Q: Induction motor with constant V/f control – starting torque at reduced frequency A 3-phase induction motor is operated with constant V/f control. Its starting torque at 50 Hz is 100 N·m. What happens to the starting torque if the supply frequency is…

Introduction / Context: With variable-frequency drives, constant V/f control keeps air-gap flux approximately constant. The starting (slip = 1) torque, however, depends not only on flux but also on the relative magnitudes of rotor resistance and standstill reactance, which scale with frequency. Hence, predicting starting torque at reduced frequency requires considering both V and f effects. Given Data / Assumptions: V/f kept constant from 50 Hz to 25 Hz. Starting torque at 50 Hz = 100 N·m. Rotor parameters not specified (R2, X2 at standstill). Concept / Approach: At start (s = 1): T_start ∝ V^2 * R2 / (R2^2 + X2^2). With constant V/f, V ∝ f and X2 ∝ f, but R2 is approximately constant. Reducing f lowers both V and X2 proportionally; the ratio R2/X2 changes, potentially moving operation closer to maximum torque condition. Depending on whether X2 dominated at 50 Hz or R2 dominated at 25 Hz, starting torque can increase, decrease, or stay close to the original value. Step-by-Step Solution: At 50 Hz: evaluate trend T50 ∝ (f^2) * R2 / (R2^2 + (k f)^2).At 25 Hz: T25 ∝ (f^2/4) * R2 / (R2^2 + (k f/2)^2).Comparing numerators and denominators shows the outcome depends on the relative size of R2 vs X2. Verification / Alternative check: Many drives texts note that with constant V/f, breakdown torque remains roughly constant, while starting torque can vary with frequency according to rotor circuit parameters and stator/rotor leakage reactances. Why Other Options Are Wrong: Always more / always less: ignores parameter dependence. “Very small”: not correct under constant V/f, which preserves flux. Common Pitfalls: Assuming T_start scales simply with V^2; overlooking the frequency dependence of X2 and the resulting change of R2/X2 at start. Final Answer: may be more or less than 100 N·m depending on motor parameters

Q: Component count – single-phase semi-converter In a single-phase semi-converter (half-controlled bridge) used for controlled rectification, how many thyristors are employed in the power circuit (excluding diodes)?

Introduction / Context: A single-phase semi-converter (also called a half-controlled bridge) combines two thyristors and two diodes in a full-bridge configuration to achieve control during one half of each cycle while providing free conduction in the other half. Recognizing the device count is fundamental to topology identification and cost/performance tradeoffs. Given Data / Assumptions: Standard single-phase semi-converter topology. Focus is on thyristor count only (diodes not counted). Concept / Approach: In the semi-converter, one leg of the bridge consists of thyristors (controllable devices) and the opposite leg consists of diodes (uncontrolled). This provides phase control in positive and negative half cycles via the controlled leg, while the diode leg completes the path without gating in the complementary half cycle. Step-by-Step Solution: Bridge has 4 devices total.Two of them are SCRs (controlled), two are diodes (uncontrolled).Therefore, number of thyristors = 2. Verification / Alternative check: Power electronics texts depict the half-controlled bridge explicitly with two SCRs and two diodes. Why Other Options Are Wrong: 4, 8, 16 imply other multi-pulse or multi-phase arrangements; they do not match the basic semi-converter. Common Pitfalls: Confusing a fully controlled bridge (4 thyristors) with the semi-converter (2 thyristors + 2 diodes). Final Answer: 2

Q: Implementing a 3-input NAND logic function using thyristors A NAND gate has three inputs and one output. Approximately how many thyristors would be required to implement this function using SCR logic building blocks (ignore auxiliary components)?

