Formula
Peak vs RMS values
I_rms = I_m / √2; V_rms = V_m / √2 (sinusoidal). Stated AC voltages (e.g. 220 V) are RMS unless noted otherwise.
-- NCERT Class 12 Physics, Ch. 7, p. 179Peak RMS AC
8 MCQs1 revision card9-step worked example
Source: NCERT Electromagnetic Induction and Alternating CurrentsPYQ coverage: NEET 2022Official key: NTA-verifiedLast updated: 23 Sep 2026
Peak RMS AC, explained for NEET
The trap in this topic is not the √2. It is knowing which number the question handed you.
A stated AC voltage is RMS unless the paper says otherwise. "A 220 V supply", "a source of 100 V", "a 12 V AC adapter" — all RMS. The peak is larger by √2. But when a question writes the source as an equation, v = 200 sin(100πt), the coefficient 200 is the peak, because that is what a sinusoid's amplitude means. Same chapter, same page, opposite convention — and a question can hand you one and ask for a quantity defined in terms of the other. Substituting a peak where RMS belongs (or the reverse) is the documented error mode here.
The definitions, from NCERT Class 12 Physics Chapter 7, Alternating Current, page 179:
I_rms = I₀/√2 and V_rms = V₀/√2, for a pure sinusoid.
RMS means root-mean-square: square the instantaneous value, average over a full cycle, take the root. The squaring is why it is not zero — a sinusoid's plain average over a cycle is zero, which is why the RMS value exists at all. The factor 1/√2 ≈ 0.707 comes from ⟨sin²⟩ = ½ over a complete cycle, and it holds only for a pure sinusoid. A square or triangular waveform has a different factor; NEET stems that say "alternating" without saying "sinusoidal" are worth a second look.
Why RMS is the quantity that gets stated: it is the DC value that would dissipate the same heat in the same resistor. Meters read it; ratings quote it; power formulas take it.
Watch-out: before you multiply or divide by √2, say out loud which of the two you were given. Half of the marks lost here go to students who knew the formula perfectly and applied it in the wrong direction.
Can you answer these Peak RMS AC MCQs?
Select an option to see the explanation. Wrong answers show why your choice was tempting — and name the exact trap it exploits.
MCQ 1Easy RecallPractice
For a sinusoidal alternating voltage of peak value V₀, the RMS value is given by
Show answer and why every option is right or wrong
Answer: C. C is correct. For a pure sinusoid, V_rms = V₀/√2, stated in NCERT Class 12 Physics, Alternating Current chapter, page 179.
Why A is wrong: A inverts the relation — multiplying by √2 converts RMS to peak, not peak to RMS, and would make the RMS value exceed the peak, which is impossible for any waveform.
Why B is wrong: B uses a factor of 2 instead of √2; the square root arises because the mean of sin² over a cycle is ½ and the root is then taken.
Why D is wrong: D is the mean value of a sinusoid over a half cycle (2V₀/π ≈ 0.637 V₀), a different average that is not the RMS value.
MCQ 2Easy RecallPractice
The RMS value of an alternating current is defined as the value of steady direct current that, in the same resistance over the same time, produces
Show answer and why every option is right or wrong
Answer: B. B is correct. The RMS (effective) value is defined by equal heating effect in the same resistance — the basis for the definition given in NCERT Class 12 Physics, Alternating Current chapter, page 179.
Why A is wrong: A confuses RMS with peak; a DC of magnitude I_rms never reaches the AC peak value, and equality of peaks is not what the definition requires.
Why C is wrong: C describes the mean current, since charge transferred is the time-integral of current; for a full sinusoidal cycle that integral is zero while the RMS value is not.
Why D is wrong: D is the plain time-average, which is zero over a complete sinusoidal cycle and therefore cannot serve as an effective value.
MCQ 3Concept TrapPractice
An AC source is specified in a question as v = 150 sin(ωt) volt. The number 150 in this expression is
Show answer and why every option is right or wrong
Answer: C. C is correct. In v = V₀ sin(ωt) the coefficient of the sine is the amplitude, that is the peak value — the standard form used in NCERT Class 12 Physics, Alternating Current chapter, page 179.
Why A is wrong: A applies the mains-labelling convention to an equation. A stated supply voltage is RMS, but a coefficient in a sinusoidal expression is the amplitude; here the RMS value would be 150/√2 ≈ 106 V.
Why B is wrong: B is wrong because the half-cycle mean of this waveform is 2(150)/π ≈ 95.5 V, not 150 V; the sine coefficient is never the mean.
