19.2Electric Current and Circuits
Resistance and Ohm’s Law — American System Physics Revision Notes & Topical Questions
10 min read 7 parts Lesson 4 of 5 1 quiz
Lesson summary
Define resistance and use R = V/I.Resistance = potential difference ÷ currentR = V/I (Ω = V/A).RearrangedV = IR and I = V/R.For a wireR = ρL/A — resistance ∝ length and ∝ 1/cross-sectional area (Supplement: proportionality).Cross-sectional area of a round wireA = πd2/4 = πr2.Powe…
- Thermistor
- A resistor has a p.d. of 9.0 V across it and a current of 0.30 A. Find its resistance
- Ohmic conductor
- Non-ohmic component
- Variable resistor (rheostat)
- Filament lamp
Part 1 of 7Key ideas
1Resistance
Resistance potential difference ÷ current
R = V/I (Ω = V/A).
2Rearranged
V = IR and I = V/R.
3For a wire
R = ρL/A — resistance ∝ length and ∝ 1/cross-sectional area (Supplement: proportionality).
4Cross-sectional area of a round wire
A = πd2/4 = πr2.
5
Power can also be written P = I2R = V2/R (W).
6
For an ohmic conductor, the gradient of an I–V graph = 1/R.
7Thermistor
Thermistor temperature ↑ → resistance ↓; LDR: light intensity ↑ → resistance ↓.
8
A filament lamp's resistance rises with temperature, so its I–V graph curves.
Part 2 of 7Learning objectives
1
Define resistance and use R = V/I.
2
Describe an experiment to find resistance with an ammeter and voltmeter.
3
Sketch and explain I–V graphs for a resistor, filament lamp and diode.
4
State how length and cross-sectional area affect the resistance of a wire.
5
Describe the behaviour of thermistors and LDRs.
Part 3 of 7Essential knowledge
1
2
Resistance measures how much a component opposes current. For a given p.d., a larger resistance means a smaller current.
3
To measure resistance, record the p.d. across the component with a voltmeter in parallel and the current through it with an ammeter in series, then calculate R = V/I.
4
In a metal, electrons collide with vibrating positive ions in the lattice. These collisions transfer energy to the ions (heating the wire) and are the cause of resistance.
5
At constant temperature a metal wire obeys Ohm's law: doubling V doubles I, so the I–V graph is a straight line through the origin and R is constant.
6
In a filament lamp, a larger current heats the filament; the ions vibrate more, collisions increase and resistance rises. So the I–V graph curves towards the V-axis (gradient decreases).
7
A diode has very high resistance in reverse and low resistance in the forward direction once a small threshold p.d. is exceeded. It can be used to convert a.c. to d.c. (rectification).
8
The resistance of a wire is directly proportional to its length: charges must travel through twice as many collisions in a wire twice as long.
9
The resistance of a wire is inversely proportional to its cross-sectional area: a thicker wire gives more paths for the charge, like a wider pipe.
10
An NTC thermistor's resistance decreases as temperature rises; it is used in thermostats and fire alarms. An LDR's resistance decreases as light intensity increases; it is used in automatic street lights.
11
Combining P = IV with V = IR gives P = I2R, which shows why transmission cables lose a lot of power as heat when the current is large.
12
When an I–V graph is plotted with I on the y-axis, a steeper line means a smaller resistance.
13
Use a low p.d. and switch off between readings so the component does not heat up during a resistance experiment.
Part 4 of 7Process & sequence
1
Connect a cell/power supply, a variable resistor, an ammeter and the component under test in series.
2
Connect a voltmeter in parallel across the component.
3
Set the variable resistor to give a small current; record V and I.
4
Adjust the variable resistor to obtain at least 6 pairs of readings over the range.
5
Reverse the supply connections and repeat to obtain negative values (important for a diode).
6
Plot a graph of current (y-axis) against p.d. (x-axis).
7
Calculate R = V/I at several points; switch off between readings to avoid heating.
Why a filament lamp's resistance increaseswrite these steps in this order; each step is a marking point.
1
The p.d. across the lamp is increased, so the current increases.
2
More energy is transferred to the filament each second, so its temperature rises.
3
Metal ions in the filament vibrate with larger amplitude.
4
Free electrons collide more often with the ions.
5
The resistance increases, so current rises less than proportionally: the I–V graph curves.
Part 5 of 7Worked examples
30 Ω
A resistor has a p.d. of 9.0 V across it and a current of 0.30 A. Find its resistance.
- R = V/I = 9.0 V ÷ 0.30 A = 30 Ω.
5.0 Ω
A wire of length 1.2 m has resistance 4.0 Ω. What is the resistance of a 3.0 m length of the same wire with double the cross-sectional area?
- R ∝ L
- 4.0 × (3.0/1.2) = 10 Ω.
- R ∝ 1/A: 10 ÷ 2 = 5.0 Ω.
12 V
A 240 Ω resistor carries a current of 0.050 A. Find the p.d. across it and the power transferred.
- V = IR = 0.050 × 240 = 12 V.
