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AXIOM PHYSICS Class 12 • Open Access

CBSE Class XII • Unit 02 (Current Electricity)[cite: 12, 13]

Chapter 03: Current Electricity

Master Revision Dossier • Transport Equations, Microscopic Ohm's Law, Cell Matrices & Bridge Laws

1. Definitive Formula & Physical Transport Matrix

Concept Mathematical Formulation SI Unit / Dim Boundary Conditions & Notes
Electric Current ($I$)[cite: 13] $$I = \frac{dQ}{dt} = n e A v_d\text{[cite: 13]}$$ $\text{Ampere (A)}$
$[\text{A}]$		 Scalar quantity (obeys algebraic addition, not vector law)[cite: 13]. Current density $\vec{J} = (I/A)\hat{n}$ is a true vector[cite: 13].
Drift Velocity ($v_d$)[cite: 13] $$\vec{v}_d = -\frac{e\vec{E}}{m}\tau \implies v_d = \frac{eE\tau}{m}\text{[cite: 13]}$$ $\text{m/s}$
$[\text{LT}^{-1}]$		 Net average velocity acquired against field $\vec{E}$[cite: 13]; order of magnitude $\approx 10^{-4}\text{ m/s}$ ($0.1\text{ mm/s}$)[cite: 13].
Mobility ($\mu$)[cite: 13] $$\mu = \frac{|\vec{v}_d|}{E} = \frac{e\tau}{m}\text{[cite: 13]}$$ $\text{m}^2/\text{V}\cdot\text{s}$
$[\text{M}^{-1}\text{T}^2\text{A}]$		 Defined as positive scalar[cite: 13]. Electron mobility $\mu_e$ is significantly higher than hole mobility $\mu_h$ ($m_e < m_h$).
Resistivity ($\rho$) & $\sigma$[cite: 13] $$\rho = \frac{m}{n e^2 \tau}, \quad \sigma = \frac{1}{\rho} = \frac{n e^2 \tau}{m}\text{[cite: 13]}$$ $\rho: \Omega\cdot\text{m}$
$\sigma: \text{S}\cdot\text{m}^{-1}$		 Independent of conductor dimensions ($l, A$)[cite: 13]; depends strictly on material and temperature[cite: 13].
Microscopic Ohm's Law[cite: 13] $$\vec{J} = \sigma \vec{E} = \frac{\vec{E}}{\rho}\text{[cite: 13]}$$ $\vec{J}: \text{A/m}^2$ Universal vector form[cite: 13]. Macroscopic scalar equivalent: $V = IR$, where $R = \rho \frac{l}{A}$[cite: 13].
Temperature Dependence[cite: 13] $$R_T = R_0[1 + \alpha(T - T_0)]\text{[cite: 13]}$$ $\alpha: \text{K}^{-1} \text{ or } ^\circ\text{C}^{-1}$ Metals: $\alpha > 0$[cite: 13]; Semiconductors: $\alpha < 0$[cite: 13]; Standard Alloys (Manganin, Nichrome): $\alpha \approx 0$[cite: 13].
Cell Terminal Voltage[cite: 13] $$V = \mathcal{E} - Ir \quad (\text{Discharge})\text{[cite: 13]}$$ $\text{Volt (V)}$ Open circuit: $V = \mathcal{E}$[cite: 13]. During battery recharge: $V = \mathcal{E} + Ir$. Internal resistance $r = (\frac{\mathcal{E}}{V}-1)R$[cite: 13].
Cells in Series[cite: 13] $$\mathcal{E}_{\text{eq}} = \sum \mathcal{E}_i, \quad r_{\text{eq}} = \sum r_i\text{[cite: 13]}$$ $\text{V}, \Omega$ If $m$ out of $n$ cells are reversed: $\mathcal{E}_{\text{eq}} = (n - 2m)\mathcal{E}$, while $r_{\text{eq}} = nr$ always.
Cells in Parallel[cite: 13] $$\mathcal{E}_{\text{eq}} = \frac{\sum \mathcal{E}_i/r_i}{\sum 1/r_i}, \quad \frac{1}{r_{\text{eq}}} = \sum \frac{1}{r_i}\text{[cite: 13]}$$ $\text{V}, \Omega$ For two cells: $\mathcal{E}_{\text{eq}} = \frac{\mathcal{E}_1 r_2 + \mathcal{E}_2 r_1}{r_1 + r_2}$, $r_{\text{eq}} = \frac{r_1 r_2}{r_1 + r_2}$[cite: 13].
Wheatstone Bridge Balance[cite: 13] $$\frac{P}{Q} = \frac{R}{S} \iff I_g = 0\text{[cite: 13]}$$ Dimensionless Galvanometer current is zero ($V_B = V_D$)[cite: 13]. Bridge retains balance if battery and detector are interchanged.

