AQA AS Level Physics Paper 1, June 2024: Question 5

8 marks · Medium difficulty · Short Answer

Analyze current, voltage, resistance, and power relationships in a filament lamp circuit containing a variable resistor, including graphical representation, power dissipation calculations, and circuit modifications.

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Question

Figure 12 shows a 9.0 V battery in series with a variable resistor R and a filament lamp L1. Figure 13 shows a graph of resistance versus current I for L1, starting at around 0.75 ohms and rising steeply above 1.0 A up to 4.0 ohms at 2.0 A. Figure 14 is an empty V-I grid for L1 from 0 to 0.50 A. Subsequent parts ask to calculate power dissipation in L1 when I = 1.9 A, calculate the resistance of R when I = 1.5 A and V = 3.3 V, and deduce the change in resistance of R when an identical lamp L2 is added in parallel.

Question text

05 Figure 12 shows a circuit for controlling the current I in a filament lamp L1.

The battery has negligible internal resistance.

Figure 12

Figure 13 shows how the resistance of L1 varies with I.

Figure 13

05.1 The current in L1 is increased from 0 to 0.50 A.

The potential difference V across L1 is 0.375 V when I is 0.50 A.

Draw, on Figure 14, a V–I graph for L1 in the current range 0 to 0.50 A.

[1 mark]

Figure 14

05.2 Calculate the power dissipated in L1 when I is 1.9 A.

[2 marks]

power dissipated = W

05.3 The variable resistor R in Figure 12 is adjusted until I is 1.5 A and V is 3.3 V.

Calculate the resistance of R.

[2 marks]

resistance of R = Ω

05.4 Figure 15 shows a second lamp L2, identical to L1, connected to the circuit.

Figure 15

R is adjusted so that the potential difference across L1 is again 3.3 V.

Deduce, without calculation, the change in the resistance of R.

[3 marks]

Mark scheme

Show the mark scheme

Question Answers Additional Comments/Guidance Mark AO

05.1 Straight-line graph through the origin and passing through Line must: 1 AO2

(0.5 A, 0.375 V)

• be drawn with a ruler.

• be close to (0.5 A, 0.375 ). Do not

allow vertical and horizontal

inaccuracy.

• pass within their line’s width of origin.

05.2 correct read-off of resistance for I = 1.9 A Allow R in range 3.5 to 3.7 Ω 2 1 × AO1

OR 1 × AO2

Use of P = I 2R for their R

Use of V=IR and P= VI for their R

𝑉𝑉2

Use of V=IR and P= for their R

𝑅𝑅

(power =) 13(.0) (W) Accept answers in range 12.6 to 13.4

Allow 1 mark for 13.7(18) (W) or 14 (W) on

answer line without supporting work.

05.3 Use of VT = V1 + V2 (V =) 9 – 3.3 or (V=) 5.7 (V) 2 1 × AO1

OR 1 × AO2

Use of V = IR for their V

(R =) 3.8 Ω

Alternative method:

𝑉𝑉

Use of R = 3.3

𝐼𝐼 Expect to see (Rlamp=) OR (Rlamp =)2.2 (Ω)

1.5

OR 9

And (RT=) OR (RT =)6 (Ω)

1.5

Use of RT = R1 + R2 for their Rlamp or their RT

(R =) 3.8 Ω

Alternative method:

Reads off Rlamp = 2.2 Ω

Allow R in range 2.1 to 2.2 Ω

𝑉𝑉

Use of R = 9

𝐼𝐼 Expect to see R= OR R =6 (Ω)

OR 1.5

Use of RT = R1 + R2 for their Rlamp or their RT

(R =) 3.8 Ω Range of R where read-off used for R is 3.8

lamp

to 3.9 Ω.

05.4 R’s value must decrease Allow any value quoted for decrease. 3 2 × AO2

Condone Max 2 for use of a calculation. 1 × AO3

• More current (in bulbs parallel section)

• Idea this requires a lower resistance for R to maintain

same pd across R (therefore same pd across bulbs). Be wary of ‘the resistance is decreasing’,

needs a definite statement that this relates to R

to score MP1.

• Total resistance (of L1 & L2) is lower

• Idea that the ratio of pd division

For example:

VR : VL is same for R and L1 & L2 combination as for R

and L1.

