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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)

  1. Check meter ranges: start with the highest range for safety (ammeter high range, voltmeter high range).
  2. Set up the circuit as above. Close switch with supply set to zero volts.
  3. Increase supply to a small non-zero voltage (e.g. 1.0 V). Wait until readings stabilise. Record V (voltmeter) and I (ammeter).
  4. 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).
  5. If readings are too small or too large, change the meter range and repeat the measurement for that step (note the range used).
  6. Take multiple readings at each step and average them if possible to reduce random error.
  7. 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)
1.00.1010.0
2.00.2010.0
3.00.3010.0
4.00.4010.0
5.00.5010.0

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.

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