Axiom Physics | Class 12 - Resistivity, Conductivity & Temperature Dependence

⚡ 01. The Spark (Analogy)

Socho ek hallway me log chal rahe hain aur floor par stationary pillars (ions) lage hain. Jab temperature badhta hai, toh yeh pillars apni jagah par violent vibration shuru kar dete hain. Ab kisi person (electron) ka bina takraye nikalna bohot mushkil ho jata hai—takkar ka samay interval ($\tau$) chhota ho jata hai, jisse Metals ki Resistance badh jati hai. Lekin Semiconductors me kahani ulati hai: wahan temperature badhne se bandhe hue log (covalent electrons) azaad hokar daudne lagte hain ($n$ exponentially badh jata hai), isliye wahan temperature badhne par Resistance ghat jati hai!

📐 02. The Core Law & Formulae

Resistivity ($\rho$) & Conductivity ($\sigma$)

Microscopic parameters determining material-specific electrical transport:

$$\rho = \frac{m}{n e^2 \tau}, \quad \sigma = \frac{1}{\rho} = \frac{n e^2 \tau}{m}$$

SI Units: Resistivity $\rho \to \Omega\cdot\text{m}$; Conductivity $\sigma \to \Omega^{-1}\cdot\text{m}^{-1} = \text{S}\cdot\text{m}^{-1}$ (Siemens per metre).

Temperature Dependence Formula

Linear approximation for moderate temperature range ($T - T_0$):

$$\rho_T = \rho_0 [1 + \alpha(T - T_0)], \quad R_T = R_0 [1 + \alpha(T - T_0)]$$

Where $\alpha$ is the temperature coefficient of resistance ($\text{K}^{-1}$ or $^\circ\text{C}^{-1}$): $\alpha = \frac{\Delta R}{R_0 \Delta T}$.

03. Interactive Visualizer: Lattice Vibration & $\rho\text{ vs }T$ Curve Engine

Operating Temperature ($T$) 300 K (27°C)

Temperature Coefficient ($\alpha$) +0.0039 /°C

Positive coefficient: $\rho$ increases with $T$[cite: 13]

Carrier Density ($n$) 8.5 × 10²⁸ m⁻³

Constant $n$; resistance governed by $\tau$[cite: 13]

Calculated Resistivity ($\rho_T$):

1.70 × 10⁻⁸ Ω·m

Mean Collision Time ($\tau$):

2.45 × 10⁻¹⁴ s

Conductivity ($\sigma = 1/\rho$):

5.88 × 10⁷ S/m

📝 04. Stepwise Derivations Blueprint[cite: 13]

Derivation A

Microscopic Origin of Resistivity ($\rho = \frac{m}{ne^2\tau}$)[cite: 13]

1. Consider a conductor carrying current density $J = \frac{I}{A}$ under electric field $E$[cite: 13].

2. Transport current relation: $I = n e A v_d \implies J = n e v_d$[cite: 13].

3. Substitute the expression for drift speed $v_d = \frac{e E \tau}{m}$[cite: 13]:

$$J = n e \left( \frac{e E \tau}{m} \right) = \left( \frac{n e^2 \tau}{m} \right) E$$

4. Comparing this with microscopic Ohm's Law in vector form $\vec{J} = \sigma \vec{E}$[cite: 13]:

$$\sigma = \frac{n e^2 \tau}{m}$$

5. Since resistivity is reciprocal of conductivity ($\rho = \frac{1}{\sigma}$)[cite: 13]:

$$\rho = \frac{m}{n e^2 \tau} \text{[cite: 13]}$$

Core Deduction: Resistivity depends inversely on two key variables: free carrier density $n$ and relaxation time $\tau$[cite: 13]. Electron mass $m$ and charge $e$ are fundamental constants[cite: 13].

Derivation B

Derivation of Temperature Coefficient of Resistance ($\alpha$)

1. Let resistance at reference temperature $T_0$ (usually $0^\circ\text{C}$ or $20^\circ\text{C}$) be $R_0$, and at temperature $T$ be $R_T$[cite: 13].

2. Experimentally, change in resistance $\Delta R = R_T - R_0$ is directly proportional to initial resistance $R_0$ and temperature rise $\Delta T = T - T_0$:

