Key Fact
Diode as rectifier
Half-wave rectifier: single diode, one half of AC cycle passes. Full-wave rectifier: 2 diodes (centre-tap) or 4 diodes (bridge). Capacitor smoothing reduces ripple.
-- NCERT Class 12 Physics, Ch. 14, p. 338Diode As Rectifier
8 MCQs1 revision card9-step worked example
Source: NCERT Electronic DevicesPYQ coverage: NEET 2022, 2023Official key: NTA-verifiedLast updated: 23 Sep 2026
Diode As Rectifier, explained for NEET
The trap that costs marks here is a single number: the output frequency. Half-wave and full-wave rectifiers are drawn on the same page, taught in the same hour, and their ripple frequencies differ by a factor of two. Aspirants who "know rectifiers" still swap them under time pressure.
Fix it by reasoning from the waveform, not from a remembered pair of answers. A half-wave rectifier passes one half-cycle and blocks the other. In one input period you get one output pulse, so f_out = f_in. A full-wave rectifier (centre-tap or bridge) redirects the negative half-cycle so it emerges with the same polarity as the positive one. In one input period you now get two output pulses, so f_out = 2 f_in. NCERT Class 12 Physics Part 2, Chapter 14, page 334 states this directly.
Count pulses per input cycle. That single habit removes the confusion permanently.
Two consequences follow from the same waveform picture, and NEET asks about both. First, the full-wave output is easier to smooth: its pulses are twice as frequent, so a capacitor filter has half the time to discharge between them and the ripple is smaller for the same capacitance. Second, a rectifier's output is unidirectional but not steady — it is pulsating DC, not DC. A question that calls raw rectifier output "steady DC" is testing whether you read carefully.
For the bridge circuit, note that two of the four diodes conduct on each half-cycle, in series, and the other two are reverse-biased. The load current flows through it in the same direction in both cases. That is the whole mechanism; the diode's own forward/reverse behaviour is the subject of the I-V characteristics lesson.
Watch out: if an Indian mains frequency of 50 Hz is given, full-wave output ripple is 100 Hz, not 50 Hz and not 25 Hz.
Can you answer these Diode As Rectifier 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
In a half-wave rectifier supplied with an AC input, how many output pulses appear across the load during one complete cycle of the input?
Show answer and why every option is right or wrong
Answer: A. A is correct. A half-wave rectifier conducts during only one half-cycle of the input and blocks the other, so exactly one pulse appears per input cycle — the basis of f_out = f_in, as described in NCERT Class 12 Physics Part 2, Chapter 14, page 334.
Why B is wrong: B is wrong because two pulses per input cycle is the full-wave result, obtained when the negative half-cycle is redirected rather than blocked; this is the half-wave/full-wave swap.
Why C is wrong: C is wrong because four pulses per cycle corresponds to no single-phase rectifier configuration in the syllabus; a bridge rectifier has four diodes but still yields two pulses per cycle.
Why D is wrong: D is wrong because a rectifier does produce output; zero pulses would mean the diode never conducts.
MCQ 2Direct ApplicationPractice
A full-wave bridge rectifier is fed from the 50 Hz Indian mains. The ripple frequency of the unfiltered output is
Show answer and why every option is right or wrong
Answer: C. C is correct. For a full-wave rectifier f_out = 2 f_in = 2 × 50 Hz = 100 Hz, because both half-cycles are delivered to the load with the same polarity (NCERT Class 12 Physics Part 2, Chapter 14, page 334).
Why A is wrong: A is wrong because halving the input frequency inverts the relation; no rectifier configuration produces an output slower than its input.
Why B is wrong: B is wrong because 50 Hz is the half-wave answer, applied here to a full-wave circuit — the classic half-wave/full-wave frequency swap.
Why D is wrong: D is wrong because 200 Hz would require four output pulses per input cycle; a single-phase full-wave rectifier delivers two.
MCQ 3Easy RecallPractice
The output of a rectifier operating without any filter is best described as
Show answer and why every option is right or wrong
Answer: B. B is correct. The unfiltered output is unidirectional but its magnitude varies between zero and the peak each pulse, which is what "pulsating DC" means (NCERT Class 12 Physics Part 2, Chapter 14, page 334).
Why A is wrong: A is wrong because steady DC requires a filter; the raw rectified waveform drops to zero between pulses, so its magnitude is not constant.
Why C is wrong: C is wrong because the output never reverses direction, so it is not AC, even though its pulses recur at twice the input frequency in the full-wave case.
