CIE iGCSE Co-ordinated Sciences-P4.2.4 Resistance- Study Notes- New Syllabus
CIE iGCSE Co-ordinated Sciences-P4.2.4 Resistance – Study Notes
CIE iGCSE Co-ordinated Sciences-P4.2.4 Resistance – Study Notes -CIE iGCSE Co-ordinated Sciences – per latest Syllabus.
Key Concepts:
Core
- Recall and use the equation for resistance R=V/I
- Describe an experiment to determine resistance using a voltmeter and an ammeter and do the appropriate calculations
Supplement
- Sketch and explain the current–voltage graph of a resistor of constant resistance
- Recall and 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
CIE iGCSE Co-Ordinated Sciences-Concise Summary Notes- All Topics
Resistance
Resistance is a measure of how much a component opposes the flow of electric current in a circuit.
Equation for Resistance:
$\mathrm{R = \dfrac{V}{I}}$
- \(\mathrm{R}\) = resistance (ohms, Ω)
- \(\mathrm{V}\) = potential difference across the component (volts, V)
- \(\mathrm{I}\) = current through the component (amperes, A)
Explanation:
- The greater the resistance, the smaller the current for the same voltage.
- A resistor limits current in a circuit by converting electrical energy into heat.
- Metals usually have low resistance; insulators have extremely high resistance.
Unit of Resistance:
- The SI unit of resistance is the ohm (Ω).
- 1 Ω means 1 volt of potential difference produces a current of 1 ampere.
Everyday Understanding:
- Thin or long wires have higher resistance than thick, short wires.
- Electric heaters and toasters rely on high-resistance wires to produce heat.
Example :
A lamp has a current of 2 A when connected to a 12 V supply. Calculate its resistance.
▶️ Answer/Explanation
Step 1: Use formula \(\mathrm{R = \dfrac{V}{I}}\).
Step 2: Substitute: \(\mathrm{R = \dfrac{12}{2} = 6 \, \Omega}\).
Final Answer: The resistance of the lamp is 6 Ω.
Aim
To determine the resistance of a given resistor using measurements of potential difference (V) and current (I).
Apparatus
- Resistor under test (unknown R)
- Variable dc supply (0–12 V) or a power supply with rheostat
- Ammeter (suitable range, e.g. 0–1 A) — connected in series
- Voltmeter (suitable range, e.g. 0–10 V or 0–20 V) — connected in parallel across the resistor
- Connecting leads, switch, low-resistance rheostat (optional)
- Meter selection tool (set ranges) and notebook for data
Circuit and Connections (important)
- Connect the ammeter in series with the resistor and supply so the whole circuit current flows through the ammeter.
- Connect the voltmeter in parallel with the resistor only (not across the supply unless measuring source voltage intentionally).
- Polarity: connect the positive terminal of voltmeter to the resistor end connected to the positive supply terminal; ammeter polarity likewise (or use the meter’s markings).
- Use a switch or rheostat to vary the voltage and therefore the current.
Procedure (step-by-step)
- Check meter ranges: start with the highest range for safety (ammeter high range, voltmeter high range).
- Set up the circuit as above. Close switch with supply set to zero volts.
- Increase supply to a small non-zero voltage (e.g. 1.0 V). Wait until readings stabilise. Record V (voltmeter) and I (ammeter).
- Increase supply in steps (e.g. 1.0 V → 2.0 V → 3.0 V …) up to a safe maximum (do not exceed resistor power rating). Record several pairs of V and I (at least 5 different readings over a useful range).
- If readings are too small or too large, change the meter range and repeat the measurement for that step (note the range used).
- Take multiple readings at each step and average them if possible to reduce random error.
- Use the data to calculate resistance by: (a) Single-reading method: \(\;R = \dfrac{V}{I}\) for each pair (take mean). (b) Graph method (recommended): plot V (vertical axis) against I (horizontal axis) and determine the slope \(\dfrac{\Delta V}{\Delta I}\) — slope = R.
Data Table (example format)
V (V) | I (A) | R = V / I (Ω) |
---|---|---|
1.0 | 0.10 | 10.0 |
2.0 | 0.20 | 10.0 |
3.0 | 0.30 | 10.0 |
4.0 | 0.40 | 10.0 |
5.0 | 0.50 | 10.0 |
Worked Example — Single-reading method (step-by-step arithmetic)
Suppose at one step you measured: \(V = 5.00\ \text{V}\) and \(I = 0.50\ \text{A}\).
Use the resistance formula \(\displaystyle R = \frac{V}{I}\).
