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Tell the tutor your qualification, exam board and current topics. If you are learning outside a school course, send a description of your goal and an example of the material you want to understand. These details change the level of explanation and the questions worth practising.

A learner may remember an equation yet still think a resistor “uses up” current. In a single closed loop, the current is the same at every point. The potential difference across a resistor and the current through it are different quantities. These ideas are part of AQA GCSE Physics electricity.

Consider this teaching example: a resistor has 6 V across it and a current of 0.20 A through it. To find its resistance:

• Start with V = IR, where V is potential difference in volts, I is current in amperes and R is resistance in ohms.
• Rearrange before substituting: R = V ÷ I.
• Calculate R = 6 ÷ 0.20 = 30 Ω.

A clear response gives both the calculation and its meaning: “The resistance is 30 Ω. The current through the resistor is 0.20 A; it does not become smaller after passing through it.”

For an online lesson, drawing the circuit beside the working makes it possible to point to where each measurement belongs. Then change the question: if the potential difference becomes 3 V across the same ohmic resistor at unchanged temperature, I = 3 ÷ 30 = 0.10 A. Explaining why the current halves checks more than copying the first calculation.

At AS and A-level, interpreting direction matters as well as finding a gradient. AQA mechanics includes motion graphs and acceleration. The following constructed example uses a straight line so the challenge is interpreting the motion.

An object’s velocity changes uniformly from −2 m/s to +4 m/s in 3 s. Choose one direction as positive. The negative starting velocity means the object is initially moving the other way.

The acceleration is the gradient: (4 − (−2)) ÷ 3 = 2 m/s². The velocity reaches zero after 1 s, then becomes positive: the object has reversed direction.

To find displacement, split the area between the velocity line and the time axis into two triangles:

• Before the reversal: ½ × 1 s × 2 m/s = 1 m, counted as −1 m because the velocity is negative.
• After the reversal: ½ × 2 s × 4 m/s = 4 m, counted as +4 m.

The displacement is −1 + 4 = +3 m. The total distance is 1 + 4 = 5 m. An answer of “3 m travelled” confuses the change in position with the length of the journey. Signed area and total distance use those areas differently.

On a shared diagram, shade the two regions and label the direction of travel before calculating. A useful follow-up is to reverse every velocity sign: the displacement becomes −3 m, while the distance remains 5 m.

Use the learner’s recent work to choose a small, specific aim. “Choose the right equation and keep units consistent in circuit questions” gives a lesson more direction than “revise electricity”. Work that aim into a schedule the learner can sustain alongside schoolwork or other commitments.

After an explanation, try a related question independently. For the resistor example, that might mean finding current instead of resistance; for motion, it might mean explaining a negative area before calculating it. Look for fewer prompts, a correct diagram, a reason for the equation choice and a final answer with appropriate units.

Agree a manageable practice task between sessions and return to any step that still breaks down. If the learner can copy a solution but cannot explain it, slow down and revisit the idea. If they can explain and apply it, move to a less familiar question. This is a suggested way to review learning, with lesson frequency adjusted to the work, the target date and the time available for practice.