Why D is wrong: D is wrong because the rectifier's whole function is to remove the reversal of direction; the output is unidirectional.
MCQ 4Direct ApplicationPractice
The unfiltered output of a rectifier is observed to have a ripple frequency exactly equal to the frequency of the AC source supplying it. The circuit is
Show answer and why every option is right or wrong
Answer: C. C is correct. f_out = f_in identifies the half-wave case, in which one half-cycle is blocked and a single pulse appears per input cycle (NCERT Class 12 Physics Part 2, Chapter 14, page 334).
Why A is wrong: A is wrong because a centre-tap full-wave rectifier delivers both half-cycles to the load, giving f_out = 2 f_in.
Why B is wrong: B is wrong because a bridge rectifier is also full-wave; its four diodes conduct in pairs on alternate half-cycles, still producing two pulses per input cycle.
Why D is wrong: D is wrong on two counts: the question specifies an unfiltered output, and a full-wave circuit would give twice the input frequency in any case.
MCQ 5CalculationPractice
A half-wave rectifier and a full-wave rectifier are each supplied from the same AC source. The interval between successive output pulses of the half-wave circuit is measured as 2.0 × 10⁻² s. The interval between successive output pulses of the full-wave circuit is
Show answer and why every option is right or wrong
Answer: B. B is correct. The half-wave interval equals the input period, so T_in = 2.0 × 10⁻² s and f_in = 50 Hz. The full-wave output has f_out = 2 f_in = 100 Hz, giving an interval of 1/100 = 1.0 × 10⁻² s (NCERT Class 12 Physics Part 2, Chapter 14, page 334).
Why A is wrong: A is wrong because it quarters the input period, which would need four output pulses per input cycle rather than two.
Why C is wrong: C is wrong because it leaves the interval unchanged, i.e. it applies the half-wave relation f_out = f_in to the full-wave circuit — the half-wave/full-wave swap expressed in the time domain.
Why D is wrong: D is wrong because it doubles the interval, halving the frequency; this inverts the full-wave relation f_out = 2 f_in.
MCQ 6Concept TrapPractice
Two identical capacitor filters are used, one across a half-wave rectifier and one across a full-wave rectifier, both fed from the same AC source and driving equal loads. The full-wave arrangement gives the smaller ripple because
Show answer and why every option is right or wrong
Answer: A. A is correct. Ripple depends on how far the capacitor discharges between pulses; a full-wave output delivers two pulses per input cycle, halving the discharge interval for the same RC (NCERT Class 12 Physics Part 2, Chapter 14, page 334).
Why B is wrong: B is wrong because both circuits charge the capacitor to the same peak of the input waveform; full-wave rectification changes the pulse rate, not the peak value.
Why C is wrong: C is wrong because it assigns the half-wave relation f_out = f_in to the full-wave circuit, and then misuses it as the reason for better smoothing — more frequent pulses, not fewer, reduce ripple.
Why D is wrong: D is wrong because the diodes are the same devices in both circuits; the difference lies in how many half-cycles reach the load.
MCQ 7Direct ApplicationPractice
In a four-diode bridge rectifier, during any one half-cycle of the input the number of diodes conducting is
Show answer and why every option is right or wrong
Answer: B. B is correct. In a bridge rectifier the diodes conduct in diagonally opposite pairs: two are forward-biased and carry the load current in series while the other two are reverse-biased, and the pairs exchange roles on the next half-cycle (NCERT Class 12 Physics Part 2, Chapter 14, page 334).
Why A is wrong: A is wrong because a single conducting diode describes the half-wave circuit; a bridge needs a complete path through two diodes for each half-cycle.
Why C is wrong: C is wrong because three conducting diodes would short one arm of the bridge; conduction is strictly in pairs.
Why D is wrong: D is wrong because all four conducting simultaneously would short the source; the reverse-biased pair blocks on each half-cycle.
MCQ 8CalculationPractice
An unfiltered rectifier output is displayed on an oscilloscope and 8 identical unidirectional pulses are counted in an interval of 4.0 × 10⁻² s. The source frequency is 100 Hz. The circuit is
Show answer and why every option is right or wrong
Answer: B. B is correct. The observed output frequency is 8 pulses ÷ (4.0 × 10⁻² s) = 200 Hz, which is twice the 100 Hz source; f_out = 2 f_in identifies a full-wave rectifier (NCERT Class 12 Physics Part 2, Chapter 14, page 334).
Why A is wrong: A is wrong because a half-wave circuit would give f_out = f_in = 100 Hz, i.e. 4 pulses in the stated interval, not 8; choosing it applies the half-wave relation to a doubled output.