Compute carefully, digit by digit:
- \(V = 5.00\ \text{V}\)
- \(I = 0.50\ \text{A}\)
- \(\displaystyle R = \frac{5.00}{0.50}.\) Divide 5.00 by 0.50: \(0.50 \times 10 = 5.0\), therefore \(5.00 \div 0.50 = 10.0.\)
- \(\boxed{R = 10.0\ \Omega}\)
Worked Example — Graph (more accurate)
- Plot V (y axis) against I (x axis) using the table data points.
- If the resistor is ohmic, points should lie on a straight line through the origin.
- Pick two well-separated points on the best-fit line, e.g. (I₁ = 0.10 A, V₁ = 1.0 V) and (I₂ = 0.50 A, V₂ = 5.0 V).
- Compute slope = ΔV / ΔI = (V₂ − V₁) / (I₂ − I₁) = (5.0 − 1.0) V / (0.50 − 0.10) A = 4.0 V / 0.40 A.
- Compute: \(4.0 \div 0.40 = 10.0\). So \(\boxed{R = 10.0\ \Omega}\).
Why the graph method is better
- Reduces random errors in individual readings.
- Using slope averages many measurements → more reliable estimate of R.
- Deviations from straight line reveal non-ohmic behaviour or experimental error.
Precautions & Practical Tips
- Always connect the ammeter in series and the voltmeter in parallel (incorrect connection can damage meters).
- Start with high ranges on meters; then reduce for better accuracy if safe.
- Do not exceed the resistor’s power rating: check \(P = V \times I\) for each reading. If \(P\) approaches or exceeds the resistor rating, stop or reduce voltage.
- Allow readings to stabilise before recording (wait a few seconds for meters to settle).
- Take several readings over a range of voltages. Avoid using only one reading—use at least 4–6 points for a good V–I graph.
How to report the result
- State measured resistance with appropriate units and significant figures, e.g. \(R = 10.0\ \Omega\).
- If using multiple readings, give average and a simple uncertainty (e.g. standard deviation or range), e.g. \(R = 10.0 \pm 0.2\ \Omega\).
Current–Voltage Graph of a Resistor (Constant Resistance)
A resistor of constant resistance obeys Ohm’s Law, which states:
$\mathrm{V = I \times R} $
- Here, \(R\) is constant (does not change with voltage or current).
- This means voltage and current are directly proportional.
Graph Features:
- The graph of current (I) against voltage (V) is a straight line through the origin.
- Slope of the line = \(\dfrac{V}{I} = R\) (constant).
- A steeper slope (less gradient) means larger resistance; a shallower slope means smaller resistance.
- The line passes through both positive and negative quadrants → resistor behaves the same way regardless of current direction.
Explanation:
- If the voltage across the resistor is doubled, the current also doubles → direct proportionality.
- This behaviour is valid as long as the resistor’s temperature remains constant (no significant heating).
Example :
A resistor has a constant resistance of 10 Ω. Sketch and describe its I–V graph.
▶️ Answer/Explanation
Step 1: From \(\mathrm{V = IR}\), rearrange: \(\mathrm{I = V / R}\).
Step 2: For R = 10 Ω: – At V = 10 V → I = 1 A. – At V = 20 V → I = 2 A. – At V = 30 V → I = 3 A. Points form a straight line.
Step 3: The line passes through the origin and has a constant slope.
Final Answer: The I–V graph is a straight line through the origin, slope = 1/R.
Resistance of a Metallic Conductor
For a metallic conductor of uniform cross-section (at constant temperature):
$ \mathrm{R \propto \dfrac{L}{A}} $
where:
- (\mathrm{R}) = resistance (Ω)
- (\mathrm{L}) = length of conductor (m)
- (\mathrm{A}) = cross-sectional area (m²)
Effect of Length:
- Resistance is directly proportional to length.
- If the length of the conductor is doubled, its resistance also doubles.
- This is because electrons must travel further and experience more collisions with ions.
Effect of Cross-Sectional Area:
- Resistance is inversely proportional to cross-sectional area.
- If the area is doubled (e.g. using a thicker wire), resistance halves.
- A larger area allows more electrons to flow simultaneously (like a wider pipe allows more water flow).
Combined Expression:
For a uniform wire of resistivity \(\rho\) (material property):
$\mathrm{R = \rho \dfrac{L}{A}} $
- This shows resistance depends on both material (ρ) and dimensions (L and A).
Example :
A copper wire of length 2 m has a resistance of 1 Ω. What will be the resistance if its length is doubled to 4 m and its cross-sectional area is also doubled?
▶️ Answer/Explanation
Step 1: Resistance is proportional to \(\dfrac{L}{A}\).
Step 2: Original ratio = \(\dfrac{2}{A}\). New ratio = \(\dfrac{4}{2A} = \dfrac{2}{A}\).
Step 3: Since the ratio is unchanged, resistance remains the same.
Final Answer: The resistance is still 1 Ω.