AQA AS Level Physics Paper 2, June 2025: Question 3

💡 Key Knowledge

• Total Power supplied by source: P_total = εI = 12I
• Power dissipated inside battery: P_internal = I²r = 0.037I²
• Power consumed by parallel lamps: Two 65 W lamps operate together: 2 × 65 W = 130 W
• Principle of Conservation of Energy: Rate of energy supplied by battery = total rate of energy delivered to components plus energy wasted in internal resistance per second.

✅ Model Answer & Marks

• Mark 1: Algebraic identification of power terms:
  • 12I is the total power supplied by the battery ( P = εI ).
  • 0.037I² is the power dissipated in the internal resistance ( P = I²r ).
• Mark 2: Identification of 130 as total power transformed by both lamps ( 2 × 65 W ), linked clearly to conservation of energy / total power in = total power out.

🧠 Exam Technique

Don't just state "energy is conserved". You must explicitly match every single term in the given equation to its physical circuit counterpart: 12I (battery), 0.037I² (internal resistance), and 130 (both lamps).

❌ Common Errors

• Stating that 130 W is the power of "a lamp" rather than both lamps combined ( 2 × 65 W ).
• Confusing energy with power (rate of energy transfer).

📐 Step-by-Step Calculation

1. Rearrange into standard quadratic form ( ax² + bx + c = 0 ):
   0.037I² - 12I + 130 = 0
   Here: a = 0.037 , b = -12 , c = 130 . [Mark 1]
2. Substitute into quadratic formula:
   I = [-(-12) ± √((-12)² - 4(0.037)(130))] / [2 × 0.037]
3. Compute discriminant and evaluate:
   √[144 - 19.24] = √124.76 ≈ 11.1696
   I = (12 - 11.1696) / 0.074 ≈ 0.8304 / 0.074 = 11.22 A
   (The other root, ~313 A, is physically unrealistic for normal headlamp operation).
4. Final unrounded value:
   I = 11.2 A (which is approximately 11 A). [Mark 2]

🧠 Exam Technique for "Show That" Questions

• Never stop at 11 A: In a "show that it is approximately 11 A" question, you must calculate and write down the unrounded value to at least 3 significant figures (e.g., 11.2 A or 11.22 A ).
• You can also score Mark 1 simply by rearranging to 0.037I² - 12I + 130 = 0 and listing a = 0.037, b = -12, c = 130 .

📐 Step-by-Step Calculation

Method 1: Terminal pd and current per lamp

1. Find terminal pd ( V ):
   V = ε - Ir = 12 - (11.22 × 0.037) = 12 - 0.415 = 11.58 V
   (Using I = 11 A gives V = 11.59 V) [Mark 1]
2. Current through lamp X ( I_X ):
   Since lamps are identical and in parallel:
   I_X = I / 2 = 11.22 / 2 = 5.61 A
3. Calculate resistance ( R_X ):
   R_X = V / I_X = 11.58 / 5.61 = 2.06 Ω
   OR using P = V²/R :
   R_X = V² / P = (11.58)² / 65 = 2.06 Ω (≈ 2.1 Ω) [Mark 2]

❌ The Internal Resistance Trap!

Many students erroneously assume the terminal pd is simply the EMF ( 12 V ) and calculate:

R = V²/P = 12² / 65 = 2.22 Ω

Because there is internal resistance ( r = 0.037 Ω ), "lost volts" occur. Terminal pd is strictly 11.58 V, not 12 V. Using 12 V earns at most 1 mark!

✅ 4-Step Structured Deduction

1. Mark 1 (Resistance of X):
   Diameter of filament X decreases → cross-sectional area A decreases. Since R = ρL / A , the resistance of lamp X ( R_X ) increases.
2. Mark 2 (Total load resistance and current):
   Since lamp X and Y are in parallel, an increase in R_X causes the combined parallel load resistance to increase. Consequently, total circuit current I decreases.
3. Mark 3 (Terminal pd / "Lost Volts"):
   With total current I decreasing, "lost volts" ( Ir ) decreases. Therefore, terminal pd ( V = ε - Ir ) across the lamps increases.
4. Mark 4 (Power of lamp Y):
   For lamp Y, R_Y is constant. Since power P_Y = V² / R_Y and V has increased, the power output of Y increases (i.e. it does not continue to operate at 65 W, it operates at > 65 W).

🧠 Examiner Commentary & Logic Flow

This is a classic 4-mark deductive chain. Ensure you follow every link in the causal chain without skipping steps:

• Potential divider alternative: You can also gain Mark 2 and 3 by arguing that the ratio of external load resistance to internal resistance increases, so load receives a larger share of the 12 V EMF.
• Top students clearly stated: "No, Y operates at greater than 65 W" with the formula P = V²/R explicitly stated.