recall and use R = ρL / A

Resistance, Resistivity & DC‑Circuit Fundamentals

Overview (Cambridge AS & A‑Level Physics 9702 – 2025‑27)

The syllabus for 10 – DC circuits requires students to be able to:

• Identify and use the practical circuit symbols for ideal and real sources, resistors, ammeters, voltmeters, switches and potentiometers (10.1).
• Explain EMF, internal resistance and the terminal‑potential equation for a real source (10.1).
• Analyse and sketch the V‑I characteristics of a metal wire (Ohmic), a filament lamp (non‑Ohmic) and a semiconductor diode (10.1).
• Apply Kirchhoff’s junction and loop laws to series‑parallel networks (10.2).
• Use the voltage‑divider formula and understand its application with thermistors, LDRs or potentiometers (10.3).
• Quantitatively assess how temperature changes affect resistance (including the temperature‑coefficient formula).
• Recall and manipulate the fundamental relation R = ρ L ⁄ A (AO1‑AO3).

Learning Objectives (AO1‑AO3)

• Recall and use the relationship R = ρ L ⁄ A.
• Identify and apply the symbols for EMF, internal resistance, potential difference and power in practical circuits.
• Analyse V‑I characteristics of a metal wire, a filament lamp and a semiconductor diode, including a quantitative filament‑lamp example.
• Apply Kirchhoff’s junction and loop laws to mixed series‑parallel networks.
• Use the voltage‑divider formula and evaluate its use with thermistors, LDRs or potentiometers.
• Quantitatively assess the effect of temperature on resistance for conductors and filament lamps.
• Solve exam‑style questions involving calculation, rearrangement and interpretation of results (AO2‑AO3).

1. Key Concepts

2. Derivation of R = ρ L ⁄ A

1. Ohm’s law for the whole conductor: V = I R.
2. Microscopic form of Ohm’s law: E = ρ J, where E is the electric field and J = I/A is the current density.
3. Because the field in a uniform cylinder is E = V/L, substitute:

   \[

   \frac{V}{L}= \rho\frac{I}{A}

   \]

4. Re‑arrange:

   \[

   V = I\frac{ρL}{A}

   \]

   Comparing with V = I R gives

   \[

   \boxed{R = \frac{ρL}{A}}.

   \]

3. Units & Typical Values

4. Factors Affecting Resistance

• Material – set by the resistivity ρ.
• Length – R ∝ L (longer → higher resistance).
• Cross‑sectional area – R ∝ 1⁄A (thicker → lower resistance).
• Temperature – for most conductors

  \[

  ρ = ρ{0}\bigl[1+α(T-T{0})\bigr]

  \]

  where α is the temperature coefficient (≈ 3.9 × 10⁻³ K⁻¹ for copper).

  For filament lamps (tungsten) α ≈ 4.5 × 10⁻³ K⁻¹, giving a rapid rise in resistance as the filament heats.

5. Practical DC‑Circuit Concepts (Syllabus 10.1)

5.1 Circuit Symbols (exact syllabus symbols)

5.2 EMF, Internal Resistance & Terminal Potential

For a source delivering current I:

\[

V_{\text{terminal}} = ε - I r

\]

• Open‑circuit (I = 0) → \(V = ε\).
• Short‑circuit (V = 0) → \(I_{\text{sc}} = ε/r\).
• The sign “‑ Ir” is used because the internal‑resistance drop opposes the EMF when the source supplies current.

5.3 V‑I Characteristics (Metal wire, Filament lamp, Diode)

• Metal wire (Ohmic) – straight line through the origin. Slope = R.
• Filament lamp (non‑Ohmic) – low resistance when cold; as the filament heats, ρ increases, giving a curve that flattens.
  

  Quantitative example: A tungsten filament has \(R_{20}=2.0 Ω\) at 20 °C and α = 4.5 × 10⁻³ K⁻¹.

