Current Electricity for NEET UG
Current electricity explains every circuit from a torch to a hospital monitor. Ohm's law, resistor combinations and Kirchhoff's rules are the tools.
PhysicsClass 1264 MCQs3 lessons12 flashcards4 memory tricks
Practise Current Electricity free
What you'll learn
- Current, resistance and Ohm's law
Use V = IR and R = ρL/A, and explain drift velocity. - Resistors in series and parallel; cells
Combine resistors and use EMF and internal resistance. - Kirchhoff's rules and the Wheatstone bridge
Apply Kirchhoff's junction and loop rules, and the balance condition of a Wheatstone bridge.
Key formulas
| Current | I = Q/t |
|---|---|
| Ohm's law | V = IR |
| Resistance of a wire | R = ρL/A |
| Drift velocity | I = neAv_d |
| Temperature dependence (metals) | R_T = R₀(1 + αΔT) |
| Series | R = R₁ + R₂ + … |
| Parallel | 1/R = 1/R₁ + 1/R₂ + … |
| Cell with internal resistance r | I = ε/(R + r), terminal voltage V = ε − Ir |
Memory tricks
Black 0, Brown 1, Red 2, Orange 3, Yellow 4, Green 5, Blue 6, Violet 7, Grey 8, White 9. Gold band ±5%, silver ±10%.
In parallel the total resistance is less than the smallest single resistance: 1/R = 1/R₁ + 1/R₂ + …. In series, R = R₁ + R₂ + ….
Electrons drift at only about a millimetre per second, but the electric field travels through the wire at nearly the speed of light, so every electron starts moving almost at once.
Junction rule (sum of currents in = sum out) is conservation of charge. Loop rule (sum of potential changes round a loop = 0) is conservation of energy.
Try 3 NEET-style questions
Q1. In a Wheatstone bridge, P = 10 Ω, Q = 20 Ω and R = 30 Ω. For the bridge to be balanced, the fourth resistance S must be
- 15 Ω
- 40 Ω
- 60 Ω
- 6 Ω
Show answer
Answer: (3) 60 Ω
Balance condition: P/Q = R/S. 10/20 = 30/S, so S = 60 Ω.
Q2. Cells of emf 6 V and 2 V, with negligible internal resistance, are connected in series with opposing polarities and a total resistance of 4 Ω. The current is
- 1 A
- 2 A
- 0.5 A
- 1.5 A
Show answer
Answer: (1) 1 A
Opposing cells: net emf = 6 − 2 = 4 V. I = 4/4 = 1 A.
Q3. Resistances of 3 Ω and 6 Ω are connected in series. Their equivalent resistance is
- 2 Ω
- 9 Ω
- 18 Ω
- 0.5 Ω
Show answer
Answer: (2) 9 Ω
R = 3 + 6 = 9 Ω.
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