Alternating Current

Physics · Class 12

Simulation · Physics · Class 12

Why 220 V mains peaks at 311 V

"Why 220 V mains peaks at 311 V" is a free, interactive Class 12 Physics simulation for NEET and JEE, built on the NCERT section "RMS values and average power" in Alternating Current. Two identical 484 Ω lamps, one on ac and one on dc, with p = i²R traced over cycles and its average marked against the dc power line. NEET has asked about this idea 2 times in 10 years of papers. JEE Main asked 37 questions from this chapter across the 42 shifts Lumi has analysed (2017–2026).

From the lesson RMS values and average power in Alternating Current. Change the values and watch what happens.

Why 220 V mains peaks at 311 VPhysics · Class 12

What it shows

NCERT §7.2

Two identical 484 Ω lamps, one on ac and one on dc, with p = i²R traced over cycles and its average marked against the dc power line.

Controls

3 to try

  • Sliderac peak voltage 0–400 V
  • Sliderdc voltage 0–400 V
  • Toggleplot i(t) / p(t)

Try this

Set the dc supply so its lamp heats exactly as fast as the ac lamp on 311 V peak (goal: 220 V ± 2 V)

Common mistake

Treating 220 V mains as the peak value, or using the peak current in P = I²R.

In NEET

10 years of papers

Asked 2 times in 10 years of NEET.

See the past questions: LCR Circuits, Alternating Current (chapter)

In JEE Main

42 shifts analysed

37 questions from Alternating Current across the 42 JEE Main shifts Lumi has analysed (2017–2026).

See the past questions

The idea behind it

NCERT §7.2

  • The power in the resistor at any instant is p = i²R = im²R sin²ωt. Since sin²ωt = ½(1 − cos 2ωt) and cos 2ωt averages to zero over a cycle, the average power is ½im²R.
  • To write this like the dc formula P = I²R, define the root mean square (rms) or effective current I = im/√2 = 0.707 im. Then P = I²R.
  • Likewise the rms voltage is V = vm/√2 = 0.707 vm, and V = IR, so P = V²/R = IV, exactly as for dc.
  • The rms current is the steady current that would heat the same resistor at the same average rate.
  • AC values are quoted as rms values unless stated otherwise. The household 220 V is an rms value; its peak is √2 × 220 = 311 V.
  • Example 7.1: a bulb rated 100 W for 220 V has R = V²/P = 484 Ω; the source's peak voltage is 311 V; the rms current is P/V = 0.454 A.

Questions

5 answers

What does the "Why 220 V mains peaks at 311 V" simulation show?
Two identical 484 Ω lamps, one on ac and one on dc, with p = i²R traced over cycles and its average marked against the dc power line.
What mistake does the "Why 220 V mains peaks at 311 V" simulation fix?
A common mistake it clears up: treating 220 V mains as the peak value, or using the peak current in P = I²R.
Has NEET asked about this idea?
Yes. NEET has asked about this idea 2 times in 10 years of papers.
Does JEE Main ask questions from Alternating Current?
Yes. JEE Main asked 37 questions from this chapter across the 42 shifts Lumi has analysed (2017–2026).
Is this simulation free?
Yes. It is free and runs in the browser on a phone or a laptop. It belongs to the lesson "RMS values and average power" in Alternating Current (Class 12 Physics).

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