Introduction / Context: Thyristor logic circuits can implement digital functions by using SCRs as latching elements with steering networks. Realizing multi-input NAND behavior often requires one controlled path per input plus a combining or inversion stage, leading to variability in device count depending on the exact implementation (series/parallel logic and reset scheme). Given Data / Assumptions: Three logic inputs; single output implementing NAND. Use of SCRs as the primary active elements. Auxiliary diodes/resistors are not counted. Concept / Approach: A 3-input NAND can be formed by combining AND then NOT, or by directly configuring SCRs so that output is inhibited only when all inputs are simultaneously asserted. Designs may require one SCR per input and a final stage to obtain the NAND inversion or to provide a latch/reset function, making the count 3 or sometimes 4 depending on whether a separate inverter/latch stage is used. Step-by-Step Solution: Allocate one SCR path per input to realize the required gating.Add an additional SCR stage when output inversion or isolation is desired.Hence, count commonly falls in the range 3–4 SCRs. Verification / Alternative check: Historic SCR logic implementations in industrial control demonstrate multiple realizations with slight variations in component count to meet noise margins and latching requirements. Why Other Options Are Wrong: 1 or 2: insufficient to realize three-input gating and inversion robustly. Exactly 3: possible but not universal; often a fourth device is added for inversion/reset. Common Pitfalls: Forgetting that NAND requires inversion after an AND-equivalent network; overlooking the need for a latch or reset path in SCR logic. Final Answer: 3 or 4

Q: Single-phase full-wave AC regulator – relation between device RMS current and load RMS current In a single-phase full-wave AC regulator using two antiparallel thyristors and a resistive load, if the RMS current in each thyristor (measured over a ful…

Introduction / Context: Device current ratings are specified in RMS terms averaged over the full mains cycle. In an antiparallel thyristor regulator with a resistive load, each thyristor conducts during alternate portions of the cycle. Relating the device RMS current to the overall load RMS current helps translate device ratings into load capability. Given Data / Assumptions: Two identical thyristors in antiparallel, resistive load. Each thyristor conducts in disjoint intervals (complementary half-cycles or phase-controlled segments). Ir is the full-cycle RMS current of each thyristor. Concept / Approach: Because the conduction intervals of the two devices do not overlap in time for a resistive load, the mean-square load current over a cycle is the sum of the mean squares of the two device currents. Thus, I_load_rms^2 = Ir^2 + Ir^2 = 2 Ir^2 → I_load_rms = √2 * Ir. Step-by-Step Solution: Let i1(t) be current through SCR1; i2(t) through SCR2; i_load(t) = i1(t) + i2(t).Because i1(t) and i2(t) occur in disjoint intervals (for resistive load), i1(t)·i2(t) = 0 for all t.Then I_load_rms^2 = (1/T)∫[i1^2 + i2^2] dt = Ir^2 + Ir^2 = 2 Ir^2.Therefore I_load_rms = √2 * Ir. Verification / Alternative check: Special case at full conduction (no phase delay) also satisfies the same relationship, confirming the factor √2 holds for symmetrical sharing. Why Other Options Are Wrong: Ir: ignores contribution from the other device. 2 * Ir: double-counts RMS magnitudes linearly; RMS components add in power (square) sense. Ir/√2: would be correct in the opposite inference (device RMS from load RMS). Common Pitfalls: Adding RMS values arithmetically instead of in quadrature; forgetting that the two device conduction intervals do not overlap for a resistive load. Final Answer: √2 * Ir

Q: Class A chopper performance metrics A class A (step-down) chopper has an input voltage of 200 V with Ton = 1.5 ms and Toff = 4.5 ms. Compute (i) the duty cycle, (ii) the average output voltage, and (iii) the ripple factor defined as Vac_rms/Vdc for …

Introduction / Context: Basic metrics for a step-down chopper include duty cycle (D), average output voltage (Vdc), and ripple factor (defined here as the ratio of the RMS value of the AC component to the DC component). For a rectangular 0–Vin waveform applied to a resistive load, these quantities can be computed directly from Ton and Toff. Given Data / Assumptions: Vin = 200 V. Ton = 1.5 ms, Toff = 4.5 ms → switching period T = 6.0 ms. Ideal switch; resistive load; no output filter. Ripple factor r = Vac_rms / Vdc. Concept / Approach: Duty cycle D = Ton/T. Average output voltage for a 0–Vin rectangular wave is Vdc = D * Vin. The total output RMS is Vrms = Vin * sqrt(D). The AC component’s RMS is Vac_rms = sqrt(Vrms^2 − Vdc^2) = Vin * sqrt(D − D^2) = Vin * sqrt(D(1 − D)). Therefore, ripple factor r = Vac_rms / Vdc = sqrt(D(1 − D)) / D = sqrt((1 − D)/D). Step-by-Step Solution: Compute D = 1.5 / (1.5 + 4.5) = 0.25.Vdc = D * Vin = 0.25 * 200 = 50 V.r = sqrt((1 − 0.25) / 0.25) = sqrt(0.75 / 0.25) = sqrt(3) ≈ 1.732. Verification / Alternative check: Limit checks: D → 1 gives r → 0 (pure DC); D → 0.5 gives r = 1; D → very small gives large r, as expected. Why Other Options Are Wrong: Any entry with D = 0.50 or Vavg = 100 V contradicts Ton/Toff values. Ripple factor = 1.0 corresponds to D = 0.5, not 0.25. Common Pitfalls: Using Vrms/Vdc instead of Vac_rms/Vdc; failing to subtract the DC component when calculating ripple. Final Answer: D = 0.25, Vavg = 50 V, ripple factor = 1.732

