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Alternating Current
Physics · Class 12 · Part 1NCERT pp. 177–1962 sets × 25 questions
Key facts and formulas for Alternating Current, 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 |
|---|
| Idea | Rule or fact |
| rms and peak | V_rms = Vₘ/√2 and I_rms = iₘ/√2; ac meters read rms |
| Averages of a sine | Half-cycle average = 2iₘ/π ≈ 0.637 iₘ; full-cycle average = 0 |
| Reactance | X_L = ωL = 2πνL; X_C = 1/(ωC); both in ohm |
| Series LCR | Z = √(R² + (X_L − X_C)²); tan φ = (X_L − X_C)/R |
| Power | P = V_rms I_rms cos φ = I²R; cos φ = R/Z; pure L or C gives P = 0 |
| Resonance | ω₀ = 1/√(LC), Z = R, current maximum, cos φ = 1; independent of R |
| Q and bandwidth | Q = ω₀L/R; Δω = R/L; V_C = V_L = Q × V at resonance |
| Transformer | V_s/V_p = N_s/N_p; η = P_out/P_in; core: laminated soft iron |
| Line loss | Loss = I²R with I = P/V, so a higher voltage cuts the loss by V² |
| LC oscillation | ω = 1/√(LC); L ↔ mass, 1/C ↔ spring constant, q ↔ x, i ↔ v |
Three sample questions
Q1. A supply is marked 110 V. Its peak voltage is nearly
- 110 V
- 155 V
- 78 V
- 220 V
Hint and solution
Hint: A voltage marking on a supply always quotes the rms value. How is the peak related to rms?
- V_rms = 110 V (marked value) — The marking on a supply is the rms value, not the peak.
- Vₘ = V_rms × √2 — For a sine wave the peak is √2 times the rms value.
- Vₘ = 110 × 1.414 ≈ 155.6 V — Multiply 110 by 1.414 to get about 155.6 volt.
- Peak ≈ 155 V — So the peak voltage is nearly 155 V. Option B.
Answer: B. The marking is rms. Peak = 110 × √2 = 155.6 V.
Q2. An alternating current is i = 4 sin(200πt) A. Its rms value and frequency are
- 4 A and 100 Hz
- 2.83 A and 200 Hz
- 2 A and 50 Hz
- 2.83 A and 100 Hz
Hint and solution
Hint: Compare with i = iₘ sin ωt to read off the peak and ω. Then use rms = peak/√2 and ν = ω/2π.
- i = iₘ sin ωt → iₘ = 4 A, ω = 200π — Compare the given current with the standard form to read the peak and the angular frequency.
- i_rms = iₘ/√2 = 4/1.414 = 2.83 A — The rms value of a sine wave is peak divided by root two.
- ν = ω/2π = 200π/2π = 100 Hz — Frequency is angular frequency divided by two pi.
- rms 2.83 A, ν = 100 Hz — So the rms current is 2.83 A and the frequency is 100 Hz. Option D.
Answer: D. i_rms = 4/√2 = 2.83 A. ω = 200π, so ν = ω/2π = 100 Hz.
Q3. A voltage v = 200 sin ωt is applied across a 100 Ω resistor. The mean power is
- 200 W
- 400 W
- 100 W
- 283 W
Hint and solution
Hint: Mean power in a resistor uses the rms voltage, not the peak. Convert first.
- V_rms = Vₘ/√2 = 200/1.414 ≈ 141 V — Convert the peak voltage to rms because average power depends on rms.
- P = V_rms²/R — Power in a resistor is rms voltage squared over resistance.
- P = (200²/2)/100 = 20000/100 = 200 W — V_rms squared is 40000 divided by 2, which is 20000. Divide by 100 ohm.
- Mean power = 200 W — So the mean power is 200 watt. Option A.
Answer: A. V_rms = 200/√2. P = V_rms²/R = (20000)/100 = 200 W.
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.