Total 8

How to answer it

Filament Lamp Circuits & Non-Ohmic Components

AQA AS Level Physics • Electric Circuits

What this question tests

This multi-part question assesses your understanding of non-ohmic components, graphical interpretation of V-I characteristics, electrical power formulas, potential divider principles, and how adding parallel components alters total resistance and circuit voltage distribution.

Part 05.1: Graphical Representation of V-I Characteristics

Draw a V-I graph for the filament lamp in the range 0 to 0.50 A.

✅ Correct Answer

A straight-line graph starting at the origin (0, 0) and passing directly through the coordinate point (0.50 A, 0.375 V) .

💡 Key Knowledge

For low currents (from 0 to 0.50 A), the resistance of the filament lamp is constant (around 0.75 ohms as seen on Figure 13). Because V = IR and R is constant, V is directly proportional to I , yielding a straight line.

🧠 Exam Technique

Always use a sharp pencil and a ruler. Ensure your line starts precisely at the origin and hits the target coordinate cleanly. Examiners penalize loose freehand drawing or lines that miss the intersection.

Mark allocation: 1 mark (AO2)

Part 05.2: Calculating Power Dissipated

Calculate the power dissipated in L₁ when I is 1.9 A. [2 marks]

📐 Step-by-Step Calculation

Read off resistance: From Figure 13, find the resistance of the lamp when current I = 1.9 A . R ≈ 3.5 to 3.7 ohms (allow 3.6 ohms as a standard median).
Select power formula: P = I²R
Substitute values: P = (1.9)² × 3.6 = 12.996 W
Apply significant figures: P = 13 W (to 2 sf, matching data given in the question).

❌ Common Errors & Pitfalls

Reading the wrong axis or misinterpreting the grid lines on Figure 13 for I = 1.9 A .
Using P = IV without realizing voltage is not directly given for I = 1.9 A (though finding V = IR first is a valid alternative route).
Failing to use appropriate significant figures, though 13.0 W is also accepted.

Mark allocation: 1 mark for correct read-off / equation use, 1 mark for correct final value (13 W).

Part 05.3: Potential Divider Analysis

Calculate the resistance of R when I is 1.5 A and V across L₁ is 3.3 V. [2 marks]

📐 Step-by-Step Calculation

Find pd across R (V_R): Using Kirchhoff's Second Law for series components:
V_R = V_total - V_lamp = 9.0 V - 3.3 V = 5.7 V
Calculate resistance of R: Using Ohm's Law R = V / I :
R = 5.7 V / 1.5 A = 3.8 ohms

💡 Alternative Method

Find total circuit resistance first: R_total = 9.0 / 1.5 = 6.0 ohms . Find lamp resistance at 1.5 A: R_lamp = 3.3 / 1.5 = 2.2 ohms . Subtract to find R: R = 6.0 - 2.2 = 3.8 ohms .

Mark allocation: 1 mark for determining voltage/resistance steps, 1 mark for final answer 3.8 ohms.

Part 05.4: Adding a Parallel Lamp

Deduce, without calculation, the change in the resistance of R when an identical lamp L₂ is added in parallel. [3 marks]

✅ Correct Answer

The value of resistor R must decrease.

🧠 Top-Level Exam Strategy (Mark Scheme Logic)

MP1: Adding L₂ in parallel increases the total current drawn through the parallel combination for the same pd, or decreases the combined resistance of the lamps.
MP2: To keep the potential difference across L₁ fixed at 3.3 V, the current through L₁ remains the same, but total circuit current increases because L₂ now draws current too.
MP3: With a larger total current flowing through R, to maintain the same share of potential difference across R (or keep lamp pd at 3.3 V with a 9.0 V supply), the resistance of R must be smaller ( R = V / I ).

❌ Common Errors

Students often write "the resistance decreases" without specifying whether they mean the lamp combination or resistor R. You must explicitly state that the resistance of variable resistor R decreases to secure full marks.

Mark allocation: 1 mark for stating R decreases, 2 marks for clear physical justification involving current/resistance changes in the parallel network.

Topics

Physics · 3.5 Electricity

Question and mark scheme from the AQA AS Level Physics examination, Paper 1, June 2024. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.