Q: In power-electronics practice, which commutation method is typically employed in an inverter that converts DC to AC?

Introduction / Context: Commutation is the process by which a switching device is turned off and current is transferred to another path. Inverters convert DC to AC and therefore must periodically switch current direction. This question probes whether an inverter relies on the AC supply for device turn-off (line commutation) or needs an internal method (forced commutation). Given Data / Assumptions: Stand-alone DC source feeding an inverter. General inverter topologies using thyristors, MOSFETs, or IGBTs. No external AC mains available for natural current zero crossings. Concept / Approach: Line (natural) commutation uses the natural zero of an AC source current/voltage to turn devices off. Forced commutation creates the required reverse bias or current zero by auxiliary circuits (LC networks, snubbers, or self-commutating transistors) because a DC source has no natural zero crossings. Modern transistor inverters are self-commutated; thyristor inverters use auxiliary forced-commutation methods when not tied to an AC grid. Step-by-Step Solution: Identify the source: a DC link has no natural zero → natural commutation is not available.Determine need: devices must be turned off actively → requires forced commutation or self-commutation.Conclude: inverters generally employ forced/self commutation. Verification / Alternative check: Grid-tied line-commutated inverters do exist (called line-commutated converters), but typical stand-alone DC-AC inverters for motor drives, UPS, and PV use forced/self-commutation (PWM with IGBTs or forced-commutated SCRs). Why Other Options Are Wrong: Line commutation requires an external AC source zero; option “both simultaneously” is not standard practice; “no commutation” contradicts switching fundamentals. Common Pitfalls: Confusing rectifiers (often line-commutated) with inverters; assuming thyristors can naturally turn off on DC without auxiliary networks. Final Answer: Forced commutation

Q: Cycloconverter output description Is it correct that the output waveform of a cycloconverter is composed of multiple segments taken from the input waveform(s)?

Introduction / Context: A cycloconverter directly synthesizes a low-frequency AC output from a higher-frequency AC input by phase-controlling segments of the input. Understanding its piece-wise construction explains the characteristic low output frequency and harmonic content. Given Data / Assumptions: Input is AC (single or three phase) at frequency f_in. Output is AC at a commanded lower frequency f_out. Thyristor bridges (positive and negative groups) are used with proper gating. Concept / Approach: By advancing or delaying firing angles, the converter selects timed “segments” of the input sinusoid(s) to approximate a sinusoid at a lower frequency. Positive group conducts during portions generating positive output; negative group during negative output. Intergroup reactors or control strategies prevent circulating current. Step-by-Step Solution: Segment selection: choose appropriate input voltage portions by gating SCRs.Polarity shaping: use positive or negative bridge to form the desired half-cycle.Frequency reduction: modulate the timing so the stitched segments complete one low-frequency cycle over many input cycles. Verification / Alternative check: Oscilloscope traces of cycloconverter outputs show stepped waveforms clearly assembled from input segments; filtering and machine inductance smooth the current. Why Other Options Are Wrong: Dependence on power factor or single-phase input is incorrect; the segmented construction is fundamental to cycloconverters for both single- and three-phase inputs. Common Pitfalls: Assuming the output is a pure sinusoid; forgetting the need for inductive loads or filters to control harmonic currents. Final Answer: True

Q: Chopper quadrants: Which class operates in quadrants I and IV of the V–I plane (forward voltage with bidirectional current)?

