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Current Electricity
Physics · Class 12 · Part 1NCERT pp. 81–1022 sets × 25 questionsFree
Key facts and formulas for Current Electricity, three worked sample questions, and two full practice sets. Every question has a hint, a reason for every option and a step-by-step solution, with a calculator or animation where one helps. English and Hinglish.
Formula sheet for this chapter
| Idea | Rule or fact |
|---|
| Current | I = dQ/dt (instantaneous); I = Q/t (average); I = n e A v_d |
| Resistance | R = ρ l / A; series R = R₁ + R₂; parallel 1/R = 1/R₁ + 1/R₂ |
| Microscopic Ohm's law | j = σ E, E = ρ j; ρ = m / (n e² τ); v_d = eEτ/m; μ = v_d/E = eτ/m |
| Temperature effect | R = R₀ [1 + α (T − T₀)]; metals: ρ rises; semiconductors: ρ falls |
| Power and heat | P = VI = I²R = V²/R; H = I²Rt; 1 kWh = 3.6×10⁶ J |
| Cell with internal resistance | V = ε − I r; I = ε/(R + r); series: nε, nr; parallel identical: ε, r/n |
| Maximum power | P is maximum when R = r; P_max = ε²/(4r); efficiency = R/(R + r) |
| Wheatstone bridge | Balanced when P/Q = R/S; no current in the galvanometer |
| Meters | Voltmeter: R = V/I_g − G in series; ammeter: S = G I_g/(I − I_g) in parallel |
| Drawn wire | Volume constant: R ∝ l² (stretch by n gives n² R) |
Three sample questions
Q1. The charge through a section of a wire is Q = 3t² + 2t (Q in coulomb, t in second). The current at t = 2 s is
- 14 A
- 16 A
- 7 A
- 10 A
Hint and solution
Hint: Current is the rate of flow of charge at that instant, so differentiate Q with respect to time.
- I = dQ/dt — Instantaneous current is the derivative of charge with respect to time.
- I = d(3t² + 2t)/dt = 6t + 2 — Differentiate each term: 3t² gives 6t and 2t gives 2.
- I(2) = 6 × 2 + 2 = 14 A — Put t = 2 s in the expression. Note that Q/t would only give the average current.
- Answer: 14 A — So the current at 2 s is 14 A. Option A.
Answer: A. I = dQ/dt = 6t + 2 = 14 A at t = 2 s. Q/t would give only the average.
Q2. In 2 s, 8 C of positive charge crosses a section forward and 4 C of negative charge also crosses it forward. The net current is
- 6 A
- 4 A
- 2 A
- 1 A
Hint and solution
Hint: Negative charge moving forward is equivalent to positive charge moving backward. Find the net forward charge first.
- Conventional current: negative charge forward = positive backward — Current direction follows positive charge. A negative charge moving forward counts as current backward.
- Net forward charge = 8 − 4 = 4 C — Subtract the opposing contribution from the 8 C.
- I = 4 C / 2 s = 2 A — Current is net charge divided by time.
- Answer: 2 A — So the net current is 2 A. Option C.
Answer: C. Net forward charge = 8 − 4 = 4 C. I = 4/2 = 2 A. Negative charge going forward reduces the forward current.
Q3. A wire of resistance 6 Ω is cut into three equal pieces, and the pieces are joined in parallel. The resistance of the combination is
- 2 Ω
- 6 Ω
- 2/3 Ω
- 18 Ω
Hint and solution
Hint: Resistance is proportional to length, so each piece has one third of 6 Ω. Then combine three equal resistors in parallel.
- R ∝ length, so piece = 6/3 = 2 Ω — Resistance is proportional to length. One third of the wire has one third of the resistance.
- 1/R = 1/2 + 1/2 + 1/2 = 3/2 — For parallel resistors add the reciprocals.
- R = 2/3 Ω — Invert 3/2 to get the equivalent resistance. It is smaller than any single piece, as expected for parallel.
- Answer: 2/3 Ω — So the combination has 2/3 Ω. Option C.
Answer: C. Each piece is 2 Ω. Three 2 Ω in parallel give 2/3 Ω.
How to use this chapter
Read the NCERT pages first, then attempt Set 1 without the hints. Note every wrong answer in a mistake list, read the reason for each option, and retry Set 2 two days later. A topic is solid when you can say why each wrong option is wrong.