Definition
AC source and waveform
EMF varying sinusoidally with time: ε(t) = ε_m sin(ωt), I(t) = I_m sin(ωt - φ). Frequency f = ω/(2π); Indian standard: f = 50 Hz.
-- NCERT Class 12 Physics, Ch. 7, p. 178Alternating Currents
8 MCQs1 revision card9-step worked example
Source: NCERT Electromagnetic Induction and Alternating CurrentsOfficial key: NTA-verifiedLast updated: 23 Sep 2026
Alternating Currents, explained for NEET
The number printed on an AC source is not the number the waveform ever reaches. Write "220 V" on a socket and the voltage there swings to about 311 V twice every cycle. Aspirants who carry the mains figure straight into a peak-value slot lose the mark before any physics begins — this is the classic peak-versus-RMS substitution error, and it is why this topic sits ahead of reactance, power and resonance in the chapter.
NCERT Class 12 Physics Chapter 7 opens (page 178) by defining alternating current as a current whose magnitude changes continuously with time and whose direction reverses periodically. Two consequences follow immediately. First, the time-average of a sinusoidal AC over a full cycle is zero — the positive and negative halves cancel exactly — so the arithmetic mean is useless as a rating. Second, heating does not cancel, because dissipated power goes as the square of the current. That is the whole reason the root-mean-square value exists: it is the DC value that would heat the same resistor at the same rate.
For a pure sinusoid the relation is I_rms = I₀/√2 and V_rms = V₀/√2, giving I_rms ≈ 0.707 I₀. Hold two conventions firmly. Any AC voltage or current quoted without qualification is RMS. AC meters read RMS. Peak values appear only when a question says "peak", "maximum", or writes the source as v = v₀ sin(ωt), where v₀ is by definition the peak.
Frequency deserves one line here because NEET uses it as a quiet discriminator. Indian mains runs at 50 Hz, so the current completes 50 cycles per second — but it passes through zero twice per cycle, i.e. 100 times per second, and a filament lamp therefore dims at 100 Hz, not 50 Hz.
Watch-out: √2 ≈ 1.414 and 1/√2 ≈ 0.707. Multiplying when you should divide moves the answer by a factor of two, and that wrong value is always on the option list.
Can you answer these Alternating Currents 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
An alternating current is best defined as a current that
Show answer and why every option is right or wrong
Answer: C. C matches the definition given at the opening of NCERT Class 12 Physics Chapter 7, page 178 — both the magnitude varies continuously and the direction reverses periodically.
Why A is wrong: A is wrong because a unidirectional current of varying magnitude is pulsating DC, not AC; the defining feature of AC is the periodic reversal of direction.
Why B is wrong: B is wrong because it describes a square-wave-like reversal of a fixed magnitude; the NCERT definition requires the magnitude to vary continuously as well.
Why D is wrong: D is wrong because it confuses the current itself with its full-cycle average. The instantaneous current is zero only at isolated instants; it is the mean over a cycle that vanishes.
MCQ 2Easy RecallPractice
For a sinusoidal alternating current, the average value taken over one complete cycle is
Show answer and why every option is right or wrong
Answer: B. B is correct: over a full cycle the positive and negative half-cycles of a sinusoid are equal and opposite, so they cancel exactly (NCERT Class 12 Physics Chapter 7, around page 179).
Why A is wrong: A is wrong because the instantaneous current equals the peak value only at two instants per cycle, not on average.
Why C is wrong: C is wrong because peak/√2 is the RMS value, not the average value. The question asks for the mean of the current itself, which is a different quantity.
Why D is wrong: D is wrong because 0.637 I₀ (= 2I₀/π) is the average over a HALF cycle. Over a full cycle the two halves cancel and the mean is zero.
MCQ 3Easy RecallPractice
An AC ammeter connected in a circuit carrying sinusoidal alternating current displays a steady non-zero reading. The quantity it indicates is the
Show answer and why every option is right or wrong
Answer: C. C is correct — AC meters are calibrated to display RMS values, which is also why a quoted AC rating is RMS unless stated otherwise (NCERT Class 12 Physics Chapter 7, page 180).
Why A is wrong: A is wrong because peak values are reported only when a question explicitly says 'peak' or 'maximum', or writes the source as v₀ sin(ωt). Treating a meter reading as peak is the peak-versus-RMS substitution error.
Why B is wrong: B is wrong because the full-cycle average of sinusoidal AC is zero, so a meter reading it would show nothing at all.
Why D is wrong: D is wrong because the instantaneous value oscillates far too rapidly at mains frequency for a pointer or display to follow; the meter responds to a time-averaged heating effect.
MCQ 4Direct ApplicationPractice
A sinusoidal alternating current has a peak value of 4.0 A. Its RMS value is closest to
Show answer and why every option is right or wrong
Answer: A. A is correct: I_rms = I₀/√2 = 4.0/1.414 = 2.8 A, applying the RMS relation for a pure sinusoid from NCERT Class 12 Physics Chapter 7, page 180.
Why B is wrong: B is wrong because it multiplies by √2 instead of dividing (4.0 × 1.414). This is the direction-of-conversion error: RMS is always SMALLER than peak for a sinusoid.
Why C is wrong: C is wrong because it treats RMS as equal to peak, dropping the √2 factor entirely.
Why D is wrong: D is wrong because it halves the peak value. The conversion factor is 1/√2 ≈ 0.707, not 1/2.
MCQ 5Direct ApplicationPractice
An AC supply is labelled 220 V. The peak voltage of this supply is closest to
Show answer and why every option is right or wrong
Answer: C. C is correct: a quoted AC rating is RMS, so V₀ = V_rms × √2 = 220 × 1.414 ≈ 311 V (NCERT Class 12 Physics Chapter 7, page 180).