Introduction / Context: DC choppers are categorized by the quadrants of operation on the voltage–current (V–I) plane. Knowing which class supports motoring and regenerative modes is essential in electric drives and power conversion. Given Data / Assumptions: Quadrant I: V > 0, I > 0 (forward motoring). Quadrant IV: V > 0, I < 0 (regeneration to the source at positive voltage). Ideal switching and proper freewheeling/feedback paths. Concept / Approach: Class A operates only in Quadrant I (step-down). Class B operates in Quadrant II (regenerative). Class C (composite of A and B) covers Quadrants I and II. Class D uses two diagonal choppers enabling current reversal while keeping output voltage positive, thus covering Quadrants I and IV. Class E spans all four quadrants with a full bridge arrangement. Step-by-Step Solution: Identify requirement: V positive, I reversible → Quadrants I and IV.Map to classes: Class D meets this condition.Answer: Class D chopper. Verification / Alternative check: Standard drive texts show Class D with two choppers and diodes arranged to allow reversal of current through the load while maintaining positive voltage at the load terminals. Why Other Options Are Wrong: Class A and B are single-quadrant; Class C is I and II; Class E is four-quadrant (I, II, III, IV), more than required. Common Pitfalls: Mixing up Class C (I & II) with Class D (I & IV); assuming all bridges are four-quadrant. Final Answer: Class D chopper

Question 1

Single-phase full-wave AC voltage regulator with resistive load Considering a purely resistive load, what is the average load current over one full cycle for a single-phase full-wave AC regulator (antiparallel SCRs with phase control)?

Accepted Answer

Introduction / Context: Full-wave AC voltage regulators using antiparallel SCRs control the RMS voltage applied to a resistive load by delaying turn-on within each half cycle. For resistive loads, current is in phase with voltage and reverses direction every half cycle, which determines the average current value over a full period. Given Data / Assumptions: Load is purely resistive. Full-wave control, triggering in each half cycle. No DC bias present in the source. Concept / Approach: Average (DC) current is the time-average of instantaneous current over a full cycle. With a resistive load on an AC source, instantaneous current i(t) = v(t)/R follows the polarity of v(t). Positive and negative half cycles are equal in magnitude and opposite in sign, so the net average over a full cycle is zero regardless of firing angle (provided conduction occurs symmetrically in both halves). Step-by-Step Solution: i(t) = v(t)/R with v(t) alternating sign each half cycle.The area under i(t) in the positive half equals the magnitude in the negative half.Therefore, I_avg over a full cycle = 0. Verification / Alternative check: Even with phase control, provided both halves are triggered equally, the symmetry ensures zero DC component; only the RMS value changes. Why Other Options Are Wrong: “Cannot be zero / always positive”: contradict symmetry for a resistive AC load. “May be sometimes zero”: it is always zero under the stated conditions. Common Pitfalls: Confusing average (DC) with RMS; a regulator changes RMS value but not the fact that the average over a full cycle is zero for resistive AC loads. Final Answer: is always zero

Question 2

UJT negative-resistance region A UJT (unijunction transistor) exhibits negative resistance between which current levels on its characteristic?

Accepted Answer

Introduction / Context: The UJT is a popular triggering device whose emitter characteristic features a negative-resistance region. Understanding where this region lies relative to the peak and valley points is important for designing relaxation oscillators and trigger circuits for thyristors. Given Data / Assumptions: Standard UJT E–I characteristic with clearly defined peak and valley points. Terminology: I_P (peak current), I_V (valley current). Concept / Approach: As the emitter voltage rises to the peak point, the UJT fires and enters a region where emitter voltage drops while emitter current increases—this is the negative-resistance region. The characteristic continues until the valley point is reached; below the valley current the device reverts to the high-resistance (off) state. Step-by-Step Solution: At I = I_P: device turns on (peak point).For I between I_P and I_V: V_E decreases as I_E increases → negative resistance region.At I = I_V: device reaches valley; decreasing current further turns it off. Verification / Alternative check: Reference UJT curves show the slope dV/dI negative between the peak and valley points (I_P < I < I_V). Why Other Options Are Wrong: “Less than valley current” or “less than peak current” do not correspond to the negative-resistance segment; these are in off-region or pre-trigger region. “None”: incorrect because the negative-resistance range is well defined between I_P and I_V. Common Pitfalls: Assuming valley current is smaller than peak current conceptually; numerically I_V is typically greater than I_P on the I–V plot of a UJT. Final Answer: Current is more than peak current but less than valley current

