Current Electricity

Master electric current, Ohm's law, resistors, Kirchhoff's laws, and electrical circuits for JEE Physics.

Current Electricity deals with moving charges and their effects. This topic forms the foundation for electrical circuits and electronics.

Overview

graph TD
    A[Current Electricity] --> B[Basic Concepts]
    A --> C[Ohm's Law]
    A --> D[Combinations]
    A --> E[Kirchhoff's Laws]
    A --> F[Instruments]
    D --> D1[Series]
    D --> D2[Parallel]
    F --> F1[Wheatstone Bridge]
    F --> F2[Potentiometer]

Electric Current

Rate of flow of charge:

$$\boxed{I = \frac{dQ}{dt} = \frac{Q}{t}}$$

Unit: Ampere (A) = Coulomb/second

Drift Velocity

$$v_d = \frac{eE\tau}{m}$$

Current in terms of drift velocity:

$$\boxed{I = nAv_d e}$$

where:

  • n = number density of electrons
  • A = cross-sectional area
  • e = electron charge

Current Density

$$J = \frac{I}{A} = nev_d$$

Mobility

$$\mu = \frac{v_d}{E} = \frac{e\tau}{m}$$

Relation: $\sigma = ne\mu$

Ohm’s Law

$$\boxed{V = IR}$$

Resistance

$$R = \frac{\rho L}{A}$$

where:

  • ρ = resistivity
  • L = length
  • A = cross-sectional area

Conductivity

$$\sigma = \frac{1}{\rho}$$

Temperature Dependence

$$R_T = R_0(1 + \alpha\Delta T)$$

where α = temperature coefficient of resistance

Materialα behavior
MetalsPositive (R increases with T)
SemiconductorsNegative (R decreases with T)
AlloysVery small
JEE Tip
For stretched/compressed wires: If a wire is stretched to n times its length, new resistance = n²R (since volume is constant: A’ = A/n)

Electrical Power and Energy

Power:

$$P = VI = I^2R = \frac{V^2}{R}$$

Energy:

$$E = Pt = VIt = I^2Rt = \frac{V^2t}{R}$$

Heat produced (Joule’s Law):

$$H = I^2Rt$$

Combination of Resistors

Series Combination

$$\boxed{R_{eq} = R_1 + R_2 + R_3 + ...}$$
  • Same current through all resistors
  • Voltage divides proportionally

Voltage divider:

$$V_1 = \frac{R_1}{R_1 + R_2}V$$

Parallel Combination

$$\boxed{\frac{1}{R_{eq}} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3} + ...}$$
  • Same voltage across all resistors
  • Current divides inversely

Current divider:

$$I_1 = \frac{R_2}{R_1 + R_2}I$$

Special Cases

Two resistors in parallel:

$$R_{eq} = \frac{R_1 R_2}{R_1 + R_2}$$

n equal resistors:

  • Series: $R_{eq} = nR$
  • Parallel: $R_{eq} = \frac{R}{n}$

EMF and Internal Resistance

Terminal voltage:

$$V = \varepsilon - Ir$$

where:

  • ε = EMF of cell
  • r = internal resistance

Cells in Series

$$\varepsilon_{eq} = \varepsilon_1 + \varepsilon_2$$ $$r_{eq} = r_1 + r_2$$

Cells in Parallel

$$\varepsilon_{eq} = \frac{\varepsilon_1/r_1 + \varepsilon_2/r_2}{1/r_1 + 1/r_2}$$ $$\frac{1}{r_{eq}} = \frac{1}{r_1} + \frac{1}{r_2}$$

For identical cells (ε, r) in parallel:

$$\varepsilon_{eq} = \varepsilon, \quad r_{eq} = \frac{r}{n}$$

Kirchhoff’s Laws

Junction Rule (KCL)

Sum of currents entering = Sum of currents leaving

$$\sum I_{in} = \sum I_{out}$$

Loop Rule (KVL)

Sum of potential differences around any closed loop = 0

$$\sum V = 0$$
graph LR
    A[Kirchhoff's Laws] --> B[KCL: Conservation of charge]
    A --> C[KVL: Conservation of energy]
Common Mistake
In KVL, going through a resistor in the direction of current causes a potential drop (-IR), while going against current causes a rise (+IR).

Wheatstone Bridge

Balanced condition:

$$\boxed{\frac{P}{Q} = \frac{R}{S}}$$

When balanced, no current flows through galvanometer.

Metre Bridge

Special case of Wheatstone bridge:

$$\frac{R}{S} = \frac{l}{100-l}$$

Potentiometer

Principle: Potential drop ∝ Length

$$\frac{V_1}{V_2} = \frac{l_1}{l_2}$$

Applications

Comparison of EMFs:

$$\frac{\varepsilon_1}{\varepsilon_2} = \frac{l_1}{l_2}$$

Internal resistance:

$$r = R\left(\frac{l_1 - l_2}{l_2}\right)$$

Advantages over voltmeter:

  • Draws no current at balance
  • More accurate
  • Can measure EMF directly

Galvanometer Conversion

To Ammeter (low resistance)

Add shunt resistance $S$ in parallel:

$$S = \frac{I_g G}{I - I_g}$$

To Voltmeter (high resistance)

Add high resistance $R$ in series:

$$R = \frac{V}{I_g} - G$$

Practice Problems

  1. A wire of resistance 10Ω is stretched to double its length. Find new resistance.

  2. Find equivalent resistance between A and B when three 6Ω resistors are connected as a triangle.

  3. In a potentiometer, a cell of EMF 2V balances at 60 cm. Another cell balances at 80 cm. Find its EMF.

  4. A galvanometer of resistance 50Ω gives full deflection for 1 mA. Convert it to (a) ammeter of range 10 A (b) voltmeter of range 10 V.

Quick Check
Why is an ammeter connected in series and a voltmeter in parallel?

Further Reading