  At 200 °C:

  \[

  R{200}=R{20}\bigl[1+α(200-20)\bigr]

  =2.0\bigl[1+4.5\times10^{-3}\times180\bigr]

  =2.0(1+0.81)=3.62 Ω.

  \]

  The V‑I graph therefore starts with a steep slope (≈2 Ω) and ends with a gentler slope (≈3.6 Ω).

• Semiconductor diode – negligible current until forward‑bias ≈ 0.6 V, then exponential rise; reverse‑bias shows a tiny leakage current until breakdown.

All sketches should be clearly labelled (axes, key points such as “break‑down voltage”, “forward voltage”).

6. Kirchhoff’s Laws (Syllabus 10.2)

6.1 Junction (First) Law

\[

\sum I{\text{in}} = \sum I{\text{out}}\qquad\text{or}\qquad\sum I = 0

\]

6.2 Loop (Second) Law

\[

\sum V{\text{rise}} - \sum V{\text{drop}} = 0

\]

(Include EMF and all IR drops, taking a consistent direction.)

6.3 Worked Example

Find the current through each resistor in the circuit below (mixed series‑parallel).

Given: ε = 12 V, r = 0.5 Ω, R₁ = 2 Ω, R₂ = 3 Ω (parallel with R₃ = 6 Ω).

1. Combine the parallel branch:

   \[

   R{23}= \frac{R{2}R{3}}{R{2}+R_{3}} = \frac{3\times6}{3+6}=2\;\Omega.

   \]

2. Total external resistance:

   \[

   RT = R{1}+R_{23}=2+2=4\;\Omega.

   \]

3. Current from the source (including internal resistance):

   \[

   I = \frac{ε}{r+R_T}= \frac{12}{0.5+4}=2.4\;\text{A}.

   \]

4. Current in the parallel branch splits in the ratio of the opposite resistances:

   \[

   I{2}=I\frac{R{3}}{R{2}+R{3}}=2.4\frac{6}{9}=1.6\;\text{A},

   \qquad

   I{3}=I\frac{R{2}}{R{2}+R{3}}=2.4\frac{3}{9}=0.8\;\text{A}.

   \]

5. Current through R₁ is the total current, 2.4 A.

7. Voltage Dividers (Syllabus 10.3)

When two resistors R₁ and R₂ are in series across a source V_in, the voltage across R₂ is

\[

V{\text{out}} = V{\text{in}}\frac{R{2}}{R{1}+R_{2}}.

\]

Applications

• Measuring an unknown resistance with a galvanometer (balance method).
• Thermistor or LDR as R₂ to obtain a temperature‑ or light‑dependent output voltage.
• Potentiometer used as a variable divider; the null‑method allows comparison of two potentials without drawing current from the circuit.

Example – LDR Voltage Divider

R₁ = 5 kΩ, LDR varies 1 kΩ (bright) → 10 kΩ (dark), V_in = 5 V.

• Bright: \(V_{\text{out}} = 5\times\frac{1}{1+5}=0.83\;V\).
• Dark: \(V_{\text{out}} = 5\times\frac{10}{10+5}=3.33\;V\).

8. Temperature Effect on Resistance (Expanded)

• Conductors (e.g., copper):

  \[

  R = R{0}\bigl[1+α(T-T{0})\bigr]

  \]

  with α ≈ 3.9 × 10⁻³ K⁻¹.

• Filament lamps (tungsten):

  same linear form, but α is larger (≈ 4.5 × 10⁻³ K⁻¹) and the temperature change is huge, so the resistance can increase by a factor of 10–15 when the filament glows.

• Semiconductors (e.g., diodes, thermistors):

  resistance decreases with temperature; for a thermistor the relation is often expressed as

  \(R = R{0}e^{β(1/T-1/T{0})}\) (β = material constant).