Question 3

Induction motor with constant V/f control – starting torque at reduced frequency A 3-phase induction motor is operated with constant V/f control. Its starting torque at 50 Hz is 100 N·m. What happens to the starting torque if the supply frequency is…

Accepted Answer

Introduction / Context: With variable-frequency drives, constant V/f control keeps air-gap flux approximately constant. The starting (slip = 1) torque, however, depends not only on flux but also on the relative magnitudes of rotor resistance and standstill reactance, which scale with frequency. Hence, predicting starting torque at reduced frequency requires considering both V and f effects. Given Data / Assumptions: V/f kept constant from 50 Hz to 25 Hz. Starting torque at 50 Hz = 100 N·m. Rotor parameters not specified (R2, X2 at standstill). Concept / Approach: At start (s = 1): T_start ∝ V^2 * R2 / (R2^2 + X2^2). With constant V/f, V ∝ f and X2 ∝ f, but R2 is approximately constant. Reducing f lowers both V and X2 proportionally; the ratio R2/X2 changes, potentially moving operation closer to maximum torque condition. Depending on whether X2 dominated at 50 Hz or R2 dominated at 25 Hz, starting torque can increase, decrease, or stay close to the original value. Step-by-Step Solution: At 50 Hz: evaluate trend T50 ∝ (f^2) * R2 / (R2^2 + (k f)^2).At 25 Hz: T25 ∝ (f^2/4) * R2 / (R2^2 + (k f/2)^2).Comparing numerators and denominators shows the outcome depends on the relative size of R2 vs X2. Verification / Alternative check: Many drives texts note that with constant V/f, breakdown torque remains roughly constant, while starting torque can vary with frequency according to rotor circuit parameters and stator/rotor leakage reactances. Why Other Options Are Wrong: Always more / always less: ignores parameter dependence. “Very small”: not correct under constant V/f, which preserves flux. Common Pitfalls: Assuming T_start scales simply with V^2; overlooking the frequency dependence of X2 and the resulting change of R2/X2 at start. Final Answer: may be more or less than 100 N·m depending on motor parameters

Question 4

Component count – single-phase semi-converter In a single-phase semi-converter (half-controlled bridge) used for controlled rectification, how many thyristors are employed in the power circuit (excluding diodes)?

Accepted Answer

Introduction / Context: A single-phase semi-converter (also called a half-controlled bridge) combines two thyristors and two diodes in a full-bridge configuration to achieve control during one half of each cycle while providing free conduction in the other half. Recognizing the device count is fundamental to topology identification and cost/performance tradeoffs. Given Data / Assumptions: Standard single-phase semi-converter topology. Focus is on thyristor count only (diodes not counted). Concept / Approach: In the semi-converter, one leg of the bridge consists of thyristors (controllable devices) and the opposite leg consists of diodes (uncontrolled). This provides phase control in positive and negative half cycles via the controlled leg, while the diode leg completes the path without gating in the complementary half cycle. Step-by-Step Solution: Bridge has 4 devices total.Two of them are SCRs (controlled), two are diodes (uncontrolled).Therefore, number of thyristors = 2. Verification / Alternative check: Power electronics texts depict the half-controlled bridge explicitly with two SCRs and two diodes. Why Other Options Are Wrong: 4, 8, 16 imply other multi-pulse or multi-phase arrangements; they do not match the basic semi-converter. Common Pitfalls: Confusing a fully controlled bridge (4 thyristors) with the semi-converter (2 thyristors + 2 diodes). Final Answer: 2

Question 5

Implementing a 3-input NAND logic function using thyristors A NAND gate has three inputs and one output. Approximately how many thyristors would be required to implement this function using SCR logic building blocks (ignore auxiliary components)?

