5 high-yield, syllabus-aligned questions on Current electricity, covering 14 individually creditable mark points. Each one shows the answer that scores full marks under our guide, which phrase earns which mark, and a common incomplete answer — so you can see the difference rather than guess at it.
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Every question and solution is reviewed by examiners versed in WAEC and JAMB, each with more than thirty years of experience.
Mark allocations are TopMarks’ own, written against the published syllabus and Chief Examiners’ reports. They are a guide to how these answers are usually credited, not an official marking scheme.
State3 marks
State Ohm's law and state two conditions under which it holds.
TopMarks model answer
Ohm's law states that the current flowing through a conductor is directly proportional to the potential difference across its ends, provided that the temperature and other physical conditions remain constant. It holds for a metallic conductor, and only while the temperature is kept constant.
Suggested marking guide
1Current directly proportional to potential difference
2The temperature must be constant
3It applies to a metallic conductor
Why it scores
V = IR is a DEFINITION of resistance and is true for everything; Ohm’s law is the much stronger claim that R stays constant, and it is only true under conditions. That is why the conditions carry marks. A filament lamp obeys V = IR at every instant and does not obey Ohm’s law at all.
A common incomplete answer
“V = IR”
What it costs: All three marks. This is the defining relationship for resistance, not a statement of Ohm’s law, and none of the conditions is given.
Three resistors of 2 Ω, 3 Ω and 6 Ω are connected in parallel across a 6 V battery of negligible internal resistance. Calculate the effective resistance of the combination and the total current drawn from the battery.
TopMarks model answer
For resistors in parallel, 1/R = 1/2 + 1/3 + 1/6 = 3/6 + 2/6 + 1/6 = 6/6 = 1. Therefore R = 1 Ω. The total current I = V ÷ R = 6 ÷ 1 = 6 A.
Suggested marking guide
1Correct parallel formula used
2Reciprocals correctly summed
3Effective resistance of 1 Ω
4Total current of 6 A
The working, line by line
11/R = 1/R₁ + 1/R₂ + 1/R₃ — Write the formula first. In parallel you add the RECIPROCALS, not the resistances — mixing up the series and parallel rules is the single biggest error in this topic.
21/R = 1/2 + 1/3 + 1/6 = 3/6 + 2/6 + 1/6 = 1 — Put everything over a common denominator so the examiner can follow it.
3R = 1 Ω — Remember to INVERT at the end. Stopping at 1/R = 1 and writing “R = 1” works here by luck; with a different set of resistors it would be wrong.
4I = V ÷ R = 6 ÷ 1 = 6 A — The question asked for two things. Answer both.
Why it scores
The effective resistance of a parallel combination is always SMALLER than the smallest individual resistor. Here the smallest is 2 Ω and the answer is 1 Ω, which is a free sanity check — if your answer comes out bigger than 2 Ω, you have added the resistances instead of the reciprocals.
A common incomplete answer
“1/R = 1/2 + 1/3 + 1/6 = 1, so R = 1 Ω.”
What it costs: One mark of four. The resistance is correct and fully worked, but the question also asked for the total current and it was never calculated.
Key ideas to include: parallel, reciprocal, effective resistance, total current.
Distinguish2 marks
Distinguish between the electromotive force of a cell and the potential difference across its terminals.
TopMarks model answer
The electromotive force of a cell is the total energy supplied by the cell in driving one coulomb of charge round the complete circuit, including through the cell itself. The potential difference across the terminals is the energy converted per coulomb in the external part of the circuit only; the difference between the two is the energy lost in overcoming the internal resistance of the cell.
Suggested marking guide
1E.m.f. covers the complete circuit
2P.d. covers the external circuit only, the difference being the internal resistance
Why it scores
Both quantities are measured in volts, which is exactly why the examiner asks the question — the unit tells you nothing about the difference. The difference is WHERE the energy is delivered: e.m.f. covers the whole circuit including the cell, p.d. covers only the outside. That is also why terminal p.d. falls as the current rises.
A common incomplete answer
“E.m.f. is the voltage of the cell and potential difference is the voltage across a resistor.”
What it costs: Both marks. It describes where each is measured but never says what either one IS — energy per unit charge — and the internal resistance, which is the whole reason they differ, is not mentioned.
Key ideas to include: energy per coulomb, internal resistance, terminal potential difference, lost volts.
State3 marks
State three factors on which the resistance of a metallic wire depends, and state how the resistance is affected by each.
TopMarks model answer
The resistance is directly proportional to the length of the wire, so a longer wire has a greater resistance. The resistance is inversely proportional to the cross-sectional area, so a thicker wire has a lower resistance. The resistance also depends on the nature of the material, that is its resistivity, and for a metal it increases as the temperature rises.
Suggested marking guide
1Length, with the direction of the effect
2Cross-sectional area, with the direction of the effect
3Nature of the material or the temperature
Why it scores
The question says “and state how”, so each factor must come with its direction. A bare list — “length, area, temperature” — answers only half the question and picks up at most one mark of three. Watch for “and state how” or “and explain”; they always double the work per item.
A common incomplete answer
“Length, cross-sectional area and temperature.”
What it costs: One mark of three. Temperature is credited because naming it is enough for that mark point, but length and area are given with no indication of whether they raise or lower the resistance.
Explain why a fuse is always connected in the live wire and not in the neutral wire.
TopMarks model answer
A fuse is a deliberately weak link that melts and breaks the circuit when the current becomes too large. It is placed in the live wire so that when it blows, the appliance is completely disconnected from the high potential of the supply. If it were placed in the neutral wire, the appliance would remain connected to the live supply after the fuse had blown, and anyone touching it could receive a fatal shock.
Suggested marking guide
1The fuse melts to break the circuit on excess current
2Placing it in the live wire isolates the appliance from the supply
Why it scores
The second mark is a safety argument, not a circuit argument. Electrically the fuse would break the circuit from either position — the current would stop either way. The reason it must be in the live wire is what remains AFTER it blows, and that is the point the examiner is testing.
A common incomplete answer
“Because the fuse melts when the current is too high and protects the appliance.”
What it costs: One mark of two. The action of the fuse is credited, but nothing explains why the LIVE wire specifically, which is the whole question.
Key ideas to include: live wire, neutral wire, excess current, isolate, earth.
Practise Current electricity on real questions
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