Current Electricity – JEECUP Study Notes
Physics – Basics of Electric Current
These notes are written for the JEECUP physics curriculum.
They give a clear, step‑by‑step walk‑through of the essential concepts, formulas, and a worked example.
1. Simple Explanation
Electric current is the flow of electric charge (mainly electrons) through a conductor.
- Charge (Q) moves under the influence of an electric field created by a potential difference (voltage, V).
- The rate at which charge flows is the current (I), measured in amperes (A).
One ampere = one coulomb of charge passing through a cross‑sectional area in one second.
(I = \dfrac{Q}{t})
2. Key Points
| Concept | Symbol | Formula | Notes |
|---|---|---|---|
| Current | (I) | ( I = \frac{Q}{t} ) | SI unit: ampere (A) |
| Voltage (Potential difference) | (V) | ( V = IR ) (Ohm’s law) | SI unit: volt (V) |
| Resistance | (R) | ( R = \frac{V}{I} ) | SI unit: ohm (Ω) |
| Power | (P) | ( P = VI = I^2R = \frac{V^2}{R} ) | SI unit: watt (W) |
| Resistivity | (\rho) | ( R = \rho \dfrac{L}{A} ) | (L): length, (A): cross‑sectional area |
| Conductivity | (\sigma) | ( \sigma = \frac{1}{\rho} ) | SI unit: siemens per metre (S·m(^{-1})) |
| Series Resistors | – | ( R_{\text{total}} = \sum R_i ) | Currents equal, voltages add |
| Parallel Resistors | – | ( \frac{1}{R_{\text{total}}} = \sum \frac{1}{R_i} ) | Voltages equal, currents divide |
| Kirchhoff’s Rules | – | Current law: (\sum I_{\text{in}} = \sum I_{\text{out}}) Voltage law: (\sum V = 0) |
For complex circuits |
| Temperature Effect | – | ( R(T) = R_0[1 + \alpha(T-T_0)] ) | (\alpha) = temperature coefficient |
Additional Concepts
- Conductors vs. Insulators – Conductors (metals) have low resistivity; insulators (plastics, rubber) have high resistivity.
- Semi‑conductors – Resistivity between conductors and insulators; can be controlled by doping and temperature.
- Current density (J) – ( J = \frac{I}{A} ) (A/m²). Indicates how densely charge flows.
- Magnetic Effects – Current in a wire produces a magnetic field (right‑hand rule).
3. Worked Example
Problem
A 12 V battery is connected to a circuit containing two resistors:
- (R_1 = 4,\Omega)
- (R_2 = 6,\Omega)
The resistors are connected in parallel.
Find:
- The equivalent resistance (R_{\text{eq}}).
- The total current supplied by the battery.
- The current through each resistor.
- The power dissipated by each resistor.
Step‑by‑Step Solution
| Step | Calculation | Result |
|---|---|---|
| 1. Equivalent resistance | (\displaystyle \frac{1}{R_{\text{eq}}} = \frac{1}{R_1} + \frac{1}{R_2}) | (\displaystyle \frac{1}{R_{\text{eq}}} = \frac{1}{4} + \frac{1}{6} = 0.25 + 0.1667 = 0.4167) ( R_{\text{eq}} = \frac{1}{0.4167} = 2.4,\Omega) |
| 2. Total current (battery) | ( I_{\text{total}} = \frac{V}{R_{\text{eq}}} = \frac{12,\text{V}}{2.4,\Omega} ) | ( I_{\text{total}} = 5,\text{A}) |
| 3. Current through each resistor | Ohm’s law: ( I_i = \frac{V}{R_i} ) | ( I_1 = \frac{12}{4} = 3,\text{A}) ( I_2 = \frac{12}{6} = 2,\text{A}) |
| 4. Power dissipated | ( P_i = I_i^2 R_i = \frac{V^2}{R_i} ) | ( P_1 = 3^2 \times 4 = 36,\text{W}) ( P_2 = 2^2 \times 6 = 24,\text{W}) |
Check
- Current conservation: ( I_{\text{total}} = I_1 + I_2 = 3,\text{A} + 2,\text{A} = 5,\text{A}) ✔️
- Power conservation: ( P_{\text{total}} = P_1 + P