Current Electricity – Class 12 (CBSE)
(Physics – Chapter 7)
Aim – Understand the flow of electric charge in conductors, the laws governing it, and how to analyse simple DC circuits.
1. Simple Explanation
Electric current is the rate at which charge flows through a conductor.
If Q coulombs of charge pass through a cross‑section in t seconds, the current I is
[ I = \frac{Q}{t}\quad(\text{A}) \quad \text{(Ampere)} ]
- Voltage (potential difference) drives the electrons.
[ V = \frac{W}{Q}\quad(\text{V})\quad \text{(Volt)} ] - Resistance (R) opposes the flow.
[ R = \frac{V}{I}\quad(\Omega) ]
The fundamental relationship among these three quantities is Ohm’s Law.
2. Key Points
| Concept | Symbol | SI Unit | Definition |
|---|---|---|---|
| Current | (I) | A | Flow of charge per second |
| Voltage | (V) | V | Work per unit charge |
| Resistance | (R) | Ω | Opposition to current |
| Conductance | (G) | S | (G = 1/R) |
| Power | (P) | W | Rate of energy transfer |
| Energy | (E) | J | Work done or heat generated |
| Electromotive Force (EMF) | (\mathcal{E}) | V | Potential difference produced by a source |
| Internal Resistance | (r) | Ω | Resistance inside the source |
2.1 Ohm’s Law
[ V = IR \quad\text{or}\quad I = \frac{V}{R}\quad\text{or}\quad R = \frac{V}{I} ]
- Linear behaviour – For a good conductor, (V) vs. (I) is a straight line, slope = (R).
2.2 Power, Energy, and Efficiency
[ P = IV = I^{2}R = \frac{V^{2}}{R}\quad(\text{W}) ] [ E = Pt = \frac{V^{2}t}{R}\quad(\text{J}) ]
- Efficiency of a source: (\eta = \frac{P_{\text{out}}}{P_{\text{in}}}\times 100%)
2.3 Kirchhoff’s Laws (for DC circuits)
- KCL (Kirchhoff’s Current Law) – Sum of currents entering a junction = sum leaving.
[ \sum I_{\text{in}} = \sum I_{\text{out}} ] - KVL (Kirchhoff’s Voltage Law) – Sum of potential differences around any closed loop = 0.
[ \sum V_{\text{rise}} = \sum V_{\text{drop}} ]
2.4 Series and Parallel Connections
| Configuration | Total Resistance | Current |
|---|---|---|
| Series | (R_{\text{eq}} = R_1+R_2+R_3+\dots) | Same through all components ((I_{\text{eq}})) |
| Parallel | (\displaystyle \frac{1}{R_{\text{eq}}}=\frac{1}{R_1}+\frac{1}{R_2}+\dots) | Splits: (I_{\text{eq}}=I_1+I_2+\dots) |
- Voltage across each parallel branch is equal.
- Current through each series element is equal.
2.5 Internal Resistance of a Source
Real sources have a small internal resistance (r).
- Open‑circuit voltage: (\mathcal{E}) (EMF)
- Terminal voltage under load (I):
[ V_{\text{term}} = \mathcal{E} - Ir ] - Power delivered to load (R):
[ P_{\text{load}} = \frac{(\mathcal{E})^{2}R}{(R+r)^{2}} ] - Maximum power transfer occurs when (R = r).
2.6 Conduction in Metals
- Electrons drift slowly, but the electric field propagates at near‑c speed.
- Resistivity (\rho) depends on temperature: (\rho(T) = \rho_0[1+\alpha(T-T_0)]).
- Resistivity decreases with cross‑section (A) and increases with length (L):
[ R = \rho \frac{L}{A} ]
3. Worked Example
Problem
A battery of EMF (12,\text{V}) and internal resistance (0.5,\Omega) is connected to a load of (4,\Omega).
(i) Find the terminal voltage and current.
(ii) Calculate power delivered to the load and the power dissipated in the internal resistance.
(iii) Determine the efficiency of the battery.
(i) Terminal voltage & current
Total resistance (R_{\text{eq}} = R_{\text{load}} + r = 4 + 0.5 = 4.5,\Omega)
[ I = \frac{\mathcal{E}}{R_{\text{eq}}} = \frac{12}{4.5} = 2.67,\text{A} ]
[ V_{\text{term}} = \mathcal{E} - Ir = 12 - 2.67 \times 0.5 = 12 - 1.335 = 10.