Question 1

Waveform terminology (digital and timing): A rectangular wave having a 50% duty cycle is correctly referred to as which standard waveform shape?

Accepted Answer

Introduction: Square and rectangular waves are ubiquitous in digital electronics, timing circuits, and pulse-width modulation. Knowing the precise terminology helps in reading datasheets, setting test equipment, and interpreting signal integrity discussions. Given Data / Assumptions: A periodic rectangular waveform is described. Its duty cycle is 50% (the high level persists for half the period). Ideal edges (negligible rise/fall time) are implied for the definition. Concept / Approach: A “square wave” is a special case of a rectangular wave in which the high and low durations are equal. That is exactly a 50% duty cycle. Other shapes—sawtooth and triangular—feature linear ramps rather than flat high/low intervals. Step-by-Step Solution: Define duty cycle: D = (time high) / (period).For a square wave, D = 0.5 (50%).Given D = 50%, the correct term is “square wave”. Verification / Alternative check: Consult standard signal definitions used in oscilloscopes and function generators: manufacturers define “square” as a rectangular wave with 50% duty cycle, confirming the classification. Why Other Options Are Wrong: ""c"" wave: Not a standard signal category. sawtooth wave: Has a linear rise and sudden fall (or vice versa), not two flat levels. triangular wave: Has linear rise and linear fall with equal slopes, not flat tops. rectified sine wave: Derived from taking the absolute value of a sine wave, not a rectangular pulse train. Common Pitfalls: Calling any rectangular wave “square” even when duty cycle ≠ 50%. Ignoring finite rise/fall time effects, which in practice slightly distort ideal shapes. Final Answer: square wave

Question 2

Single-phase induction motor starting characteristics: Which statement is correct for a basic single-phase induction motor (without special starting aids)?

Accepted Answer

Introduction: Single-phase induction motors are widely used in household and light-industrial applications. Their starting behavior and speed regulation differ markedly from three-phase machines and synchronous motors. Understanding these distinctions is important for selection, control, and troubleshooting. Given Data / Assumptions: Motor type: single-phase induction motor (SPIM). No explicit mention of auxiliary starting means (start winding, capacitor, shaded pole). Steady operation on single-phase supply with typical slip characteristics. Concept / Approach: A plain single-phase induction motor without an auxiliary starting mechanism is not inherently self-starting because a single-phase stator field alone does not produce an initial rotating magnetic field. In practice, SPIMs include auxiliary windings/capacitors or shaded poles to create a phase shift for starting. Once running, they behave like induction machines with slip; therefore they do not operate at a perfectly fixed speed (speed varies slightly with load). Reliability comparisons to “three-phase synchronous motors” are context-dependent and not a general truth. Step-by-Step Reasoning: Self-starting claim: False for a bare SPIM; true only when equipped with starting aids such as capacitor-start or shaded-pole designs.Fixed speed claim: Induction motors always run with slip; speed is near but below synchronous speed and load-dependent.Reliability claim: Not an inherent attribute; depends on design, quality, and application. No universal statement applies. Verification / Alternative check: Standard motor theory shows that a single-phase sinusoidal field decomposes into two counter-rotating fields of equal magnitude, producing zero starting torque without asymmetry or auxiliary windings—explaining the need for starting mechanisms. Why Other Options Are Wrong: is self-starting: Only when augmented with start windings/capacitors; not strictly true in the generic statement. operates at a fixed speed: Induction motors have slip; speed changes with load. is less reliable than a three-phase synchronous motor: Not a fundamental rule. starts by itself without any auxiliary means under all conditions: Overly absolute and incorrect. Common Pitfalls: Confusing induction and synchronous operation with respect to speed constancy. Assuming “single-phase induction motor” implies a specific starting method without stating it. Final Answer: none of the above

Question 3

Average of a rectangular (pulse) waveform with DC baseline: A rectangular wave has a baseline (low level) of 20 V, a peak-to-peak value of 50 V, and a duty cycle of 20%. What is its average voltage over one full period?

