Core · Practice questions · Resistance

Voltage, current, resistance.

Six original Cambridge-style questions on R = V ÷ I: defining the ohm, the three rearrangements, reading a current from voltage and resistance, and why a hot filament breaks the simple pattern.

Original questions All questions on this page are original work, written in the Cambridge IGCSE style. They are not from past papers. They test the same concepts and skills the syllabus rewards.

How to handle resistance questions

R = V ÷ I, every time.

Match the pair: use the voltage across a component with the current through that same component.
Three forms: R = V ÷ I, V = IR, I = V ÷ R.
Units: volts, amperes and ohms; one ohm is one volt per ampere.
Hold conditions steady: the simple proportion holds only while the temperature is constant.

Analysis

[2 marks]

Write the equation that defines resistance, name each quantity, and state the unit of each.

R = V ÷ I, where V is the potential difference (volts), I is the current (amperes) and R is the resistance (ohms). ✓
A resistance of 1 Ω passes a current of 1 A when 1 V is across it. ✓

Calculation

[2 marks]

A current of 0.40 A flows through a component when the potential difference across it is 6.0 V. Calculate its resistance.

R = V ÷ I = 6.0 ÷ 0.40

R = 15 Ω

Calculation

[2 marks]

A 24 Ω resistor is connected across a 12 V supply. Calculate the current through it.

I = V ÷ R = 12 ÷ 24

I = 0.50 A

Calculation

[2 marks]

A heating element has a resistance of 30 Ω and carries a current of 8.0 A. Calculate the potential difference across it.

V = IR = 8.0 × 30

V = 240 V

Calculation

[3 marks]

A fixed resistor is kept at a constant temperature. When the potential difference across it is 4.0 V, the current is 0.25 A.

(a) Calculate its resistance.
(b) The potential difference is increased to 8.0 V. Calculate the new current.

(a) R = V ÷ I = 4.0 ÷ 0.25

R = 16 Ω

(b) I = V ÷ R = 8.0 ÷ 16

I = 0.50 A

double the voltage, double the current, since R is unchanged

Analysis

[3 marks]

A student measures the current through a filament lamp at several voltages and finds that the value of V ÷ I is larger at high voltages than at low voltages.

(a) State what happens to the resistance of the lamp as the voltage increases.
(b) Explain why, in terms of the filament.

(a) The resistance increases as the voltage increases. ✓
(b) A larger voltage drives a larger current, which heats the filament so its temperature rises. ✓
The hotter metal resists the current more, so R = V ÷ I gives a larger value. ✓

so the current is no longer proportional to the voltage

Mark this once you have attempted all six and checked your working. It records a Practiced badge on the topic and adds a one-time bonus. Revealing the solutions alone does not count.