IGCSE Physics: Electrical Quantities Practice Questions
Current is the rate of flow of electric charge, measured in amperes. Potential difference is the work done per unit charge, measured in volts. Resistance is potential difference divided by current, measured in ohms.
Sub-topic 4.2 of Cambridge IGCSE Physics 0625 supplies four equations that every circuit question depends on. Definitions are examined as often as calculations, and vague definitions score nothing. The questions below cover both.
What you need to know for Electrical Quantities
- CurrentThe rate of flow of electric charge. I = Q divided by t, with charge in coulombs and time in seconds. In a metal the charge is carried by delocalised electrons.
- Potential differenceThe work done per unit charge passing between two points. V = W divided by Q. One volt is one joule per coulomb.
- ResistanceR = V divided by I, measured in ohms. Resistance opposes the flow of current.
- Ohm's lawFor a metallic conductor at constant temperature, the current through it is directly proportional to the potential difference across it, so a graph of I against V is a straight line through the origin.
- Electrical power and energyP = IV, and energy transferred E = IVt. Power can also be written P = I2R using V = IR.
- Resistance of a wireResistance increases with length and decreases as cross sectional area increases. A longer, thinner wire has a higher resistance.
IGCSE Physics Electrical Quantities questions and answers
4 exam-style questions written to the 0625 syllabus. Try each one on paper first, then open the worked answer to check your method against the marks.
Define electric current and calculate the charge that flows when a current of 0.40 A passes for 5.0 minutes.
Show the worked answer
- Current is the rate of flow of electric charge, measured in amperes. One ampere is one coulomb per second.
- Convert the time to seconds: 5.0 minutes = 5.0 x 60 = 300 s.
- Rearrange I = Q divided by t to give Q = I x t.
- Q = 0.40 x 300 = 120 C.
A 12 V supply drives a current of 0.50 A through a lamp. Calculate the resistance of the lamp and the power it dissipates.
Show the worked answer
- Use R = V divided by I.
- R = 12 divided by 0.50 = 24 ohms.
- Use P = IV.
- P = 0.50 x 12 = 6.0 W.
Describe how the resistance of a metal wire depends on its length and its cross sectional area, and explain the effect of each.
Show the worked answer
- Resistance increases in proportion to length. Doubling the length doubles the resistance.
- A longer wire means the electrons travel further and undergo more collisions with the vibrating metal ions, so more opposition to the flow.
- Resistance decreases as cross sectional area increases. Doubling the area halves the resistance.
- A thicker wire provides more paths through which the electrons can flow at any instant, so the same potential difference drives a larger current.
A student plots current against potential difference for a filament lamp. The graph is a curve that becomes less steep as the potential difference increases. Explain the shape.
Show the worked answer
- A straight line through the origin would mean constant resistance. This graph curves, so the resistance is changing.
- As the potential difference increases, more current flows and more energy is transferred in the filament, so its temperature rises.
- The metal ions in the filament vibrate more vigorously at the higher temperature, so the electrons collide with them more often.
- The resistance therefore increases with temperature, so each additional volt produces a smaller increase in current and the graph becomes less steep. A filament lamp does not obey Ohm's law.
Common mistakes in this topic
- Leaving time in minutes or hours in charge and energy calculations.
- Confusing the equations for resistance and power.
- Saying a filament lamp obeys Ohm's law. It does not, because its temperature changes.
- Defining current as the flow of electricity. Define it as the rate of flow of charge.
- Forgetting units, particularly the ohm and the coulomb.
Exam tips
- Learn the four equations and which quantities they link, then choose based on what the question gives you.
- Convert all times to seconds before substituting.
- For any I against V graph, first ask whether the line is straight through the origin. If not, the resistance is changing.
- Explain resistance changes in terms of electron collisions with vibrating metal ions.
- Check your answer's unit. If it should be ohms and the arithmetic gave watts, the equation is inverted.
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Electrical Quantities FAQs
What is electric current?
Current is the rate of flow of electric charge, calculated as charge divided by time and measured in amperes. One ampere is one coulomb of charge passing a point each second. In a metal conductor the charge is carried by free delocalised electrons.
What is potential difference?
Potential difference is the work done per unit charge passing between two points in a circuit, measured in volts. One volt is one joule of energy transferred per coulomb of charge. It is measured with a voltmeter connected in parallel across the component.
Does a filament lamp obey Ohm's law?
No. Ohm's law applies to a metallic conductor at constant temperature. As current through a filament lamp increases, the filament heats up and its resistance rises, so a graph of current against potential difference curves rather than forming a straight line through the origin.
How does the resistance of a wire depend on its dimensions?
Resistance is directly proportional to length, because a longer wire means more collisions between electrons and vibrating metal ions. Resistance is inversely proportional to cross sectional area, because a thicker wire offers more paths for electrons to flow through at any moment.
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Written to the published Cambridge IGCSE Physics (0625) syllabus. Check your school entry code and syllabus year, because Core and Extended candidates are assessed on different content. Last reviewed 2026-08-12.