- P = I2R = 0.0502 × 240 = 0.60 W .What the symbols mean: V = potential difference (V) · I = current (A) · R = resistance (Ω)
- (check: IV = 0.050 × 12 = 0.60 W)
- P = I2R = 0.0502 × 240 = 0.60 W .
Part 6 of 7Summary
1
Resistance is R = V/I, measured in ohms.
2
Ohm's law: I ∝ V at constant temperature for a metallic conductor.
3
A longer wire has more resistance; a thicker wire has less resistance.
4
A filament lamp's resistance increases as it heats up.
5
Thermistors and LDRs have resistance that falls as temperature or light increases.
Part 7 of 7Compare: Compare
| Component | Shape of I–V graph | How resistance behaves | Reason |
|---|---|---|---|
| Fixed resistor (constant T) | Straight line through origin | Constant | Obeys Ohm's law |
| Filament lamp | Curve that flattens at higher V | Increases as V (and I) increase | Filament gets hotter; ions vibrate more and impede electrons |
| Diode | Near zero current in reverse; current rises sharply above about 0.6 V forward | Very high in reverse, low in forward | Allows current in one direction only |
| Thermistor | Curve bending upwards | Decreases as temperature rises | More free charge carriers when hotter |
| LDR | — | Decreases as light intensity rises | Light releases more charge carriers |
| Change to wire | Effect on R |
|---|---|
| Double the length | R doubles (R ∝ L) |
| Double the cross-sectional area | R halves (R ∝ 1/A) |
| Double the diameter | Area ×4, so R becomes one quarter |
| Heat a metal wire | R increases |
Resistance of a wireR = ρL/A, where ρ is resistivity (Ω m).
Key terms
Command terms
Command terms
Identify / Describe (FRQ)
A short, specific statement; describing requires relevant characteristics, not a single word.
Explain / Justify (FRQ)
Give the cause-and-effect reasoning or evidence; a restated fact does not score.
Calculate
Show setup, substitution and the answer with units.
Predict
State the expected outcome; add a reason only if asked to justify.
MCQ technique
Read every option; eliminate distractors that are true statements but do not answer the question; watch for NOT/EXCEPT.
Exam technique
Keywords the mark scheme looks for in this lesson
Keywords the mark scheme looks for in this lesson
Resistance · Ohm (Ω) · Ohm's law · Ohmic conductor · Non-ohmic component · Resistor · Variable resistor (rheostat) · Filament lamp · Diode · Light-dependent resistor (LDR) · Thermistor (NTC) · Resistivity · Conductor / insulator
Learn each definition in the Key words tab exactly; a missing keyword in a definition usually loses the mark.
How to get full marks (examiner guidance)
1
Answer exactly what is asked; extra correct material does not earn credit, and contradictory statements can cancel credit (AP FRQ).
2
Use precise vocabulary, never “it/stuff/things”.
3
Show units and working in calculations.
4
In MCQs, predict the answer before reading the options, then eliminate.
Words to avoid
1
resistance is how hard it is for current to flowresistance is the p.d. across a component divided by the current through it (R = V/I).
2
Ohm's law is V = IRcurrent is directly proportional to p.d. for a metallic conductor at constant temperature.
3
a thicker wire has more resistanceresistance is inversely proportional to cross-sectional area, so a thicker wire has less resistance.
4
a thermistor's resistance goes up with heatthe resistance of an (NTC) thermistor decreases as temperature increases.
5
the lamp's resistance increases because current increasesthe filament's temperature rises, ions vibrate more, so resistance increases.
Instant feedbackCheck yourself
Q1. A current of 2.0 A flows through a 6.0 Ω resistor. What is the p.d. across it?
- A. 3.0 V
- B. 0.33 V
- C. 12 V
- D. 8.0 V
Show answer
C. V = IR = 2.0 × 6.0 = 12 V.
Q2. Which component has resistance that decreases as the light gets brighter?
- A. LDR
- B. Thermistor
- C. Diode
- D. Filament lamp
Show answer
A. A light-dependent resistor's resistance falls as light intensity increases.
Q3. A wire is replaced by one of the same material, same length but twice the diameter. Its resistance becomes
1
A. twice as large
2
B. half as large
3
C. four times as large
4
D. one quarter as large
Show answer
D. Doubling diameter makes the area four times larger; R ∝ 1/A so R is one quarter.
Q4. Which statement is Ohm's law?
1
A. Resistance increases with temperature
2
B. Current is directly proportional to p.d. for a metallic conductor at constant temperature
3
C. Power equals current times p.d.
4
D. Current is the same everywhere in a series circuit
Show answer
B. Ohm's law relates current and p.d. under constant temperature.
Q5. What is the shape of the I–V graph for a filament lamp?
1
A. Straight line through origin
2
B. Horizontal line
3
C. Curve with decreasing gradient as V increases
4
D. Zero current in one direction
Show answer
C. As the filament heats up its resistance rises, so the gradient (1/R) decreases.
Q6. A 12 V supply gives a current of 0.40 A in a resistor. What is the resistance?
- A. 4.8 Ω
- B. 30 Ω
- C. 0.033 Ω
- D. 12.4 Ω
Show answer
B. R = V/I = 12 ÷ 0.40 = 30 Ω.
Marked practiceQuizzes for this lesson
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