Current Electricity & Ohm’s Law
(AP EAMCET – Physics – Chapter 6)
Goal: Understand the flow of electric charge, how it is measured, and how voltage, current, and resistance relate to each other.
Key Skills:
- Use Ohm’s law to calculate any of the three quantities.
- Analyze simple series/parallel circuits.
- Convert between SI units and common engineering units.
- Apply power formulas.
1. Simple Explanation
| Symbol | Quantity | Unit | Meaning |
|---|---|---|---|
| (I) | Current | A (ampere) | Rate of charge flow |
| (V) | Voltage (potential difference) | V (volt) | Energy per unit charge |
| (R) | Resistance | Ω (ohm) | Opposition to current |
- Current is the flow of electrons through a conductor.
- Ohm’s Law states that the voltage across a conductor is directly proportional to the current flowing through it, with the resistance as the proportionality constant.
Ohm’s Law (in SI):
[ V = I , R ]
This law applies to Ohmic materials (where (R) remains constant over the operating range).
2. Key Points
| Concept | Details |
|---|---|
| Electrical Current | (I = \frac{dQ}{dt}). One ampere equals (1 \text{ C/s}). |
| Resistance | Depends on material, length (L), cross‑section (A), and temperature. For a uniform wire: (\displaystyle R = \rho,\frac{L}{A}) where (\rho) is resistivity. |
| Ohm’s Law | (V = I R). Rearranged: (I = \frac{V}{R}) and (R = \frac{V}{I}). |
| Power Dissipated | (P = V I = I^2 R = \frac{V^2}{R}). |
| Series Circuits | Resistors add: (R_{\text{eq}} = R_1 + R_2 + \dots). Current same in all branches. |
| Parallel Circuits | Voltages across each branch equal. Equivalent resistance: (\displaystyle \frac{1}{R_{\text{eq}}} = \frac{1}{R_1} + \frac{1}{R_2} + \dots). |
| Temperature Effect | (\displaystyle R = R_0[1 + \alpha (T - T_0)]). (\alpha) is temperature coefficient. |
| Units Conversion | 1 kΩ = 1 000 Ω, 1 mA = 0.001 A, 1 kW = 1 000 W. |
| Data Tables | Resistivity values: Cu ≈ 1.68 × 10⁻⁸ Ω·m, Al ≈ 2.82 × 10⁻⁸ Ω·m. |
| Common Problems | - Find missing quantity using Ohm’s law. - Compute equivalent resistance in series/parallel. - Calculate power loss in a wire. |
3. Worked Example
Problem (typical of AP EAMCET)
A 2 m long copper wire of diameter 1 mm is connected across a 12 V battery. Calculate (i) its resistance, (ii) the current flowing, and (iii) the power dissipated.
Step‑by‑Step
-
Find cross‑sectional area (A).
Diameter (d = 1) mm = (1\times10^{-3}) m.
[ A = \pi \left(\frac{d}{2}\right)^2 = \pi \left(\frac{1\times10^{-3}}{2}\right)^2 = \pi (5\times10^{-4})^2 = \pi \times 25\times10^{-8} \approx 7.85\times10^{-7},\text{m}^2 ] -
Resistivity (\rho) of copper.
(\rho_{\text{Cu}} = 1.68\times10^{-8},\Omega\cdot\text{m}). -
Compute resistance (R).
[ R = \rho,\frac{L}{A} = 1.68\times10^{-8},\frac{2}{7.85\times10^{-7}} \approx 1.68\times10^{-8}\times 2.55\times10^{6} \approx 0.0428,\Omega ] Take (R \approx 0.043,\Omega). -
Current (I) using Ohm’s law.
[ I = \frac{V}{R} = \frac{12}{0.043} \approx 279,\text{A} ] (Very high due to low resistance; realistic only for short, thick wires.) -
Power dissipated (P).
[ P = V I = 12 \times 279 \approx 3348,\text{W} ] Or [ P = I^2 R = (279)^2 \times 0.043 \approx 3348,\text{W} ]
Answer:
- (R \approx 0.043,\Omega)
- (I \approx 279,\text{A})
- (P \approx 3.35,\text{kW})
4. Quick Revision Summary
| Concept | Formula | Units | Typical Values |
|---|---|---|---|
| Current | (I = \frac{dQ}{dt}) | A | 1 A = 1 C/s |
| Resistance | (R = |