9. Example Calculations Using R = ρL/A

1. Copper wire – L = 2.0 m, d = 1.0 mm, ρ = 1.68 × 10⁻⁸ Ω·m.

   \(A = \pi(d/2)^2 = 7.85\times10^{-7}\,\text{m}^2\).

   \(R = \dfrac{1.68\times10^{-8}\times2.0}{7.85\times10^{-7}} \approx 0.043\;\Omega.\)

2. Nickel‑chrome resistor – ρ = 1.10 × 10⁻⁶ Ω·m, L = 5.0 cm, A = 2.0 × 10⁻⁸ m².

   \(R = \dfrac{1.10\times10^{-6}\times0.05}{2.0\times10^{-8}} = 2.75\;\Omega.\)

3. Three‑wire copper bundle (parallel) – each wire: L = 1 m, A = 0.5 mm² = 5.0 × 10⁻⁷ m², ρ = 2.0 × 10⁻⁸ Ω·m.

   Individual \(R = ρL/A = 2.0\times10^{-8}\times1 / 5.0\times10^{-7}=0.04\;\Omega\).

   Parallel total \(R_{\text{eq}} = R/3 = 0.013\;\Omega.\)

10. Common Mistakes & How to Avoid Them

• Confusing ρ and R – ρ is a material constant; R changes with length and area. Always start with ρ → R using the geometry.
• Using diameter instead of area – Convert diameter d to area with \(A = \pi d^{2}/4\) before substituting.
• Neglecting temperature effects – For high‑current circuits calculate the temperature rise and adjust ρ or R with the appropriate α.
• Wrong sign in the terminal‑potential equation – For a delivering source the internal‑resistance drop is subtracted: \(V = ε - I r\). For a charging source the sign reverses.
• Inconsistent current directions in Kirchhoff problems – Choose a direction for each loop and stick to it; opposite directions simply give negative values.
• Assuming all components are Ohmic – Remember that filament lamps and diodes are non‑Ohmic; their V‑I graphs are required for accurate analysis.

11. Practice Questions (Exam‑style)

1. A nichrome wire (ρ = 1.10 × 10⁻⁶ Ω·m) is 5.0 cm long and has a cross‑sectional area of 2.0 × 10⁻⁸ m². Calculate its resistance. (Use R = ρL/A)
2. Two wires of the same material have resistances R₁ = 4.0 Ω and R₂ = 9.0 Ω. If R₁ is twice as long as R₂, find the ratio of their cross‑sectional areas A₁/A₂.
3. Explain how the resistance of a copper conductor changes when its temperature is increased from 20 °C to 70 °C, given α = 3.9 × 10⁻³ K⁻¹. Include a short calculation.
4. A 12 V battery has an EMF of 12.6 V and internal resistance 0.2 Ω. It is connected to a lamp of resistance 6 Ω. Determine:

   • the terminal voltage across the lamp,
   • the current through the circuit, and
   • the power dissipated in the lamp.

5. In the circuit of Section 6.3 (R₁ = 2 Ω, R₂ = 3 Ω, R₃ = 6 Ω, ε = 12 V, r = 0.5 Ω):

   • Calculate the total power supplied by the source.
   • Calculate the power dissipated in each resistor.

6. A voltage divider consists of R₁ = 10 kΩ and a thermistor R₂ that varies from 5 kΩ (cold) to 20 kΩ (hot). The input voltage is 9 V. Determine the output voltage for the two extreme temperatures and comment on its suitability for a temperature‑sensor circuit.
7. For a copper wire of length 1.0 m and diameter 0.5 mm, calculate the resistance at 20 °C and at 80 °C. Use ρ₂₀ = 1.68 × 10⁻⁸ Ω·m and α = 3.9 × 10⁻³ K⁻¹.
8. Quantitative filament‑lamp question: A tungsten filament has R = 2.0 Ω at 20 °C and α = 4.5 × 10⁻³ K⁻¹. What is its resistance when the filament temperature reaches 250 °C? How does this affect the slope of its V‑I graph?