Resistance and Ohm's law

Key ideas

Resistance measures how much a component opposes the flow of current. A bigger resistance means less current for the same voltage.

Ohm's law

$V = IR$

$V$ — voltage across the component (V)
$I$ — current through it (A)
$R$ — resistance (Ω, ohms)

Rearranged: $R = \dfrac{V}{I}$ and $I = \dfrac{V}{R}$ .

Power and resistance

Combining $P = IV$ with $V = IR$ gives two more useful forms:

$P = I^2R \qquad P = \frac{V^2}{R}$

These are handy when you know the resistance and only one of current or voltage.

:::tip $V$ in Ohm's law is the voltage across that one component, and $I$ is the current through that same component — not the supply values, unless the component is the only thing in the circuit. In multi-component circuits, match each $V$ , $I$ and $R$ to the same component. :::

Worked ExampleResistance and power of a heater element

A heater element carries $4.0$ A when connected to a $230$ V supply. Find its resistance and its power.

Step 1 — Resistance from Ohm's law

$R = \frac{V}{I} = \frac{230}{4.0} = 57.5 \;\Omega$

Step 2 — Power

$P = IV = 4.0 \times 230 = 920 \text{ W}$

Practice question

A $12$ V supply drives $0.30$ A through a resistor. Find its resistance.

Worked solution: $R = V/I = 12/0.30 = 40\ \Omega$ .

Test yourself

Practice by grade

One question each at Achieved, Merit and Excellence. Have a go, then compare with the model answer.

Achieved

A resistor carries a current of $0.50$ A when $6.0$ V is applied across it.

Calculate its resistance.

Merit

A $15\ \Omega$ resistor is connected to a $9.0$ V supply.

Calculate the current through it and the power it dissipates.

Excellence

Two resistors, $10\ \Omega$ and $20\ \Omega$ , are each connected in turn across the same $12$ V supply. Explain fully, with calculations, which dissipates more power and why.

Resistance and Ohm's law

Covers AS91173

Key ideas

Resistance measures how much a component opposes the flow of current. A bigger resistance means less current for the same voltage.

Ohm's law

$V = IR$

$V$ — voltage across the component (V)
$I$ — current through it (A)
$R$ — resistance (Ω, ohms)

Rearranged: $R = \dfrac{V}{I}$ and $I = \dfrac{V}{R}$ .

Power and resistance

Combining $P = IV$ with $V = IR$ gives two more useful forms:

$P = I^2R \qquad P = \frac{V^2}{R}$

These are handy when you know the resistance and only one of current or voltage.

Worked ExampleResistance and power of a heater element

A heater element carries $4.0$ A when connected to a $230$ V supply. Find its resistance and its power.

Step 1 — Resistance from Ohm's law

$R = \frac{V}{I} = \frac{230}{4.0} = 57.5 \;\Omega$

Step 2 — Power

$P = IV = 4.0 \times 230 = 920 \text{ W}$

Practice question

A $12$ V supply drives $0.30$ A through a resistor. Find its resistance.

Worked solution: $R = V/I = 12/0.30 = 40\ \Omega$ .

Test yourself

Practice by grade

One question each at Achieved, Merit and Excellence. Have a go, then compare with the model answer.

Achieved

A resistor carries a current of $0.50$ A when $6.0$ V is applied across it.

Calculate its resistance.

Merit

A $15\ \Omega$ resistor is connected to a $9.0$ V supply.

Calculate the current through it and the power it dissipates.

Excellence

Two resistors, $10\ \Omega$ and $20\ \Omega$ , are each connected in turn across the same $12$ V supply. Explain fully, with calculations, which dissipates more power and why.