Electrical Resistance – Lecture 2 Comprehensive Notes

Introduction to Electrical Resistance

• In conductive materials, free electrons act as charge carriers.

• Application of an external voltage causes electrons to drift, but

  • Electrons continually collide with lattice atoms/other electrons.

  • These collisions convert part of the electrical energy into heat, an effect analogous to mechanical friction.

• The cumulative opposition to the flow of charge created by these collisions is called electrical resistance (R).

  • Greater opposition ⇒ smaller current for the same applied voltage (per $I = \dfrac{V}{R}$).

• Understanding resistance is critical for:

  • Safe conductor sizing (heat dissipation, fire prevention).

  • Energy efficiency (minimising $I^{2}R$ losses in power systems).

  • Component design (e.g.
    resistors for biasing, voltage dividers).

Definition & Physical Basis of Resistance

• Resistance (R):

  • "The property of a substance that opposes/restricts the flow of electric current through it."

• Mechanism:

  • Electron–lattice friction ⇢ heat generation.

  • Comparable to fluid friction in pipes or mechanical friction between solids.

• Classification of materials by ease of electron flow:

  • Good conductors: metals (Ag > Cu > Al), acids, salt solutions.

  • Poor conductors / Insulators: bakelite, mica, glass, rubber, PVC, dry wood, paper, cotton, mineral oil, ceramics.

  • Conductor quality linked to number of free / loosely-bound electrons in atomic structure.

• Ethical / practical impact:

  • Choosing proper insulation prevents shock hazards.

  • Material selection balances conductivity, cost, weight, corrosion, sustainability.

Unit of Resistance & Symbols

• Symbol: R.

• SI Unit: ohm (Ω), named after Georg Simon Ohm.

  • Definition: A conductor has $1\,\Omega$ of resistance if a current of $1\,A$ flows when a voltage of $1\,V$ is applied across it.

• Practical device: Resistor

  • Two-terminal passive component designed to possess a predetermined R.

  • Circuit symbol shown in Fig.
    2.1 (zig-zag or rectangular box depending on IEC/ANSI).

The Law of Resistance (Dependence on Length, Area, Material & Temperature)

1. Dependence on Length (L)

• $R \propto L$ (directly proportional).

  • Doubling length doubles collision path ⇒ doubles R.

• L measured in metres (m) or feet (ft).

2. Dependence on Cross-Sectional Area (A)

• $R \propto \dfrac{1}{A}$ (inversely proportional).

  • Larger area offers more pathways (like a wider pipe for water).

  • Doubling A halves effective collisions ⇒ halves R.

• A measured in square metres (m²) or circular mils (CM) (imperial).

Circular Mil (CM) System

• mil = $1/1000$ inch.

• 1 CM: Area of a circle with $d = 1$ mil.

  • $1\,\text{CM} = \dfrac{\pi}{4}$ square mils.

• Conversions and formulae:

  • $1\,\text{sq mil} = \dfrac{4}{\pi}\,\text{CM}$.

  • For wire diameter $n$ mils: $A<em>{sq\,mil} = \dfrac{\pi}{4}n^{2}$ ; $A</em>{CM} = n^{2}$.

  • If $d$ in mils ⟹ $A_{CM} = d^{2}$ (handy mental rule).

3. Dependence on Material (Resistivity – ρ)

• Different atomic structures alter collision severity.

  • Example: Silver ⇒ more free electrons than copper ⇒ lower R for same dimensions.

• Factor introduced via resistivity (ρ), a material constant.

  • Units: $\Omega·m$ (SI) or $\Omega·\text{cmil}/\text{ft}$ (imperial).

  • High ρ ⇒ good insulator; low ρ ⇒ good conductor.

• Typical ρ values at $20\,^{\circ}!\mathrm{C}$ (Table 2.1)

  • Silver: $1.63×10^{-8}\,\Omega·m$ (best conductor).

  • Copper (annealed): $1.72×10^{-8}\,\Omega·m$.

  • Aluminum: $2.83×10^{-8}\,\Omega·m$.

  • Glass, mica, etc.
    > $10^{10} – 10^{14}\,\Omega·m$ (insulators).

4. Dependence on Temperature

• Rising T ⇒ more lattice vibrations & extra free electrons ⇒ more collisions ⇒ R increases almost linearly for metals.

• Below very low T metals can reach superconductivity (R≈0) – not covered here but underpins cryogenic tech & quantum computing ethics.