2. Core Derivations Blueprint (Step-by-Step Board Proofs)[cite: 13]

Blueprint 3.1: Relation between Electric Current and Drift Velocity ($I = n e A v_d$)[cite: 13]

1. Consider a uniform conductor of cross-sectional area $A$, length $l$, and free electron density $n$[cite: 13].

2. Under electric field $\vec{E}$, electrons drift with average speed $v_d$[cite: 13]. In time interval $\Delta t$, distance covered is $\Delta x = v_d \Delta t$[cite: 13].

3. Total volume of electrons crossing the plane in time $\Delta t$[cite: 13]: $\Delta V = A \cdot \Delta x = A (v_d \Delta t)$.

4. Total charge transported across the area[cite: 13]:

$$\Delta Q = (\text{Total Electrons}) \times e = (n \cdot \Delta V) e = (n A v_d \Delta t) e$$

5. Dividing by time interval $\Delta t$ yields the fundamental relation[cite: 13]:

$$I = \frac{\Delta Q}{\Delta t} = n e A v_d \implies \vec{J} = \frac{I}{A} = n e \vec{v}_d\text{[cite: 13]}$$

Blueprint 3.2: Microscopic Deduction of Ohm's Law and Resistivity ($V = IR$, $\rho = \frac{m}{ne^2\tau}$)[cite: 13]

1. Drift velocity expression under acceleration $\vec{a} = -\frac{e\vec{E}}{m}$ with mean relaxation time $\tau$[cite: 13]: $v_d = \frac{e E \tau}{m}$.

2. Substitute $v_d$ into the transport current equation $I = n e A v_d$[cite: 13]:

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

3. In a conductor of length $l$ across potential difference $V$, uniform electric field is $E = \frac{V}{l}$[cite: 13]:

$$I = \left(\frac{n e^2 \tau A}{m l}\right) V \implies V = \left(\frac{m}{n e^2 \tau} \frac{l}{A}\right) I$$

4. At constant physical temperature, the bracketed term is a constant defined as Resistance $R$[cite: 13]:

$$V = R I \quad \text{where } R = \rho \frac{l}{A} \quad \text{and} \quad \rho = \frac{m}{n e^2 \tau}\text{[cite: 13]}$$

Blueprint 3.3: Balanced Wheatstone Bridge Condition via Kirchhoff's Rules ($P/Q = R/S$)[cite: 13]

1. Apply KVL to Loop 1 ($ABDA$) containing resistors $P, R$ and galvanometer $G$:

$$-I_1 P - I_g G + I_2 R = 0 \implies I_1 P + I_g G = I_2 R \quad \text{--- [Eq. 1]}$$

2. Apply KVL to Loop 2 ($BCDB$) containing resistors $Q, S$ and galvanometer $G$:

$$-(I_1 - I_g)Q + (I_2 + I_g)S + I_g G = 0 \quad \text{--- [Eq. 2]}$$

3. At balanced condition, galvanometer current is zero: $I_g = 0$ ($V_B = V_D$)[cite: 13].

$$\text{From Eq. 1: } I_1 P = I_2 R \qquad \text{From Eq. 2: } I_1 Q = I_2 S$$

4. Dividing the two resulting equations yields the balancing condition[cite: 13]:

$$\frac{I_1 P}{I_1 Q} = \frac{I_2 R}{I_2 S} \implies \frac{P}{Q} = \frac{R}{S}\text{[cite: 13]}$$

3. High-Yield "Red Alert" Exam Trap Matrix

Master Trap 1: Temperature Dependence & Conduction Mechanisms[cite: 13]

Class of Material Thermal Coeff ($\alpha$) Microscopic Governing Factor Exam Trap Point
Metals (Cu, Al)[cite: 13] Positive ($\alpha > 0$)[cite: 13] $n$ is fixed; lattice vibrations shorten relaxation time $\tau$ ($\rho \propto 1/\tau$)[cite: 13]. When cooled to $0\text{ K}$, resistivity approaches near zero (superconductivity).
Semiconductors (Si, Ge)[cite: 13] Negative ($\alpha < 0$)[cite: 13] Thermal generation makes $n$ surge exponentially, dominating any drop in $\tau$[cite: 13]. At $0\text{ K}$, semiconductors behave as perfect insulators ($R \to \infty$).
Alloys (Manganin, Constantan)[cite: 13] Nearly Zero ($\alpha \approx 0$)[cite: 13] High structural disorder causes extreme scattering regardless of temperature[cite: 13]. Chosen exclusively for standard resistance boxes and bridge wires[cite: 13].

