Current Electricity & Network Theorems: Master Notes

Chapter Overview

Current electricity bridges microscopic electron transport and macroscopic circuit theory. In IIT-JEE (Advanced), the most crucial tools are nodal analysis, symmetry reduction in 3D resistor grids (Cube Networks), Potentiometer sensitivity balancing, and galvanometer modifications.

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1. Microscopic Conduction & Ohm's Law

$$\overrightarrow{J}=\sigma \overrightarrow{E}=\frac{\overrightarrow{E}}{\rho }=nq{\overrightarrow{v}}_d$$$$\text{Drift Velocity: }{\overrightarrow{v}}_d=-\frac{e\overrightarrow{E}\tau }{m_e}\approx {10}^{-4}\text{ m/s}$$$$\text{Resistivity: }\rho =\frac{m_e}{n{e}^2\tau },R=\rho \frac{L}{A}$$

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2. 3D Resistor Cube Symmetry Network (12 Equal Resistors $R$)

Multi-Mode DiagramRotational Dynamics: Spool on Incline & Pure Rolling

Option 1: Publication-Grade Scientific Vector SVG

Free-body force decomposition on incline showing normal force N, static friction fs, tension T, and Instantaneous Axis of Rotation C.

Linear Acceleration: $a_{\text{cm}} = \frac{g\sin\theta}{1 + I_{\text{cm}}/(MR^2)} = \frac{2}{3}g\sin\theta$

Critical Tension Angle: $\phi_c = \arccos(r/R)$

Option 2: Interactive 60 FPS Canvas Physics Engine

Real-time 60 FPS pure rolling simulation on incline.

Incline Angle θ: 25 °

Acceleration ($a_{\text{cm}}$):2.76 m/s²

Option 3: Interactive 3D Spatial Viewport

3D rotatable gyroscope disc.

+X

+Y

+Z

👆 Click and drag anywhere to rotate in 3D space

Azimuth Angle (ψ):35°

Elevation Angle (θ):-22°

Precession Equation:$\vec{\tau} = \frac{d\vec{L}}{dt}$

Option 5: Reactive Parametric Physics with Live Calculus

Parametric angle sliders with live torque calculation.

Incline Angle θ: 25 °

Pull Angle φ: 45 °

Critical Angle ($\phi_c$):$62.5°$

Active Mode:Option 1 (Print & Derivation Vector SVG)Switch modes using the tabs above to explore different visualizations.

Terminals Connected Across	Equivalent Resistance $R_{\text{eq}}$	Symmetry / Path Reasoning
Body Diagonal ($A⟷C^{\prime }$)	$\boxed{{\displaystyle \frac{5}{6}R}}$	Current splits $1/3\to 1/6\to 1/3$ across 3 layers
Face Diagonal ($A⟷C$)	$\boxed{{\displaystyle \frac{3}{4}R}}$	Plane of symmetry through $AC{C}^{\prime }{A}^{\prime }$
Single Edge ($A⟷B$)	$\boxed{{\displaystyle \frac{7}{12}R}}$	Superposition of source $+I$ at $A$ and sink $-I$ at $B$

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3. Potentiometer & Measurement Instruments

3.1 Potentiometer Principle

A potentiometer draws zero current from the test circuit at the null point, acting as an ideal voltmeter of infinite resistance.

• Potential Gradient: $k=\frac{V_{\text{wire}}}{L_{\text{wire}}}=\left(\frac{{\mathcal{E}}_0}{R_{\text{wire}}+R_h+r_0}\right)\frac{R_{\text{wire}}}{L_{\text{wire}}}$
• Comparing EMFs of Two Cells: $\frac{{\mathcal{E}}_1}{{\mathcal{E}}_2}=\frac{{\ell }_1}{{\ell }_2}$
• Internal Resistance of a Primary Cell:$$\boxed{{\displaystyle r=R\left(\frac{{\ell }_1-{\ell }_2}{{\ell }_2}\right)}}$$(where ${\ell }_1$ is open-circuit balancing length, ${\ell }_2$ is shunted length with external resistor $R$).

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3.2 Galvanometer Conversion Rules

• To Ammeter (Full scale deflection $I$): Connect small shunt $S$ in parallel:$$S=\frac{I_g{R}_g}{I-I_g}$$
• To Voltmeter (Full range $V$): Connect high resistance $R_s$ in series:$$R_s=\frac{V}{I_g}-R_g$$

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4. Authentic Previous Years Questions (PYQs)

PYQ 1: JEE Advanced 2022 — Non-Uniform Conical Resistor

Question:
A truncated solid metal cone of length $L$ has base radius $r_1$ and apex radius $r_2$ ($r_1>r_2$). The resistivity of the metal is $\rho$. Find the total electrical resistance $R$ between its two flat circular ends.

Step-by-Step Solution:

1. Slice the cone into thin circular disks of thickness $dx$ at distance $x\in [0,L]$ from the smaller radius end ($x=0,r=r_2$).
2. Radius of disk at distance $x$:$$r(x)=r_2+\left(\frac{r_1-r_2}{L}\right)x$$
3. Cross-sectional area of disk:$$A(x)=\pi [r(x){]}^2$$
4. Resistance of differential slice:$$dR=\rho \frac{dx}{A(x)}=\frac{\rho }{\pi }\frac{dx}{{[r_2+\left(\frac{r_1-r_2}{L}\right)x]}^2}$$
5. Integrate from $x=0$ to $x=L$:$$R=\frac{\rho }{\pi }{\int }_0^L\frac{dx}{{[r_2+\left(\frac{r_1-r_2}{L}\right)x]}^2}=\frac{\rho }{\pi }\left(\frac{L}{r_1-r_2}\right){[-\frac{1}{r(x)}]}_{r_2}^{r_1}$$$$R=\frac{\rho L}{\pi (r_1-r_2)}(\frac{1}{r_2}-\frac{1}{r_1})=\frac{\rho L}{\pi (r_1-r_2)}\left(\frac{r_1-r_2}{r_1{r}_2}\right)=\boxed{{\displaystyle \frac{\rho L}{\pi r_1{r}_2}}}$$

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PYQ 2: JEE Advanced 2020 — Maximum Power Transfer Theorem

Question:
A variable load resistor $R_L$ is connected across a network having open-circuit Thevenin voltage $V_{\text{th}}=12\text{ V}$ and internal resistance $R_{\text{th}}=6\mathrm{\Omega }$. Find:

1. The value of $R_L$ for which maximum power is transferred to the load.
2. The maximum power $P_{max}$ delivered to $R_L$.

Step-by-Step Solution:

• 1. Maximum Power Condition: By the Maximum Power Transfer Theorem:$$R_L=R_{\text{th}}=\boxed{{\displaystyle 6\mathrm{\Omega }}}$$
• 2. Maximum Power Delivered:$$P_{max}=\frac{V_{\text{th}}^2}{4R_{\text{th}}}=\frac{{12}^2}{4\times 6}=\frac{144}{24}=\boxed{{\displaystyle 6.0\text{ W}}}$$

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5. High-Yield Formula Sheet

Entity	Formula	Notes
Truncated Cone Resistance	$R=\frac{\rho L}{\pi r_1{r}_2}$	Geometric mean area
Cube Resistor Diagonal	$R_{A{C}^{\prime }}=\frac{5}{6}R$	3D symmetry
Potentiometer Cell Internal $r$	$r=R\left(\frac{{\ell }_1-{\ell }_2}{{\ell }_2}\right)$	Open vs closed circuit
Max Power Transfer	$P_{max}=\frac{{\mathcal{E}}^2}{4r}$	When $R_L=r$