Current Electricity – Study Notes
(MHT CET – Physics)
1. Simple Explanation
Current electricity deals with the flow of electric charge (electrons) in conductors.
- Electric charge (
q) moves under the influence of an electric field. - Electric current (
I) is the rate of charge flow:
[ I = \frac{dq}{dt}\quad(\text{units: A = C s}^{-1}) ] - The direction of current is taken from positive to negative (conventional current).
2. Key Points
| Concept | Formula / Definition | Units | Notes |
|---|---|---|---|
| Electric Current | (I = \frac{dq}{dt}) | 1 A = 1 C s⁻¹ | Constant current means uniform charge flow. |
| Voltage (Potential Difference) | (V = \frac{W}{q}) | 1 V = 1 J C⁻¹ | Drives current. |
| Ohm’s Law | (V = IR) | R in Ω | Holds for ohmic conductors. |
| Resistivity | (R = \rho\frac{l}{A}) | ρ in Ω m | l = length, A = cross‑section. |
| Power | (P = IV = I^2R = \frac{V^2}{R}) | 1 W = 1 J s⁻¹ | Energy per unit time. |
| Series Circuit | (R_{\text{eq}} = R_1+R_2+\dots) | – | Same current through all elements. |
| Parallel Circuit | (\frac{1}{R_{\text{eq}}} = \frac{1}{R_1}+\frac{1}{R_2}+…) | – | Same voltage across all elements. |
| Kirchhoff’s Laws | KCL: ΣI = 0 (node) KVL: ΣV = 0 (loop) |
– | Fundamental for circuit analysis. |
| Conductance | (G = \frac{1}{R}) | 1 S = Ω⁻¹ | Useful in parallel calculations. |
| Temperature Effect | (R(T) = R_0[1+\alpha (T-T_0)]) | α for metals | Resistance increases with temperature. |
3. Worked Example
Problem
A 12 V battery supplies a circuit containing:
- (R_1 = 4,\Omega) (series)
- (R_2 = 6,\Omega) (series)
- A light bulb (R_3 = 12,\Omega) (parallel with (R_2))
Find:
- Total resistance
- Current from the battery
- Voltage across each component
- Power dissipated by the bulb
Solution
-
Total resistance
Parallel part first:
[ \frac{1}{R_{23}} = \frac{1}{R_2} + \frac{1}{R_3} = \frac{1}{6} + \frac{1}{12} = \frac{2+1}{12} = \frac{3}{12} \Rightarrow R_{23} = 4,\Omega ]Now series with (R_1):
[ R_{\text{eq}} = R_1 + R_{23} = 4 + 4 = 8,\Omega ] -
Battery current (Ohm’s law)
[ I = \frac{V}{R_{\text{eq}}} = \frac{12}{8} = 1.5,\text{A} ]
-
Voltage across each part
Voltage across series combination (V_{12} = IR_1 = 1.5 \times 4 = 6,V)
The remaining voltage across the parallel part:
[ V_{23} = V - V_{12} = 12 - 6 = 6,V ]Since (R_2) and (R_3) are in parallel, both drop 6 V.
-
Power in bulb
[ I_3 = \frac{V_{23}}{R_3} = \frac{6}{12} = 0.5,\text{A} ] [ P_3 = I_3^2 R_3 = (0.5)^2 \times 12 = 3,\text{W} ]
(Alternatively (P_3 = V_{23}^2 / R_3 = 6^2 / 12 = 3,\text{W}))
Answer
- (R_{\text{eq}} = 8,\Omega)
- Battery current (I = 1.5,\text{A})
- Voltage across bulb (V = 6,\text{V})
- Power dissipated by bulb (P = 3,\text{W})
4. Quick Revision Summary
- Current (I = dq/dt).
- Ohm’s Law (V = IR) – holds for ohmic materials.
- Resistivity (R = \rho l / A).
- Power (P = IV = I^2R = V^2/R).
- Series: (R_{\text{eq}} = \sum R); same (I).
- Parallel: (1/R_{\text{eq}} = \sum 1/R); same (V).
- Kirchhoff: KCL (sum of currents at a node = 0); KVL (sum of voltages in a loop = 0).
- Temperature: (R(T) = R_0[1+\alpha (T-T_0)]).
- Typical MHT‑CET questions
- Find current or voltage in