⚠️ Trap 2: Terminal Voltage vs EMF Boundary States[cite: 13]

• Discharging (Delivering power): $V = \mathcal{E} - Ir < \mathcal{E}$[cite: 13]
• Open Circuit ($I = 0$): $V = \mathcal{E}$[cite: 13]
• Charging (Battery Charger active): $\mathbf{V = \mathcal{E} + Ir > \mathcal{E}}$
• Short Circuit ($R = 0$): $V = 0$, $I_{\text{sc}} = \mathcal{E}/r$[cite: 13]

⚠️ Trap 3: Wire Stretching Rules ($R \propto l^2$)

Volume $V = A \cdot l$ is conserved during stretching[cite: 13]. Thus, $A \propto 1/l \implies R = \rho \frac{l}{A} \propto l^2 \propto \frac{1}{r^4}$.
• For small fractional changes ($< 5\%$): $\frac{\Delta R}{R} \approx 2\left(\frac{\Delta l}{l}\right)$.
• If length is doubled ($l' = 2l$): $R' = 4R$ ($\mathbf{300\%}$ increase, NOT $100\%$)!

4. Handcrafted Exemplars (Analytical Numerical Solutions)

Problem 1: Platinum Thermometer Temperature Calibration[cite: 13] Exemplar 01

Resistance of a platinum thermometer is $5.00\text{ }\Omega$ at ice point ($0^\circ\text{C}$) and $5.39\text{ }\Omega$ at steam point ($100^\circ\text{C}$)[cite: 13]. When placed in a hot bath, resistance is $5.795\text{ }\Omega$. Determine bath temperature[cite: 13].

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

$$t = \frac{R_t - R_0}{R_{100} - R_0} \times 100^\circ\text{C} = \frac{5.795 - 5.00}{5.39 - 5.00} \times 100 = \frac{0.795}{0.39} \times 100$$
$$t = 2.03846 \times 100 \approx 345.65^\circ\text{C}$$
Problem 2: Non-Identical Cells in Parallel with Back-Charging[cite: 13] Exemplar 02

Two cells $\mathcal{E}_1 = 2\text{ V}, r_1 = 1\text{ }\Omega$ and $\mathcal{E}_2 = 1\text{ V}, r_2 = 2\text{ }\Omega$ are connected in parallel and feed an external resistor $R = 10\text{ }\Omega$[cite: 13]. Find terminal voltage and branch currents[cite: 13].

Solution:

1. Equivalent cell parameters[cite: 13]:

$$r_{\text{eq}} = \frac{1 \times 2}{1 + 2} = \frac{2}{3}\text{ }\Omega, \quad \mathcal{E}_{\text{eq}} = \frac{\frac{2}{1} + \frac{1}{2}}{\frac{1}{1} + \frac{1}{2}} = \frac{2.5}{1.5} = \frac{5}{3}\text{ V}$$

2. Total load current and terminal potential difference[cite: 13]:

$$I = \frac{\mathcal{E}_{\text{eq}}}{R + r_{\text{eq}}} = \frac{5/3}{10 + 2/3} = \frac{5}{32}\text{ A} \implies V = IR = \frac{50}{32} = 1.5625\text{ V}$$

3. Individual branch currents: $I_1 = \frac{2 - 1.5625}{1} = +0.4375\text{ A}$ (discharging), $I_2 = \frac{1 - 1.5625}{2} = -0.28125\text{ A}$ (charging!).

$$V = 1.5625\text{ Volts}, \quad I_1 = 0.438\text{ A (Discharge)}, \quad I_2 = -0.281\text{ A (Recharging)}$$

5. Printable Pocket Flashcard Grid

Transport Current

$I = n e A v_d$[cite: 13]

Resistivity Formula

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

Discharging Cell

$V = \mathcal{E} - Ir$[cite: 13]

Wheatstone Null

$\frac{P}{Q} = \frac{R}{S} \iff I_g=0$[cite: 13]