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Original topic practice

Resistance: apply the explanation

Try six fresh questions without your notes. Write a reason or calculation before choosing an answer. Use the feedback to find the explanation you need to revisit.

How to use this practice

You can keep your own tally of correct first selections, with six self-check points available. No total or answers are saved by this page. Retries are for learning; these are not official marks or a grade prediction. Written working and partial method credit are not automatically assessed. Completing this practice does not mark the lesson mastered.

Reason about the proposed circuits on paper. These questions do not demonstrate supervised practical competence. Actual measurements require teacher-supervised, suitable low-voltage equipment; never build a mains circuit.

Specialist review has not been completed. The questions and data are original; difficulty labels describe task demands, not exam-board ratings.

Scope and curriculum

Grade 8 is StudyForA’s suggested placement, not an official grade requirement. This practice develops resistance as a voltage/current ratio in England KS3 current electricity. Combined resistances and parallel networks remain further study.

Foundation · Convert milliamperes before dividing

A component has 6.6 V across it while 220 mA passes through it. Use these readings for this operating point.

Question 1

What is its resistance?

Check the reasoning

30 Ω

  1. Convert 220 ÷ 1000 = 0.220 A.
  2. R = 6.6 ÷ 0.220 = 30 Ω. This reading alone does not establish constant resistance at other voltages.
Revisit: Converting current and comparing readings

Apply · Use readings for the intended resistor

R1 and R2 form one steady series path across a supply. An ideal ammeter reads 0.30 A. The supply terminal voltage is 6.0 V, and an ideal voltmeter across R2 reads 1.8 V. Neglect wire voltage drops.

Question 2

Which calculation gives the resistance of R2 alone?

Check the reasoning

1.8 ÷ 0.30 = 6 Ω

  1. Identify the component named in the question: R2.
  2. Use the voltage across R2 and current through R2: R = 1.8 ÷ 0.30 = 6 Ω.
  3. Supply readings would refer to the whole path; the number of series resistors does not split the current.
Revisit: Measuring across and through the same component

Apply · Compare unequal-voltage cases

At the stated operating points, X has 3.2 V across it and carries 80 mA. Y has 7.2 V across it and carries 120 mA. The two components are measured separately.

Question 3

Which resistance comparison is justified?

Check the reasoning

X is 40 Ω and Y is 60 Ω; Y has greater resistance.

  1. X: 80 mA = 0.080 A; R = 3.2 ÷ 0.080 = 40 Ω.
  2. Y: 120 mA = 0.120 A; R = 7.2 ÷ 0.120 = 60 Ω.
  3. Compare the ratios, not the currents alone; Y has 1.5 times X’s resistance at these operating points.
Revisit: Why the voltage must be held fixed for a current-only comparison

Apply · State the condition behind a prediction

A fixed resistor carries 0.070 A with 2.8 V across it. In a paper model, its temperature and resistance remain unchanged when the potential difference becomes 4.8 V.

Question 4

What current does the model predict at 4.8 V?

Check the reasoning

0.12 A, provided the stated constant-resistance condition holds.

  1. Calculate the original resistance: 2.8 ÷ 0.070 = 40 Ω.
  2. Use the given model: I = 4.8 ÷ 40 = 0.12 A (120 mA).
  3. The prediction relies on unchanged resistance; one initial reading cannot establish that condition by itself.
Revisit: Conditions for constant resistance

Evaluate · Test a model against a data pattern

These constructed values illustrate three operating points for one component; they are not laboratory measurements. Treat the values as exact for this paper calculation. Temperature was not controlled.

Constructed voltage and current values
Across component (V)Through component (A)
1.00.05
2.00.08
3.00.10

Question 5

Which conclusion follows from calculating the three resistance ratios?

Check the reasoning

The ratios are 20 Ω, 25 Ω and 30 Ω; a constant-resistance model does not fit these values.

  1. Calculate matching ratios: 1.0 ÷ 0.05 = 20 Ω; 2.0 ÷ 0.08 = 25 Ω; 3.0 ÷ 0.10 = 30 Ω.
  2. These ratios differ, so one constant-resistance value does not describe the supplied data.
  3. The data alone do not identify the cause. For a real investigation, monitor temperature, check instruments and compare repeat readings under controlled conditions.
Revisit: Using ratios without overstating the evidence

Evaluate · Choose a valid measurement plan

A teacher asks for a paper plan to estimate the resistance of a low-voltage resistor. Points P and Q are its two ends. Assume ideal meters for the calculation; actual apparatus must be suitable and teacher-supervised.

Question 6

Which plan measures the required quantities while addressing heating?

Check the reasoning

Place the ammeter in series and voltmeter across P and Q; use brief low-current readings and check for temperature drift.

  1. Ammeter in series; voltmeter in parallel across P and Q. Use correct sockets and ranges, and switch off before rewiring.
  2. Use appropriate low currents and brief readings within apparatus ratings. Allow cooling if needed and compare a repeated low setting after higher settings to look for drift.
  3. R = V ÷ I applies to matching readings. Written planning is not a demonstration of practical competence.
Revisit: Meter placement and control of conditions

Choose your next step

Explore the resistance mind map

Revisit any explanation you could not justify, then return to the question and explain why the other options fail. The lesson’s recall check remains separate.

Return to the resistance lesson · Back to Physics and technology