Accepted Answer

Introduction: Pulse and rectangular waveforms often ride on a DC baseline. Computing the average (DC) component is essential for biasing, measurement, and filter design. When duty cycle and peak-to-peak amplitude are given, you can reconstruct the high and low levels and take a time-weighted average. Given Data / Assumptions: Baseline (low level) Vlow = 20 V. Peak-to-peak Vpp = 50 V → Vhigh = Vlow + Vpp = 70 V. Duty cycle D = 20% = 0.2 (fraction of period at Vhigh). Ideal rectangular shape; negligible rise/fall time. Concept / Approach: The average over one period is the time-weighted sum of the levels: Vavg = D * Vhigh + (1 − D) * Vlow. This directly incorporates both the DC baseline and the fraction of time spent at the upper level. Step-by-Step Solution: Compute Vhigh: Vhigh = 20 V + 50 V = 70 V.Write average formula: Vavg = D * Vhigh + (1 − D) * Vlow.Substitute numbers: Vavg = 0.2 * 70 + 0.8 * 20.Calculate: Vavg = 14 + 16 = 30 V. Verification / Alternative check: As a check, note that Vavg must lie between 20 V and 70 V and be closer to 20 V because the signal stays low 80% of the time; 30 V satisfies both conditions. Why Other Options Are Wrong: 26 V / 40 V / 45 V: Do not match the weighted average using the given duty cycle. 20 V: Would imply D = 0% or no pulses above baseline, which is untrue here. Common Pitfalls: Confusing Vpp with amplitude about zero rather than about the given baseline. Using arithmetic mean of 20 V and 70 V (45 V) without weighting by duty cycle. Final Answer: 30 V

Question 4

Relating half-wave average to rms: If the half-cycle average voltage of a sine wave is 12 V (i.e., average of a half-wave-rectified sine over one full period), what is the corresponding rms voltage of the half-wave-rectified waveform?

Accepted Answer

Introduction: Different “averages” and “effective” (rms) values are used for sine-derived waveforms. For rectified signals, the average (mean) and rms are not the same. Converting between them requires remembering the constants that relate peak, average, and rms for half-wave rectified sinusoids. Given Data / Assumptions: Half-wave-rectified sine wave. Half-cycle average (over a full period) is 12 V. Ideal rectification and sinusoid. Concept / Approach: For a half-wave-rectified sine: average value Vavg_half = Vp / π. Its rms value is Vrms_half = Vp / 2. Therefore, if we first find Vp from the average, we can compute the rms directly. Step-by-Step Solution: Use average relation: Vavg_half = Vp / π.Given Vavg_half = 12 V → Vp = 12 * π ≈ 37.70 V.RMS for half-wave rectified: Vrms_half = Vp / 2 ≈ 37.70 / 2 = 18.85 V.Rounded to two decimals, 18.84–18.85 V matches option 18.84 V. Verification / Alternative check: Cross-relations: For full-wave rectified, Vrms_full = Vp / √2 and Vavg_full = 2 * Vp / π; those do not apply here. The half-wave constants used are standard identities and produce a self-consistent result. Why Other Options Are Wrong: 13.33 V / 7.64 V / 8.48 V: Do not follow the Vp / 2 formula once Vp is inferred from the given average. 24.00 V: Would imply Vp = 48 V and average = 15.28 V, inconsistent with the 12 V average. Common Pitfalls: Using Vrms = Vp / √2 (valid for unrectified sine or full-wave, not half-wave rms). Confusing average of rectified with average of the original sinusoid (which is zero over a full period). Final Answer: 18.84 V

Question 5

Effective (rms) value of a sinusoid: The effective value of a sine wave, expressed relative to its peak value, is equal to which of the following?