Combined Mathematical Formulation (Eq.

2.1)

$R = \rho\,\dfrac{L}{A}$

• Variables:

  • $\rho$: resistivity.

  • $L$: length.

  • $A$: cross-sectional area $= \pi r^{2} = \dfrac{\pi}{4}d^{2}$ for circular conductors.

Significance:

• Forms basis for conductor sizing, fault-current calculations, thermal analysis.

• Used in manufacturing quality control (verify alloy purity via measured ρ).

Worked Examples on Geometry & Material

1. Ex 2.1 – CM area from diameter

   • $d = 0.0159\,\text{in} = 15.9\,\text{mil}$

   • $A_{CM} = d^{2} = (15.9)^{2} = 252.81\,CM$.

2. Ex 2.2 – R of 1 km Al conductor

   • $d = 0.5\,cm = 0.005\,m$.

   • $A = \dfrac{\pi}{4}d^{2}$.

   • $\rho_{Al} = 2.83×10^{-8}\,\Omega·m$.

   • $R = \rho L/A = 1.4413\,\Omega$.

3. Ex 2.3 – 1.5 mile Cu (imperial units)

   • Use $\rho = 10.37\,\Omega·CM/ft$, $d = 0.1\,in = 100\,mil$.

   • $A_{CM} = 10{,}000\,CM$.

   • $L = 1.5\,mi = 7920\,ft$.

   • $R \approx 8.213\,\Omega$.

4. Ex 2.4 – 7-strand hard-drawn Cu

   • Each strand $d=0.1\,cm$ ⇒ $A_{strand} = \dfrac{\pi}{4}d^{2}$.

   • Total area $= 7A_{strand}$.

   • $\rho = 1.77×10^{-8}\,\Omega·m$.

   • $L = 800\,m$.

   • $R \approx 2.576\,\Omega$.

5. Ex 2.5 – Equivalent Al to replace Cu

   • Given $A_{Cu}=500\,MCM$, same $R$ & $L$.

   • $A<em>{Al} = A</em>{Cu}\dfrac{\rho<em>{Al}}{\rho</em>{Cu}} = 500\,\text{MCM}\times\dfrac{17}{10.37} = 819.7\,MCM$.

Effect of Drawing / Die-Casting (Constant Volume)

• When a wire is drawn, its volume (V) stays constant, but $L$ increases & $d$/$A$ decrease.

• Algebraic consequences:

  • $V = A L \Longrightarrow A = V/L$.

  • Substitute into $R=\rho L/A$ ⇒ $R \propto L^{2}$ for constant V.

  • Alternatively, $R\propto 1/d^{4}$ (since $A \propto d^{2}$).

• Real-world relevance:

  • Predicting winding resistances after manufacturing stages.

  • Quality assessment of drawn conductors.

Example 2.6 – Drawn Wire Resistance

• Original: $L<em>{1}=6\,km,\ d</em>{1}=11.7\,mm,\ R_{1}=0.031\,\Omega$.

• Drawn: $d_{2}=5\,mm$.

• $R<em>{2} = R</em>{1}\left(\dfrac{d<em>{1}}{d</em>{2}}\right)^{4} = 0.031\left(\dfrac{11.7}{5}\right)^{4} \approx 0.929\,\Omega$.

Temperature Dependence – Graphical & Algebraic Treatment

• Linear region (metals): $R = mT + b$.

• From similar triangles in Fig 2.2:

  • $\dfrac{R<em>{2}}{T + T</em>{2}} = \dfrac{R<em>{1}}{T + T</em>{1}}$.

  • Rearranged (Eq 2.2): $R<em>{2} = R</em>{1} \dfrac{T + T<em>{2}}{T + T</em>{1}}$ where $T = \text{temperature intercept (absolute)}$.

• More convenient form uses temperature coefficient α:

  • Slope $m = \dfrac{\Delta R}{\Delta T}$, $\alpha = \dfrac{m}{R_{1}}$.

  • Derived (Eq 2.3): $R<em>{2} = R</em>{1}\big[1 + \alpha<em>{1}(T</em>{2} - T_{1})\big]$.

  • Special case from 0 °C: $R<em>{T} = R</em>{0}\big[1 + \alpha_{0}T\big]$.

• Units of $\alpha$: $(^{\circ}C)^{-1}$.