Why A is wrong: A is wrong because it assumes the labelled value is already the peak. The stated value of an AC supply is the RMS value unless the question says otherwise.
Why B is wrong: B is wrong because it divides by √2 instead of multiplying — the conversion has been run backwards, treating 220 V as the peak and solving for RMS.
Why D is wrong: D is wrong because it doubles the RMS value. The peak-to-RMS factor is √2 ≈ 1.414, not 2.
MCQ 6Direct ApplicationPractice
An alternating current is described by i = 5.0 sin(100πt), with i in amperes and t in seconds. The RMS current and the frequency of the supply are respectively
Show answer and why every option is right or wrong
Answer: D. D is correct: the coefficient 5.0 A is the peak, so I_rms = 5.0/√2 = 3.5 A; and ω = 100π rad/s gives f = ω/2π = 50 Hz (NCERT Class 12 Physics Chapter 7, page 180).
Why A is wrong: A is wrong on the current: in the form i₀ sin(ωt) the coefficient is by definition the peak value, not the RMS value.
Why B is wrong: B is wrong on the frequency: it reads the numerical coefficient 100 from 100πt directly as the frequency, forgetting that 100π is ω and that f = ω/2π.
Why C is wrong: C is wrong because it multiplies the peak by √2 (5.0 × 1.414) rather than dividing — the conversion is applied in the wrong direction.
MCQ 7Concept TrapPractice
A resistor dissipates heat at a certain average rate when carrying sinusoidal alternating current. The RMS value of that current is defined as the value of the
Show answer and why every option is right or wrong
Answer: A. A is correct — the RMS value is defined by equal heating effect, which is exactly why it survives when the plain average does not (NCERT Class 12 Physics Chapter 7, page 179).
Why B is wrong: B is wrong because that instant gives the peak current, not the RMS current. Peak describes a single moment; RMS describes a full-cycle effect.
Why C is wrong: C is wrong because it uses equal charge transfer as the criterion. Charge transfer over a full cycle of AC is zero, since the halves cancel — it is the SQUARE of the current, which never goes negative, that gives a usable rating.
Why D is wrong: D is wrong because the half-cycle mean is 0.637 I₀, a different quantity from the RMS value 0.707 I₀; defining RMS by averaging the current rather than its square loses the heating connection.
MCQ 8CalculationPractice
An alternating current source operating at 50 Hz drives a lamp. The number of times the instantaneous current in the lamp becomes zero in one second is
Show answer and why every option is right or wrong
Answer: C. C is correct: a sinusoid crosses zero twice per cycle (once entering the negative half, once returning), and there are 50 cycles per second, giving 2 × 50 = 100 (NCERT Class 12 Physics Chapter 7, page 178, where the periodic reversal of direction is defined).
Why A is wrong: A is wrong because it halves the frequency rather than doubling it, counting one zero every two cycles.
Why B is wrong: B is wrong because it counts one zero per cycle. Each full sinusoidal cycle passes through zero twice, since the direction reverses twice — this second step is the one most often skipped.
Why D is wrong: D is wrong because it applies a factor of four. There are two direction reversals per cycle, not four, so the count is 2f and not 4f.
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Alternating Currents: quick recall before you leave
How do you solve a Alternating Currents question? A worked example
- 1
Given
An alternating voltage source is written as v = 3.11 × 10² sin(100πt) volts, with t in seconds. The source supplies a resistor of resistance 1.10 × 10² Ω.
- 2
Required
The RMS voltage of the source, the frequency, and the RMS current through the resistor.
- 3
Concept
The coefficient in front of the sine is the peak (maximum) voltage by definition — the sine factor never exceeds 1, so v reaches v₀ when sin(ωt) = 1. The RMS value is the smaller, heating-equivalent value obtained by dividing by √2. The coefficient of t inside the sine is the angular frequency ω, related to the ordinary frequency by ω = 2πf.
- 4
Formula
V_rms = V₀/√2 · f = ω/2π · I_rms = V_rms/R
- 5
Substitution
V₀ = 3.11 × 10² V (read directly off the sine coefficient)
ω = 100π rad/s (read off the coefficient of t)
V_rms = (3.11 × 10²)/√2
f = 100π/(2π)
I_rms = V_rms/(1.10 × 10²) - 6
Calculation
V_rms = 311/1.414 = 2.20 × 10² V
f = 100/2 = 5.0 × 10¹ Hz
I_rms = 220/110 = 2.00 A
The constants π and √2 are exact mathematical constants, and the 2 in ω = 2πf is an exact counting factor. None of them contributes to the significant-figure count — the three significant figures in the answers come from the given data, 3.11 × 10² V and 1.10 × 10² Ω. - 7
Final answer
V_rms = 2.20 × 10² V, f = 5.0 × 10¹ Hz, I_rms = 2.00 A.
- 8
Common trap
The peak-versus-RMS substitution error. Two forms appear here. First, feeding 311 V straight into I = V/R, which returns 2.83 A — the peak current, not the RMS current the question asks for. Second, reading "100" from 100πt as the frequency and answering 100 Hz; 100π is ω, and f = ω/2π = 50 Hz. Both wrong values will be sitting on the option list.
- 9
Similar NEET-style question
An alternating current is given by i = 1.41 sin(314t) A. Find its RMS value and the frequency of the supply. *(Take the coefficient as the peak; ω = 314 rad/s ≈ 100π.)*
What to remember before solving Alternating Currents questions
More in Electromagnetic Induction and Alternating Currents: 4 exam traps and mistakes · 11 formulas · 6 question patterns from its other lessons.
Alternating Currents questions from past NEET papers
No question in our NEET 2020–2025 set targets this topic directly.
All 20 past-paper questions from Electromagnetic Induction and Alternating Currents →
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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