Accepted Answer

Introduction: Root-mean-square (rms) value is the DC-equivalent heating value of an AC waveform. For a pure sine wave, rms has a fixed ratio to peak (maximum) and to peak-to-peak, which is essential when converting between datasheet ratings, scope measurements, and power calculations. Given Data / Assumptions: Ideal sinusoidal waveform. We want rms in terms of peak. Concept / Approach: For a pure sine: Vrms = Vp / √2. Numerically, 1 / √2 ≈ 0.707106. Also, sin 45° = √2 / 2 ≈ 0.707106. Thus 0.707 of peak and sin 45° of peak are equal statements. Note that 0.636 of peak is the average of a full-wave rectified sine, not the rms of a pure sine. Step-by-Step Solution: Start with Vrms = Vp / √2.Compute numeric factor: 1 / √2 ≈ 0.707.Recognize trigonometric identity: sin 45° = √2 / 2 ≈ 0.707.Therefore Vrms is both 0.707 * Vp and sin 45° * Vp. Verification / Alternative check: Check with Vp = 10 V: Vrms = 10 / √2 ≈ 7.071 V; sin 45° * 10 V gives the same, confirming equivalence. Why Other Options Are Wrong: 0.636 of peak voltage: This is the average of a full-wave rectified sine (2/π), not rms. 0.5 of peak voltage: No standard sinusoidal relation yields this as rms. Common Pitfalls: Mixing up average and rms for rectified signals. Confusing peak with peak-to-peak (Vp-p = 2 * Vp). Final Answer: both 0.707 of peak voltage and sin 45° of peak voltage

Question 6

Alternator fundamentals (AC generation principle): Most practical alternators generate electricity by which physical arrangement?

Accepted Answer

Introduction: Alternators convert mechanical energy to AC electrical energy. Their mechanical and magnetic architectures determine efficiency, reliability, and how power is collected. Understanding which part rotates—field or armature—matters for practical machine design and power delivery to loads. Given Data / Assumptions: Industrial and automotive alternators as common references. Desire to produce substantial current at moderate voltage safely and efficiently. Concept / Approach: Modern alternators typically use a rotating magnetic field (rotor) and a stationary armature (stator). Keeping the armature windings stationary simplifies routing high currents out of the machine (solid terminals, no brushes for load current) while only the field excitation requires slip rings or brushless excitation systems. Step-by-Step Reasoning: Armature-as-stator: Stationary windings handle large currents with fixed terminals.Rotor: Provides a rotating magnetic field via DC-excited electromagnets or permanent magnets.Induction: Changing magnetic flux through stationary coils induces AC per Faraday’s law. Verification / Alternative check: Automotive alternators and large grid alternators follow this architecture. Brushless exciters remove the need for slip rings even for field excitation, further improving reliability. Why Other Options Are Wrong: a coil rotating within a magnetic field: Historically used in small generators, but most practical power alternators use rotating field with stationary armature. a permanent magnet rotating within a varying electromagnetic field: Not the mainstream architecture for high-power alternators. none of the above / moving the stator while the rotor is stationary: Contradict established machine design. Common Pitfalls: Equating small hand-cranked generators with large alternator designs; scale changes the optimal architecture. Confusing DC generators (commutators) with AC alternators (slip rings/brushless excitation). Final Answer: a magnetic field rotating around fixed windings

Question 7

AC machines concept check: In both induction motors and synchronous AC motors used in industry, which statement correctly describes the behavior of the stator's magnetic field and overall speed characteristics?

Accepted Answer

Introduction: AC motors—both induction and synchronous—are cornerstone machines in electric drives. Despite differences in how their rotors operate, they share a fundamental principle: the stator windings create a rotating magnetic field when energized by a balanced AC supply. This question checks whether you recognize that common feature and avoid confusing it with rotor-specific details or application-dependent speed steadiness. Given Data / Assumptions: Three-phase (or appropriately driven single-phase with auxiliary means) AC supply energizes stator windings. Balanced sinusoidal currents produce spatially and temporally varying magnetomotive forces. No special starting or control schemes are being considered. Concept / Approach: With a balanced three-phase supply, the stator currents are phase-shifted by 120 electrical degrees. The superposition of their magnetic fields yields a resultant stator field that rotates at the synchronous speed Ns = 120 * f / P (f in Hz, P in poles). In synchronous motors, the rotor locks onto this speed via DC-excited field (or permanent magnets). In induction motors, rotor currents produced by slip cause torque, and the rotor runs just below Ns, but the stator field itself still rotates at Ns. Step-by-Step Solution: Identify commonality: Both machine types use a rotating stator field set by supply frequency and pole count.Compute synchronous speed if needed: Ns = 120 * f / P.Relate rotor behavior: synchronous rotor locks to Ns; induction rotor follows with small slip.Conclude: The correct universal statement is that the stator magnetic field rotates. Verification / Alternative check: Phasor diagrams and space-vector models show a constant-magnitude vector rotating at electrical angular speed ωe = 2 * pi * f * (2/P). Practical tests (stroboscope) confirm the rotating field and resultant torque production principles in both machines. Why Other Options Are Wrong: The operating speed is very steady: Often true for synchronous machines, but induction motors vary with load (slip); it is not a universal statement. The stator magnetic field is stationary: Incorrect; it rotates. The "squirrel cage" forms the rotor: True for many induction motors, not for synchronous motors; not common to both. Common Pitfalls: Confusing rotor behavior (slip vs lock) with stator-field behavior; assuming “steady speed” applies equally to both across all loads; overlooking the role of pole count and frequency. Final Answer: the stator magnetic field rotates