Accepted Answer

Introduction / Context: Thyristor logic circuits can implement digital functions by using SCRs as latching elements with steering networks. Realizing multi-input NAND behavior often requires one controlled path per input plus a combining or inversion stage, leading to variability in device count depending on the exact implementation (series/parallel logic and reset scheme). Given Data / Assumptions: Three logic inputs; single output implementing NAND. Use of SCRs as the primary active elements. Auxiliary diodes/resistors are not counted. Concept / Approach: A 3-input NAND can be formed by combining AND then NOT, or by directly configuring SCRs so that output is inhibited only when all inputs are simultaneously asserted. Designs may require one SCR per input and a final stage to obtain the NAND inversion or to provide a latch/reset function, making the count 3 or sometimes 4 depending on whether a separate inverter/latch stage is used. Step-by-Step Solution: Allocate one SCR path per input to realize the required gating.Add an additional SCR stage when output inversion or isolation is desired.Hence, count commonly falls in the range 3–4 SCRs. Verification / Alternative check: Historic SCR logic implementations in industrial control demonstrate multiple realizations with slight variations in component count to meet noise margins and latching requirements. Why Other Options Are Wrong: 1 or 2: insufficient to realize three-input gating and inversion robustly. Exactly 3: possible but not universal; often a fourth device is added for inversion/reset. Common Pitfalls: Forgetting that NAND requires inversion after an AND-equivalent network; overlooking the need for a latch or reset path in SCR logic. Final Answer: 3 or 4

Question 6

Single-phase full-wave AC regulator – relation between device RMS current and load RMS current In a single-phase full-wave AC regulator using two antiparallel thyristors and a resistive load, if the RMS current in each thyristor (measured over a ful…

Accepted Answer

Introduction / Context: Device current ratings are specified in RMS terms averaged over the full mains cycle. In an antiparallel thyristor regulator with a resistive load, each thyristor conducts during alternate portions of the cycle. Relating the device RMS current to the overall load RMS current helps translate device ratings into load capability. Given Data / Assumptions: Two identical thyristors in antiparallel, resistive load. Each thyristor conducts in disjoint intervals (complementary half-cycles or phase-controlled segments). Ir is the full-cycle RMS current of each thyristor. Concept / Approach: Because the conduction intervals of the two devices do not overlap in time for a resistive load, the mean-square load current over a cycle is the sum of the mean squares of the two device currents. Thus, I_load_rms^2 = Ir^2 + Ir^2 = 2 Ir^2 → I_load_rms = √2 * Ir. Step-by-Step Solution: Let i1(t) be current through SCR1; i2(t) through SCR2; i_load(t) = i1(t) + i2(t).Because i1(t) and i2(t) occur in disjoint intervals (for resistive load), i1(t)·i2(t) = 0 for all t.Then I_load_rms^2 = (1/T)∫[i1^2 + i2^2] dt = Ir^2 + Ir^2 = 2 Ir^2.Therefore I_load_rms = √2 * Ir. Verification / Alternative check: Special case at full conduction (no phase delay) also satisfies the same relationship, confirming the factor √2 holds for symmetrical sharing. Why Other Options Are Wrong: Ir: ignores contribution from the other device. 2 * Ir: double-counts RMS magnitudes linearly; RMS components add in power (square) sense. Ir/√2: would be correct in the opposite inference (device RMS from load RMS). Common Pitfalls: Adding RMS values arithmetically instead of in quadrature; forgetting that the two device conduction intervals do not overlap for a resistive load. Final Answer: √2 * Ir

Question 7

Class A chopper performance metrics A class A (step-down) chopper has an input voltage of 200 V with Ton = 1.5 ms and Toff = 4.5 ms. Compute (i) the duty cycle, (ii) the average output voltage, and (iii) the ripple factor defined as Vac_rms/Vdc for …

Accepted Answer

Introduction / Context: Basic metrics for a step-down chopper include duty cycle (D), average output voltage (Vdc), and ripple factor (defined here as the ratio of the RMS value of the AC component to the DC component). For a rectangular 0–Vin waveform applied to a resistive load, these quantities can be computed directly from Ton and Toff. Given Data / Assumptions: Vin = 200 V. Ton = 1.5 ms, Toff = 4.5 ms → switching period T = 6.0 ms. Ideal switch; resistive load; no output filter. Ripple factor r = Vac_rms / Vdc. Concept / Approach: Duty cycle D = Ton/T. Average output voltage for a 0–Vin rectangular wave is Vdc = D * Vin. The total output RMS is Vrms = Vin * sqrt(D). The AC component’s RMS is Vac_rms = sqrt(Vrms^2 − Vdc^2) = Vin * sqrt(D − D^2) = Vin * sqrt(D(1 − D)). Therefore, ripple factor r = Vac_rms / Vdc = sqrt(D(1 − D)) / D = sqrt((1 − D)/D). Step-by-Step Solution: Compute D = 1.5 / (1.5 + 4.5) = 0.25.Vdc = D * Vin = 0.25 * 200 = 50 V.r = sqrt((1 − 0.25) / 0.25) = sqrt(0.75 / 0.25) = sqrt(3) ≈ 1.732. Verification / Alternative check: Limit checks: D → 1 gives r → 0 (pure DC); D → 0.5 gives r = 1; D → very small gives large r, as expected. Why Other Options Are Wrong: Any entry with D = 0.50 or Vavg = 100 V contradicts Ton/Toff values. Ripple factor = 1.0 corresponds to D = 0.5, not 0.25. Common Pitfalls: Using Vrms/Vdc instead of Vac_rms/Vdc; failing to subtract the DC component when calculating ripple. Final Answer: D = 0.25, Vavg = 50 V, ripple factor = 1.732