$$\Delta R \propto R_0 \Delta T \implies \Delta R = \alpha R_0 \Delta T$$

3. Where proportionality constant $\alpha$ is the fractional change in resistance per unit change in temperature:

$$\alpha = \frac{\Delta R}{R_0 \Delta T} = \frac{R_T - R_0}{R_0 (T - T_0)}$$

4. Rearranging gives the standard temperature variation equation[cite: 13]:

$$R_T = R_0 [1 + \alpha(T - T_0)]$$

⚠️ 05. Red Alert Exam Trap: Materials Comparison Matrix

Material Category	Sign & Magnitude of $\alpha$	Dominant Microscopic Factor	Board Exam Favorite Question
Metals (Copper, Silver)[cite: 13]	Positive ($\alpha > 0$)[cite: 13] $\approx +4 \times 10^{-3}\text{ K}^{-1}$	$n$ is constant. Increasing $T$ increases thermal vibration of ions, sharply reducing $\tau$ ($\rho \propto 1/\tau$)[cite: 13].	Why does copper have non-linear $\rho\text{ vs }T$ near $0\text{ K}$? (Lattice quantum oscillations)[cite: 13].
Alloys (Nichrome, Manganin, Constantan)[cite: 13]	Nearly ZERO ($\alpha \approx 0$) $\approx 10^{-5}\text{ K}^{-1}$	High disorder causes short $\tau$ even at $0\text{ K}$; heating adds almost no additional collision impedance[cite: 13].	Why are standard resistance coils made of Manganin/Constantan? (Resistance stays constant with temperature)[cite: 13].
Semiconductors (Silicon, Germanium)[cite: 13]	NEGATIVE ($\alpha < 0$)[cite: 13] Thermistor response	Thermal generation of electron-hole pairs causes $n$ to increase exponentially ($n \propto e^{-E_g/2k_BT}$), completely overtaking the decrease in $\tau$[cite: 13].	What happens to resistance of Si and Cu when cooled from $300\text{ K}$ to $77\text{ K}$? (Cu drops; Si rises towards $\infty$)[cite: 13]!

💡 06. Handcrafted Exemplars

Board Classic Level

Problem: The resistance of a platinum wire of a platinum resistance thermometer at the ice point ($0^\circ\text{C}$) is $5.00\text{ }\Omega$ and at steam point ($100^\circ\text{C}$) is $5.39\text{ }\Omega$. When the thermometer is inserted in a hot bath, the resistance becomes $5.795\text{ }\Omega$. Calculate the temperature of the bath[cite: 13].

Solution:

1. Given: $R_0 = 5.00\text{ }\Omega$, $R_{100} = 5.39\text{ }\Omega$, $R_t = 5.795\text{ }\Omega$[cite: 13].

2. Find temperature coefficient $\alpha$ from steam point data[cite: 13]:

$$R_{100} = R_0(1 + \alpha \cdot 100) \implies \alpha = \frac{R_{100} - R_0}{100 \times R_0} = \frac{5.39 - 5.00}{100 \times 5.00} = \frac{0.39}{500} = 7.8 \times 10^{-4}\text{ }^\circ\text{C}^{-1}$$

3. Formula for unknown bath temperature $t$[cite: 13]:

$$t = \frac{R_t - R_0}{R_{100} - R_0} \times 100 = \frac{5.795 - 5.00}{5.39 - 5.00} \times 100$$

$$t = \frac{0.795}{0.39} \times 100 \approx 345.65^\circ\text{C}$$

JEE/NEET Competitive Trap

Problem: A carbon resistor and a nichrome wire resistor are connected in series. Find the ratio of their resistances at $0^\circ\text{C}$ such that their combined equivalent resistance remains completely independent of temperature over a small range. (Given: $\alpha_{\text{carbon}} = -0.0005\text{ }^\circ\text{C}^{-1}$, $\alpha_{\text{nichrome}} = +0.0001\text{ }^\circ\text{C}^{-1}$).

Solution:

1. Equivalent series resistance at temperature $t$: $R_{\text{eq}}(t) = R_1(t) + R_2(t)$[cite: 13].

$$R_{\text{eq}}(t) = R_{01}(1 + \alpha_1 t) + R_{02}(1 + \alpha_2 t) = (R_{01} + R_{02}) + (R_{01}\alpha_1 + R_{02}\alpha_2)t$$

2. For $R_{\text{eq}}$ to be independent of temperature $t$, the coefficient of $t$ must vanish identically:

$$R_{01}\alpha_1 + R_{02}\alpha_2 = 0 \implies \frac{R_{01}}{R_{02}} = -\frac{\alpha_2}{\alpha_1}$$

3. Substitute values of $\alpha$:

$$\frac{R_{0,\text{carbon}}}{R_{0,\text{nichrome}}} = -\frac{+0.0001}{-0.0005} = \frac{1}{5}$$

Exam Trap Note: This condition works strictly because carbon has a negative temperature coefficient ($\alpha < 0$)[cite: 13], which compensates the positive temperature coefficient of nichrome.

🗂️ 07. Quick Revision Flashcard Core Formulae

Resistivity Formula

$\rho = \frac{m}{n e^2 \tau}$[cite: 13]

Metals ($T \uparrow$)[cite: 13]

$\tau \downarrow \implies \rho \uparrow$ ($\alpha > 0$)[cite: 13]

Semiconductors ($T \uparrow$)[cite: 13]

$n \uparrow\uparrow \implies \rho \downarrow$ ($\alpha < 0$)[cite: 13]

Standard Alloys

$\alpha \approx 0$ (Manganin)[cite: 13]