Ohm's Law & Resistance

IGCSE Edexcel Physics
2.9–2.13 Resistance, Ohm's law and V = IR calculations
Key Concepts: Ohm's law: voltage is proportional to current for a metallic conductor at constant temperature. $V = IR$. Resistance is measured in ohms (Ω). An ohmic conductor produces a straight-line I-V graph through the origin.

Section A — Ohm's Law

1. State Ohm's law. [2]
2. Write the equation linking voltage ($V$), current ($I$) and resistance ($R$), defining each symbol and its unit. [3]
3. Explain what happens to the current through a resistor if the voltage is doubled and the resistance stays the same. [2]

Section B — Calculations

4. A resistor has a voltage of 9 V across it and a current of 3 A through it. Calculate the resistance. [2]
5. A resistor of $8\,\Omega$ carries a current of $0.5\,\text{A}$. Calculate the voltage across it. [2]
6. A lamp with resistance $6\,\Omega$ is connected to a 12 V supply. Calculate the current. [2]
7. A $4\,\Omega$ resistor is connected to a 12 V supply. Calculate the current. [2]
8. A heater takes 3.0 A from a 230 V supply. Calculate the resistance of the heater. [2]

Section C — Resistance in Context

9. A student investigates resistance by measuring current for different voltages across a resistor. Describe the graph they would expect if the resistor is ohmic, and explain what the gradient represents. [3]
10. Two identical resistors of $10\,\Omega$ are connected in series to a 20 V supply. Calculate: [4]

a) The total resistance

b) The current in the circuit

c) The voltage across one resistor

Total marks: 24

Mark Scheme

1. For a metallic conductor at constant temperature [1], current is directly proportional to voltage [1] [2]
2. $V = IR$ [1]; $V$ = voltage in volts (V), $I$ = current in amperes (A) [1], $R$ = resistance in ohms (Ω) [1] [3]
3. Current doubles [1]; because $I = V/R$ and $R$ is constant, so $I$ is proportional to $V$ [1] [2]
4. $R = V/I = 9/3 = 3\,\Omega$ [2]
5. $V = IR = 0.5 \times 8 = 4\,\text{V}$ [2]
6. $I = V/R = 12/6 = 2\,\text{A}$ [2]
7. $I = V/R = 12/4 = 3\,\text{A}$ [2]
8. $R = V/I = 230/3.0 \approx 76.7\,\Omega$ [2]
9. Straight line through the origin [1]; constant gradient = constant resistance [1]; gradient = $1/R$ (reciprocal of resistance) [1] [3]
10. a) $R_{total} = 10 + 10 = 20\,\Omega$ [1]; b) $I = V/R = 20/20 = 1\,\text{A}$ [2]; c) $V = IR = 1 \times 10 = 10\,\text{V}$ [1] [4]