Question 8

In power-electronics practice, which commutation method is typically employed in an inverter that converts DC to AC?

Accepted Answer

Introduction / Context: Commutation is the process by which a switching device is turned off and current is transferred to another path. Inverters convert DC to AC and therefore must periodically switch current direction. This question probes whether an inverter relies on the AC supply for device turn-off (line commutation) or needs an internal method (forced commutation). Given Data / Assumptions: Stand-alone DC source feeding an inverter. General inverter topologies using thyristors, MOSFETs, or IGBTs. No external AC mains available for natural current zero crossings. Concept / Approach: Line (natural) commutation uses the natural zero of an AC source current/voltage to turn devices off. Forced commutation creates the required reverse bias or current zero by auxiliary circuits (LC networks, snubbers, or self-commutating transistors) because a DC source has no natural zero crossings. Modern transistor inverters are self-commutated; thyristor inverters use auxiliary forced-commutation methods when not tied to an AC grid. Step-by-Step Solution: Identify the source: a DC link has no natural zero → natural commutation is not available.Determine need: devices must be turned off actively → requires forced commutation or self-commutation.Conclude: inverters generally employ forced/self commutation. Verification / Alternative check: Grid-tied line-commutated inverters do exist (called line-commutated converters), but typical stand-alone DC-AC inverters for motor drives, UPS, and PV use forced/self-commutation (PWM with IGBTs or forced-commutated SCRs). Why Other Options Are Wrong: Line commutation requires an external AC source zero; option “both simultaneously” is not standard practice; “no commutation” contradicts switching fundamentals. Common Pitfalls: Confusing rectifiers (often line-commutated) with inverters; assuming thyristors can naturally turn off on DC without auxiliary networks. Final Answer: Forced commutation

Question 9

Cycloconverter output description Is it correct that the output waveform of a cycloconverter is composed of multiple segments taken from the input waveform(s)?

Accepted Answer

Introduction / Context: A cycloconverter directly synthesizes a low-frequency AC output from a higher-frequency AC input by phase-controlling segments of the input. Understanding its piece-wise construction explains the characteristic low output frequency and harmonic content. Given Data / Assumptions: Input is AC (single or three phase) at frequency f_in. Output is AC at a commanded lower frequency f_out. Thyristor bridges (positive and negative groups) are used with proper gating. Concept / Approach: By advancing or delaying firing angles, the converter selects timed “segments” of the input sinusoid(s) to approximate a sinusoid at a lower frequency. Positive group conducts during portions generating positive output; negative group during negative output. Intergroup reactors or control strategies prevent circulating current. Step-by-Step Solution: Segment selection: choose appropriate input voltage portions by gating SCRs.Polarity shaping: use positive or negative bridge to form the desired half-cycle.Frequency reduction: modulate the timing so the stitched segments complete one low-frequency cycle over many input cycles. Verification / Alternative check: Oscilloscope traces of cycloconverter outputs show stepped waveforms clearly assembled from input segments; filtering and machine inductance smooth the current. Why Other Options Are Wrong: Dependence on power factor or single-phase input is incorrect; the segmented construction is fundamental to cycloconverters for both single- and three-phase inputs. Common Pitfalls: Assuming the output is a pure sinusoid; forgetting the need for inductive loads or filters to control harmonic currents. Final Answer: True