Why D is wrong: D would require the coefficient to span both extremes; the waveform runs from +150 V to −150 V, so peak-to-peak is 3.0 × 10² V, twice the coefficient.
MCQ 4Direct ApplicationPractice
A sinusoidal alternating voltage has an RMS value of 1.5 × 10² V. Its peak value is closest to
Show answer and why every option is right or wrong
Answer: A. A is correct. V₀ = √2 V_rms = 1.414 × 1.5 × 10² ≈ 2.1 × 10² V, applying the relation from NCERT Class 12 Physics, Alternating Current chapter, page 179.
Why B is wrong: B treats peak and RMS as equal, the documented peak-versus-RMS substitution error; the peak of a sinusoid always exceeds its RMS value.
Why C is wrong: C divides by √2 instead of multiplying, converting in the wrong direction: 1.5 × 10²/√2 ≈ 1.1 × 10² V is the RMS of a 1.5 × 10² V peak, not the peak of a 1.5 × 10² V RMS.
Why D is wrong: D doubles the RMS value, using a factor of 2 where the sinusoidal relation requires √2.
MCQ 5Direct ApplicationPractice
An alternating current is described by i = 8.0 sin(ωt + π/3) ampere. The RMS current is closest to
Show answer and why every option is right or wrong
Answer: D. D is correct. The amplitude is 8.0 A, so I_rms = 8.0/√2 ≈ 5.7 A; the phase constant π/3 shifts the waveform in time and does not change its amplitude. Relation from NCERT Class 12 Physics, Alternating Current chapter, page 179.
Why A is wrong: A reports the peak value itself, the peak-versus-RMS substitution error; 8.0 A is the amplitude, not the effective value.
Why B is wrong: B halves the amplitude, using 2 in place of √2 in the conversion.
Why C is wrong: C multiplies by √2 rather than dividing, so it converts in the wrong direction and returns a value larger than the peak — impossible for an RMS value.
MCQ 6Concept TrapPractice
The relation I_rms = I₀/√2 may be applied to
Show answer and why every option is right or wrong
Answer: B. B is correct. The factor 1/√2 follows from the mean of sin² over a complete cycle being ½, so it is specific to the sinusoidal waveform — the condition attached to the formula in NCERT Class 12 Physics, Alternating Current chapter, page 179.
Why A is wrong: A generalises a sinusoid-specific result; the RMS-to-peak ratio depends on waveform shape, and a triangular wave, for instance, gives I₀/√3.
Why C is wrong: C names the one common waveform for which the factor is definitely not 1/√2: a symmetric square wave spends all its time at ±I₀, so its RMS value equals I₀.
Why D is wrong: D drops the periodicity that the cycle-average in the derivation depends on, and also generalises beyond the sine.
MCQ 7CalculationPractice
A resistor of 40 Ω is connected across a sinusoidal AC source described by v = 160 sin(100πt) volt. The RMS current through the resistor is closest to
Show answer and why every option is right or wrong
Answer: C. C is correct. The coefficient 160 is the peak voltage, so V_rms = 160/√2 ≈ 113 V and I_rms = 113/40 ≈ 2.8 A. Both steps use the relation from NCERT Class 12 Physics, Alternating Current chapter, page 179.
Why A is wrong: A divides the peak voltage by the resistance without converting to RMS first: 160/40 = 4.0 A is the peak current, and reporting it as RMS is the peak-versus-RMS substitution error.
Why B is wrong: B multiplies by √2 instead of dividing, so 4.0 × 1.414 ≈ 5.7 A exceeds even the peak current — the conversion applied in the wrong direction.
Why D is wrong: D divides by 2 rather than √2 after finding the peak current, giving 4.0/2 = 2.0 A.
MCQ 8CalculationPractice
Two sinusoidal AC sources are available. Source P is labelled "2.0 × 10² V" on its casing. Source Q is described by the expression v = 2.0 × 10² sin(ωt) volt. Each is connected in turn across the same resistor. Compared with source P, source Q delivers
Show answer and why every option is right or wrong
Answer: D. D is correct. P's label is an RMS value, so V_rms,P = 2.0 × 10² V; Q's coefficient is a peak value, so V_rms,Q = 2.0 × 10²/√2. Average power in a resistor goes as V_rms², giving a ratio of (1/√2)² = ½.
Why A is wrong: A takes both quoted numbers to mean the same quantity. A label on a supply is RMS; a coefficient in a sinusoidal expression is the peak — this is exactly the substitution the topic warns about.