Question 8

Signal spectra: If a periodic waveform has a fundamental frequency of 400 Hz, what is the frequency of its fourth harmonic (i.e., the component at 4 * fundamental)?

Accepted Answer

Introduction: Harmonics are integer multiples of a waveform's fundamental frequency and are central to understanding spectra, distortion, and filtering. This question tests your ability to map an nth harmonic to its absolute frequency given the fundamental, a frequent task in signal analysis and power quality work. Given Data / Assumptions: Fundamental frequency f1 = 400 Hz. We seek the fourth harmonic (n = 4). Ideal periodic signal with exact harmonic integer multiples. Concept / Approach: By definition, the nth harmonic has frequency fn = n * f1. For n = 4, simply multiply the fundamental by 4. Converting between Hz and kHz (1 kHz = 1000 Hz) provides a convenient representation for larger values. Step-by-Step Solution: Write fn = n * f1.Substitute: n = 4, f1 = 400 Hz → fn = 4 * 400 = 1600 Hz.Express in kHz: 1600 Hz = 1.6 kHz.Select the option matching 1.6 kHz. Verification / Alternative check: Check units: 400 Hz × 4 = 1600 Hz (not 4000 Hz). Cross-check that 1.6 kHz equals 1600 Hz, confirming consistency. Why Other Options Are Wrong: 100 Hz: This is one-fourth of the fundamental, not the fourth harmonic. 4 kHz: Would be 10 * fundamental; incorrect multiple. 4 Hz: Far below the fundamental; not a harmonic. Common Pitfalls: Confusing “4th harmonic” with “fundamental divided by 4”; mis-converting Hz to kHz; forgetting that harmonic numbers are integers greater than or equal to 1. Final Answer: 1.6 kHz

Question 9

Sine-wave evaluation: For a 6 V peak sinusoid v(t) = 6 * sin(θ), what is the instantaneous peak voltage at an electrical angle of 250° (report polarity and magnitude)?

Accepted Answer

Introduction: Evaluating sinusoidal signals at specific phase angles is a routine task in AC circuit analysis and signal processing. This item reinforces the relationship between angle, sign, and amplitude, particularly in quadrants where the sine function is negative. Given Data / Assumptions: Peak amplitude Vp = 6 V. Instantaneous form: v(θ) = Vp * sin(θ). Angle θ = 250 degrees. Concept / Approach: The sine of an angle in degrees can be related to a known reference angle using quadrant identities. Since 250° = 180° + 70°, the value sin(250°) = −sin(70°). Knowing sin(70°) ≈ 0.9397 allows a quick numerical evaluation. The polarity is negative in the third quadrant for sine. Step-by-Step Solution: Write v(θ) = 6 * sin(250°).Use identity: sin(250°) = −sin(70°).Compute sin(70°) ≈ 0.9397.v(250°) ≈ 6 * (−0.9397) ≈ −5.6382 V.Round to two decimals: −5.64 V. Verification / Alternative check: Calculator check using radians: 250° * pi / 180 ≈ 4.3633 rad; sin(4.3633) ≈ −0.9397; result matches. Why Other Options Are Wrong: +5.64 V: Correct magnitude but wrong sign; sine is negative at 250°. +26.13 V or −26.13 V: Exceed the peak amplitude of 6 V; impossible for a 6 V peak sine. Common Pitfalls: Confusing peak with RMS (6/√2 ≈ 4.24 V), using degrees/radians mismatched, or selecting the correct magnitude but not the correct polarity based on the quadrant. Final Answer: –5.64 V

Alternating Current and Voltage Questions