• Ethical & practical notes:

  • Thermal runaway in resistors (positive α) vs.
    carbon/thermistor (negative α) sensor design.

  • Accurate temperature compensation vital for precision instrumentation.

Worked Temperature Examples

1. Ex 2.7 – Al wire at 35 °C

   • Compute $R_{1}$ at 20 °C as before (1.4413 Ω).

   • Use Eq 2.2 with tabulated T intercept (-228 °C for Al): $R_{35} ≈ 1.5285\,Ω$.

2. Ex 2.8 – 7-strand Cu at 40 °C

   • $R_{1}=2.5756\,Ω$ at 20 °C.

   • T intercept for hard-drawn Cu = −241 °C.

   • $R_{40} ≈ 2.773\,Ω$.

3. Ex 2.9 – Coil 100 Ω at 0 °C → 70 °C

   • $\alpha_{0,Copper}=0.0043/^{\circ}C$.

   • $R_{70}=100[1+0.0043×70] = 130.1\,Ω$.

4. Ex 2.10 – 10 Ω at 20 °C → 100 °C

   • $\alpha_{20}=0.004/^{\circ}C$.

   • $R_{100}=10[1+0.004(80)] = 13.2\,Ω$.

Self-Test / Exercises (Selected Answers)

1. 7-strand Cu, 1.5 km, d=0.15 cm ⇒ $R=2.304\,Ω$.

2. Same material & length, $A<em>{1}=2\,mm^{2}$, $R</em>{1}=300\,Ω$.

   • $R<em>{2}=R</em>{1}\dfrac{A<em>{1}}{A</em>{2}}=300\times\dfrac{2}{5}=120\,Ω$.

3. Drawn wire length ×2.5 ⇒ $R\propto L^{2} ⇒ R_{2}=0.8×2.5^{2}=5\,Ω$.

4. Find T when 7-strand Cu reaches 3 Ω ⇒ $T≈96.9^{\circ}C$.

5. Coil R: 18 Ω at 20 °C & 20 Ω at 50 °C.

   • Determine α, then R=21 Ω ⇒ temperature rise 50 °C above ambient 15 °C.

6. Field winding 120 Ω at 15 °C to 140 Ω.

   • With $\alpha_{0}=0.004$ ⇒ $T≈59.2^{\circ}C$.

7. 7-strand Al, 1.2 mi, R=2 Ω @35 °C ⇒ required strand diameter $\approx0.0903\,in$.

Connections, Significance & Broader Context

• Joule Heating: $P=I^{2}R$ emphasises why lower R conductors are preferred for power transmission (efficiency & climate-impact reduction).

• Material Trade-Offs: Aluminum vs.
  Copper debates (cost, weight, resistance) in grid infrastructure.

• Temperature Coefficient Use-Cases

  • PTC thermistors for circuit protection.

  • NTC thermistors for temperature sensing & inrush-current limiting.

• Nanotechnology & Superconductors: Research aims at materials with near-zero ρ at room T, raising ethical issues about resource allocation & disruptive economic effects.

• Safety Standards: Understanding R essential for IEC/NFPA codes—incorrect calculations can cause overheating, fires, or electrocution.

Key Formulae Quick Reference

• $R = \rho\dfrac{L}{A}$.

• $A<em>{CM} = d</em>{mil}^{2}$.

• $R<em>{2}=R</em>{1}\left(\dfrac{d<em>{1}}{d</em>{2}}\right)^{4}$ (constant volume wire drawing).

• Linear temperature model:

  • $R<em>{2}=R</em>{1}\left[1+\alpha<em>{1}(T</em>{2}-T_{1})\right]$.

  • From 0 °C: $R<em>{T}=R</em>{0}(1+\alpha_{0}T)$.

• $\alpha = \dfrac{R<em>{2}-R</em>{1}}{R<em>{1}(T</em>{2}-T_{1})}$.

Study Tips

• Memorise proportionalities: $R\propto L$, $R\propto 1/A$, $R\propto \rho$.

• Switching unit systems (SI ↔ imperial) often causes errors—track units scrupulously.

• Draw resistor-temperature graphs to internalise linear behaviour & intercept concept.

• Practise CM calculations; they are common in North-American design problems.

• Re-work examples with different numbers to reinforce algebraic manipulation.

• Relate formulas to lab experiments—measure resistance vs.
  wire length and validate $R\propto L$.