4.2.4 Resistance

Learning objectives

4.2.4.1Resistance: R = V/I• Recall/use resistance: R = V/I 4.2.4.2An experiment to determine resistance• Describe an experiment to determine resistance using a voltmeter and an ammeter and do the appropriate calculations 4.2.4.3State, qualitatively, the relationship• State, qualitatively, the relationship of the resistance of a metallic wire to its length and to its cross-sectional area 4.2.4.4And explain the current–voltage graphs• Sketch and explain the current–voltage graphs for a resistor of constant resistance, a filament lamp and a diode 4.2.4.5Following relationship for a metallic• Recall/use the following relationship for a metallic electrical conductor: (a) resistance is directly proportional to length (b) resistance is inversely proportional to cross-sectional area

Resistance compares the potential difference across a component with the current through that same component. A larger resistance needs a larger p.d. to produce the same current.

R = \frac{V}{I}

Quantity	Symbol	Unit
resistance	$R$	ohm, $\Omega$
potential difference across the component	$V$	volt, V
current through the component	$I$	ampere, A

For a resistor with 6.0 V across it and 0.30 A through it: $R = 6.0 / 0.30 = 20\,\Omega$ . The same relationship rearranges to $V = IR$ and $I = V/R$ .

Pair readings from the same component. Convert milliamperes to amperes before dividing: 4.0 mA = 0.0040 A. Do not substitute the supply p.d. if only part of the circuit is being measured.

Determine a component's resistance by measuring the current through it and the potential difference across it, then calculating $R = V/I$ .

Equipment	Connection and purpose
ammeter	in series with the component, so it measures the current through it
voltmeter	in parallel across the component, so it measures its p.d.
variable resistor	in series, to change the current safely and obtain several reading pairs

Build the circuit with the switch open and select meter ranges above the expected readings.
Close the switch briefly; record $I$ and $V$ together.
Adjust the variable resistor and repeat for several pairs.
For each pair, calculate $R = V/I$ with V, A and $\Omega$ .

If one reading pair is 2.4 V and 0.20 A, then $R = 2.4/0.20 = 12\,\Omega$ . Similar values across repeated low-current readings support a constant resistance.

Open the switch between readings and keep the current low when temperature must stay nearly constant: heating a metal component can change its resistance. An ammeter in parallel or a voltmeter in series is the wrong arrangement.

For metallic wires of the same material at the same temperature, resistance increases with length and decreases with cross-sectional area.

Change made on its own	Effect on resistance	Physical picture
make the wire longer	increases	charge carriers travel through more material and undergo more interactions
make the wire shorter	decreases	the conducting path is shorter
increase cross-sectional area	decreases	more conducting paths are available side by side
decrease cross-sectional area	increases	fewer conducting paths are available side by side

A long, thin wire therefore tends to have a larger resistance than a short, thick wire of the same metal. A larger diameter means a larger cross-sectional area, so it lowers resistance when length is unchanged.

Keep material and temperature controlled before attributing a change to length or area. If length and area both change, the two effects may oppose each other; a qualitative guess is then unsafe without comparing their factors.

On an $I$ – $V$ graph with current $I$ on the vertical axis and potential difference $V$ on the horizontal axis, the shape shows whether $V/I$ stays constant and whether current can flow in both directions.

Component	Shape to sketch	Explanation
constant-resistance resistor	straight line through the origin, continuing with the same gradient for positive and negative values	$I$ is directly proportional to $V$ , so $V/I$ is constant
filament lamp	curve through the origin that becomes less steep as $\|V\|$ increases; symmetric in the opposite direction	larger current heats the filament, its resistance increases, and each extra volt produces a smaller current increase
diode	almost zero current in reverse and at small forward p.d.; then a steep forward rise	it conducts appreciably only in its forward direction after sufficient forward p.d.

For the straight resistor line, gradient $\Delta I/\Delta V = 1/R$ : a steeper line represents a smaller constant resistance. For a curved device, compare $V/I$ at the stated operating point instead of treating the whole graph as one constant gradient.

Label both axes before describing steepness. If the axes are swapped, the gradient statements reverse. A diode graph is directional; a filament-lamp graph is curved in both positive and negative directions.

For metallic conductors made from the same material and kept at the same temperature, resistance is directly proportional to length and inversely proportional to cross-sectional area.

R \propto \frac{L}{A}

\frac{R_2}{R_1} = \frac{L_2}{L_1}\times\frac{A_1}{A_2}

A wire has $R_1=0.14\,\Omega$ , $L_1=2.0$ m and $A_1=0.40$ mm². For the same material with $L_2=3.0$ m and $A_2=0.90$ mm²: $R_2=0.14\times(3.0/2.0)\times(0.40/0.90)=0.093\,\Omega$ .

For a circular wire, $A = \pi d^2/4$ . Doubling diameter makes the area four times larger, so resistance becomes one quarter when length is unchanged. Combine simultaneous changes by multiplying the length factor by the inverse area factor.

The inverse relationship is with area, not diameter. Keep area units consistent within the ratio; they cancel. This comparison does not hold if material or temperature also changes.

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—4.2.5 Electrical energy and electrical powerRead

4.2.4 Resistance

Learning objectives

Calculate resistance from p.d. and current

Determine resistance with an ammeter and voltmeter

Predict how wire dimensions change resistance

Read current–voltage graphs for three components

Calculate resistance changes from wire dimensions

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