Resistance and Ohm's Law

Resistance measures how much a component opposes the flow of electric charge. The higher the resistance, the harder it is for charge to flow, so the smaller the current for a given push.

The rule that ties current, potential difference and resistance together is:

The current through a component depends on both its resistance and the potential difference across it. For a given potential difference, greater resistance gives a smaller current.

Resistance is measured in ohms, symbol Ω (the Greek letter omega). One ohm is one volt per ampere: a component has a resistance of 1 Ω if a potential difference of 1 V drives a current of 1 A through it.

Everything in a circuit has some resistance, including the connecting wires, but in most problems the wires' resistance is small enough to ignore.

Potential difference, current and resistance are linked by:

$V=IR$

where $V$ is potential difference in volts (V), $I$ is current in amperes (A) and $R$ is resistance in ohms (Ω).

You must recall and apply this equation. It is not given on the Physics equation sheet.

Rearranged forms let you find whichever quantity is missing:

$I=\frac{V}{R}R=\frac{V}{I}$

To measure the resistance of a component, you set up a circuit with an ammeter in series (to read the current $I$) and a voltmeter in parallel across the component (to read the potential difference $V$), then calculate $R=V/I$.

A resistor has a current of 0.25 A flowing through it when the potential difference across it is 6 V. Calculate its resistance.

Step 1 — Rearrange V = IR to make R the subject.

$R=\frac{V}{I}$

Step 2 — Substitute the values.

$R=\frac{6}{0.25}$

Step 3 — Calculate.

$R=24\Omega$

The resistance is 24 Ω.

Check by working backwards: $V=IR=0.25\times 24=6$ V, which matches the given potential difference.

A second quick example: if a 12 Ω resistor carries a current of 0.5 A, the potential difference across it is $V=IR=0.5\times 12=6$ V.

Aim: investigate how the length of a wire affects its resistance at constant temperature.

Apparatus: a length of resistance wire (such as constantan) taped alongside a metre ruler, a cell or power supply, an ammeter, a voltmeter, a switch and connecting leads with a crocodile clip (the flying lead).

Method:

1. Connect the ammeter in series with the wire and the voltmeter in parallel across the length of wire being tested.
2. Clip the flying lead onto the wire to set the tested length, starting at, say, 10 cm.
3. Close the switch, record the current $I$ and potential difference $V$, then open the switch again.
4. Calculate the resistance using $R=V/I$.
5. Repeat for a range of lengths (for example 20, 30, 40 … 100 cm) and plot resistance against length.

Variable	Type	Detail
Length of wire	Independent	The quantity you deliberately change
Resistance (from V and I)	Dependent	The quantity you measure/calculate
Wire material and thickness	Control	Same wire throughout
Temperature	Control	Keep the wire cool

Why open the switch between readings: a current warms the wire, and a hotter wire has a higher resistance, which would spoil the results. Keeping the switch closed only briefly keeps the temperature (a control variable) roughly constant.

Expected result: the graph of resistance against length is a straight line through the origin. Resistance is directly proportional to length: doubling the length doubles the resistance.

Required practical 3 also investigates how combining resistors changes the total resistance. Using identical resistors, you build a series combination and a parallel combination and compare their overall resistance.

Series: the resistors are joined end to end so charge flows through each in turn. The resistances add:

$R_{total}=R_1+R_2$

Adding a resistor in series increases the total resistance, because the charge meets more opposition on its single path.

Parallel: the resistors are connected side by side, giving the charge more than one path. This decreases the total resistance to a value less than the smallest individual resistor, because a second path makes it easier overall for charge to flow.

(Note) For AQA you are not required to calculate the total resistance of two resistors in parallel. You only need to explain, qualitatively, that adding resistors in parallel lowers the total resistance.

Worked example (series): two resistors of 10 Ω and 15 Ω are connected in series.

$R_{total}=10+15=25\Omega$

The combination has a resistance of 25 Ω, larger than either resistor on its own.

1. Confusing "greater resistance" with "greater current"

Greater resistance means a smaller current for the same potential difference, not a larger one. Resistance opposes flow.

2. Reading the resistance graph as current vs length

In Required practical 3 you plot resistance against length, giving a straight line through the origin. Do not confuse this with an I–V graph.

3. Letting the wire heat up

If the wire warms, its resistance rises and the length graph curves. Keep readings brief and the current low so temperature stays constant.

4. Getting the meter connections wrong

To measure resistance you need the ammeter in series (reads current) and the voltmeter in parallel across the component (reads potential difference). Then $R=V/I$.

5. Trying to calculate parallel resistance

AQA does not require you to calculate the total resistance of two parallel resistors. State only that it is less than the smallest resistor. Attempting a full calculation wastes time and is not credited.