Why C is wrong: C is wrong because 200 Hz output from a 100 Hz source is exactly the standard full-wave result.
Why D is wrong: D is wrong because a filter smooths the pulses towards steady DC rather than doubling their number, and the half-wave rate would still be 100 Hz.
Free NEET study resources
Get a structured 30-day Mechanics plan and a complete formula booklet — delivered to your inbox instantly.
Diode As Rectifier: quick recall before you leave
How do you solve a Diode As Rectifier question? A worked example
- 1
Given
• Input AC frequency to the rectifier, f_in = 50 Hz (exact, stated mains value)• Circuit: centre-tap full-wave rectifier, no filter• The same source is later applied to a half-wave rectifier
- 2
Required
• The ripple frequency of the full-wave output• The time interval between successive pulses of the half-wave output
- 3
Concept
A rectifier delivers one output pulse for each input half-cycle it passes to the load. A half-wave circuit passes one half-cycle per input period; a full-wave circuit passes both, inverting the negative one so that it reaches the load with the same polarity. The output frequency is therefore fixed by pulses per input cycle, not by the diode count.
- 4
Formula
f_out = f_in (half-wave); f_out = 2 f_in (full-wave); and T = 1/f.
- 5
Substitution
• Full-wave: f_out = 2 × 50 Hz• Half-wave: f_out = 50 Hz, so T_out = 1/(50 Hz)
- 6
Calculation
• Full-wave: f_out = 100 Hz• Half-wave: T_out = 1/50 = 2.0 × 10⁻² s
The factor 2 in f_out = 2 f_in is a counting integer — it counts the two half-cycles delivered per input period — and the mains value 50 Hz is a defined exact figure here. Neither limits the significant figures of the answer; the two-significant-figure form 2.0 × 10⁻² s is written to state the precision claimed, not to record a rounding of measured data. - 7
Final answer
Full-wave ripple frequency = 1.0 × 10² Hz. Half-wave pulse interval = 2.0 × 10⁻² s.
- 8
Common trap
The swap: writing 50 Hz for the full-wave output or 100 Hz for the half-wave. This is the distractor type
swaps-half-full-waverecorded for this pattern, and it survives revision because both numbers are "correct answers" for the topic — only the circuit assignment differs. Recover by counting pulses per input cycle before writing any number. A second version of the same error appears in the time domain, where the half-wave interval is quoted as 1.0 × 10⁻² s. - 9
Similar NEET-style question
A bridge rectifier without a filter is connected to an AC source. An oscilloscope shows successive output pulses separated by 5.0 × 10⁻³ s. The frequency of the AC source is (a) 50 Hz (b) 100 Hz (c) 200 Hz (d) 400 Hz. *(Answer: (b) — f_out = 1/(5.0 × 10⁻³ s) = 200 Hz, and for a full-wave bridge f_in = f_out/2 = 100 Hz.)*
What to remember before solving Diode As Rectifier questions
Which Diode As Rectifier formulas do you need for NEET?
1 formula — click to collapse
Rectifier output frequency
Output ripple frequency relative to input AC frequency.
| Symbol | Quantity | SI Unit |
|---|---|---|
| f_in | input AC frequency | Hz |
| f_out | output ripple | Hz |
Valid when
- Single-phase AC input
- No filter (raw rectified output)
Where do students lose marks on Diode As Rectifier?
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: concept gap
Correction
Half-wave: f_out = f_in (only one half-cycle passes). Full-wave: f_out = 2 f_in (both half-cycles flipped to same polarity).
More in Electronic Devices: 2 exam traps and mistakes · 2 formulas · 2 question patterns from its other lessons.
Diode As Rectifier questions from past NEET papers
2 questions from NEET 2022, 2023. Answers verified against NTA official keys. — click to collapse
NEET 2023
1p-n junction diodes
2Capacitor
3Load resistance
4A centre-tapped transformer
NTA Answer: Option 2(final)
NEET 2022
In half wave rectification, if the input frequency is 60 Hz, then the output frequency would be
1120 Hz
2Zero
330 Hz
460 Hz
NTA Answer: Option 4(final)
How does NEET ask about Diode As Rectifier?
1 recurring pattern from past papers — click to collapse
Determine bias direction or rectified output. Half vs full wave rectifier output frequency.
InterpretationEasy
Common distractors
swaps half full wave
Confuses rectifier types
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.
Test yourself on this topic with real past-paper questions:
Practice this topic →