energy for every coulomb of charge (or 5.30 J of energy per coulomb of charge).

Alternative phrasing: The work done in moving 1 coulomb of charge whole way round the circuit.

💡 Key Knowledge

• Emf definition: Work done / energy transferred per unit charge.
• Unit check: Volts (V) are equivalent to Joules per Coulomb ( J C⁻¹ ).
• Do not confuse emf with potential difference (pd)—emf refers to energy transfer into electrical energy from another form, whereas pd is energy transferred from electrical energy into other forms.

📐 Step-by-Step Calculation

1. Find pd across the 640 Ω resistor:
   5.30 V - 1.05 V = 4.25 V (since the parallel combination and the 640 Ω resistor share the total 5.30 V emf).
2. Calculate current through the circuit:
   I = V / R = 4.25 / 640 = 6.64 × 10⁻³ A
3. Convert to milliamperes:
   6.64 mA ≈ 7 mA (shows completion of proof).

❌ Common Errors

• Assuming the voltmeter has infinite resistance and treating the 320 Ω resistor in isolation.
• Failing to subtract the 1.05 V from the total 5.30 V emf to find the voltage across the remaining series resistor.

📐 Step-by-Step Calculation

1. Find current through the 320 Ω resistor branch ( I₁ ):
   I₁ = V / R = 1.05 / 320 = 3.28 × 10⁻³ A
2. Find total circuit current ( I_T ):
   Using I_T from 06.2 (~ 6.64 × 10⁻³ A ).
3. Find voltmeter current ( I_2 ) using Kirchhoff's First Law:
   I_2 = I_T - I₁ = 6.64 × 10⁻³ - 3.28 × 10⁻³ = 3.36 × 10⁻³ A
4. Calculate voltmeter resistance ( R ):
   R = V / I₂ = 1.05 / (3.36 × 10⁻³) ≈ 312.5 Ω ≈ 300 Ω

🧠 Exam Technique & Alternative Methods

You can also solve this by finding the combined parallel resistance of the 320 Ω resistor and R first using total circuit resistance ( R_T = 5.30 / 6.64mA ≈ 798 Ω ), subtracting 640 Ω to get the parallel block resistance (~ 158 Ω ), and then applying the parallel resistor formula: 1/R_parallel = 1/320 + 1/R .

📐 Step-by-Step Calculation

1. Select power equation:
   P = V² / R (or P = I²R or P = IV ).
2. Substitute values:
   Using exact/calculated values: P = (1.05)² / 312.5
   (Examiners also accept using rounded values like R = 300 Ω, giving ~0.037 W or ~0.090 W depending on path).
   Using precise values: P = 1.05 × 3.36 × 10⁻³ = 0.035 W (or using exact intermediate values giving around 0.035 W - 0.037 W ).

❌ Common Errors

Using the resistance of the resistor (320 Ω) instead of the voltmeter's calculated resistance ( R ) when calculating the power dissipated inside the voltmeter itself.

✅ Correct Answer

The total resistance of the circuit changes when the voltmeter changes position. Consequently, the total current changes, altering the ratio of the voltage dropped across each resistor component.

💡 Key Knowledge

Because the voltmeter is non-ideal (it has a finite resistance R ), placing it in parallel with a resistor decreases the total resistance of that section. Moving the voltmeter changes which resistor is in parallel, altering the overall circuit resistance, total current, and potential divider fractions.