Why B is wrong: B inverts the comparison: source Q has the smaller RMS value, so it delivers less power, not more.
Why C is wrong: C applies the √2 to the power directly. The voltage ratio is 1/√2, but power depends on the square of the RMS voltage, so the power ratio is ½, not 1/√2.
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Peak RMS AC: quick recall before you leave
How do you solve a Peak RMS AC question? A worked example
Pattern: peak ↔ RMS conversion for sinusoidal AC (the in-scope PYQ pattern for this topic; observed 2022).
- 1
Given
• An alternating current: i = 6.0 sin(314t) A, with t in seconds.• Resistance in the circuit: R = 25 Ω.
- 2
Required
• The RMS current.• The average power dissipated in the resistor.
- 3
Concept
The coefficient of the sine in an expression for instantaneous current is the amplitude, that is the peak current I₀. RMS is the effective value — the steady DC that would heat the same resistor at the same rate — and for a sinusoid it is smaller than the peak by a factor √2. Because RMS is defined by equal heating, average power in a pure resistor is computed from the RMS current, never the peak.
- 4
Formula
I_rms = I₀/√2
P_avg = I_rms² R (pure resistance) - 5
Substitution
I₀ = 6.0 A (read directly as the sine coefficient)
I_rms = 6.0 / √2
Then P_avg = (I_rms)² × 25 - 6
Calculation
I_rms = 6.0 / 1.4142 = 4.2426 A ≈ 4.2 A
I_rms² = (6.0)²/2 = 36/2 = 18.0 A²
P_avg = 18.0 × 25 = 450 W ≈ 4.5 × 10² W
The 2 in I₀²/2 comes from the exact relation (√2)² = 2, and √2 is a mathematical constant — neither contributes to the significant-figure count. The given data carry two significant figures, so both answers are reported to two. - 7
Final answer
I_rms ≈ 4.2 A; P_avg ≈ 4.5 × 10² W.
- 8
Common trap
Using the peak current in the power formula: 6.0² × 25 = 900 W, exactly twice the correct answer. The factor-of-2 error looks plausible and a doubled value is a standard distractor. The reverse slip — reading 6.0 A as an RMS value already and multiplying it by √2 to "find" 8.5 A — is the same substitution error running the other way. Before either operation, name which quantity the question gave you: a sine coefficient is a peak, a stated supply rating is RMS.
- 9
Similar NEET-style question
An AC voltage v = 141 sin(100πt) volt is applied across a 20 Ω resistor. Find the RMS voltage, the RMS current and the average power dissipated. *(Answers: ≈ 1.0 × 10² V, ≈ 5.0 A, ≈ 5.0 × 10² W.)*
What to remember before solving Peak RMS AC questions
Which Peak RMS AC formulas do you need for NEET?
1 formula — click to collapse
RMS values of sinusoidal AC
RMS values for pure sinusoid. Stated AC voltages are RMS unless noted.
| Symbol | Quantity | SI Unit |
|---|---|---|
| I_rms | RMS current | A |
| I_peak | peak current | A |
| V_rms | RMS voltage | V |
| V_peak | peak voltage | V |
Valid when
- Pure sinusoid
- Steady state
Where do students lose marks on Peak RMS AC?
These are the exact patterns that cause wrong answers in NEET. Each trap includes when it triggers and how to avoid it.
1 item — click to collapse
Root cause: unit error
Correction
RMS = peak/√2 for sinusoidal AC. Stated AC voltages (e.g. 220 V mains) are RMS unless explicitly noted as peak. Power formulas use RMS.
More in Electromagnetic Induction and Alternating Currents: 3 exam traps and mistakes · 10 formulas · 5 question patterns from its other lessons.
Peak RMS AC questions from past NEET papers
1 question from NEET 2022. Answers verified against NTA official keys. — click to collapse
NEET 2022
The peak voltage of the ac source is equal to
11/ 2 times the rms value of the ac source
2The value of voltage supplied to the circuit
3The rms value of the ac source
42 times the rms value of the ac source
NTA Answer: Option 4(final)
All 20 past-paper questions from Electromagnetic Induction and Alternating Currents →
How does NEET ask about Peak RMS AC?
1 recurring pattern from past papers — click to collapse
Convert between peak and RMS for sinusoidal AC: factor sqrt(2).
Direct ApplicationEasy
Common distractors
forgets sqrt 2
Treats peak = RMS
Sources
Page numbers are the ones printed in the current NCERT textbook (2023 rationalised edition), unless marked pre-2023. The books are free at ncert.nic.in.
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