Question 10

Half-wave controlled rectifier (positive half-cycle firing only) An SCR conducts only in the positive half-cycle. With firing angle α = 80°, the RMS anode current is 25 A. If α is increased to 160°, the expected RMS current will be:

Accepted Answer

Introduction / Context: In a single-phase half-wave controlled rectifier (SCR triggered only on positive half-cycles), the firing angle α determines the conduction interval within each positive half-cycle. Increasing α reduces the conduction width, decreasing both average and RMS currents. This question asks for a qualitative comparison at α = 80° versus α = 160°. Given Data / Assumptions: SCR fires in positive half-cycles only (half-wave control). RMS anode current at α = 80° is 25 A. Load is assumed unchanged (commonly resistive or predominantly resistive for this comparison). No change in supply amplitude/frequency. Concept / Approach: RMS current depends on the square of instantaneous current over the conduction interval. As α increases, the conduction window (α to π) narrows. Going from α = 80° (100° conduction) to α = 160° (20° conduction) dramatically reduces the area under i^2(t), so the RMS value falls much more than by a simple factor of 2; hence it will be less than 12.5 A (half of the original), not merely less than 25 A. Step-by-Step Solution: For half-wave resistive load: I_rms^2 ∝ ∫_α^π sin^2 θ dθ.Compute trend: ∫_80°^180° sin^2 θ dθ versus ∫_160°^180° sin^2 θ dθ → the latter is much smaller because the integration range is only 20° near the zero crossing.Therefore I_rms(160°) ≪ I_rms(80°), clearly below 12.5 A. Verification / Alternative check: Limiting behavior: as α → 180°, conduction interval tends to zero, and RMS current tends to zero; thus increasing α from 80° to 160° must decrease RMS current below half the former value. Why Other Options Are Wrong: 25 A assumes no change with α; 12.5 A assumes linear halving with α, which is incorrect because RMS depends on the integral of sin^2 over the conduction window; “Less than 25 A” is true but not as informative as “Less than 12.5 A”. Common Pitfalls: Equating proportional change in α with proportional change in RMS; confusing average with RMS values; ignoring that conduction near π contributes very little to RMS due to low instantaneous values. Final Answer: Less than 12.5 A

Question 11

Chopper quadrants: Which class operates in quadrants I and IV of the V–I plane (forward voltage with bidirectional current)?

Accepted Answer

Introduction / Context: DC choppers are categorized by the quadrants of operation on the voltage–current (V–I) plane. Knowing which class supports motoring and regenerative modes is essential in electric drives and power conversion. Given Data / Assumptions: Quadrant I: V > 0, I > 0 (forward motoring). Quadrant IV: V > 0, I < 0 (regeneration to the source at positive voltage). Ideal switching and proper freewheeling/feedback paths. Concept / Approach: Class A operates only in Quadrant I (step-down). Class B operates in Quadrant II (regenerative). Class C (composite of A and B) covers Quadrants I and II. Class D uses two diagonal choppers enabling current reversal while keeping output voltage positive, thus covering Quadrants I and IV. Class E spans all four quadrants with a full bridge arrangement. Step-by-Step Solution: Identify requirement: V positive, I reversible → Quadrants I and IV.Map to classes: Class D meets this condition.Answer: Class D chopper. Verification / Alternative check: Standard drive texts show Class D with two choppers and diodes arranged to allow reversal of current through the load while maintaining positive voltage at the load terminals. Why Other Options Are Wrong: Class A and B are single-quadrant; Class C is I and II; Class E is four-quadrant (I, II, III, IV), more than required. Common Pitfalls: Mixing up Class C (I & II) with Class D (I & IV); assuming all bridges are four-quadrant. Final Answer: Class D chopper

Question 12

Assertion–Reason on converter quadrants and cost Assertion (A): A fully controlled bridge converter can operate in the first and fourth quadrants. Reason (R): A semi-converter is cheaper than a full converter.

Accepted Answer

Introduction / Context: Quadrant operation describes capability for power flow direction and voltage/current polarities. Fully controlled bridges (using four thyristors) can perform rectification and inversion, enabling operation in multiple quadrants. Semi-converters (mixed diodes/thyristors) are cheaper but have reduced controllability. Given Data / Assumptions: Fully controlled bridge with appropriate firing control and load that can return power (e.g., DC motor). Sufficient source conditions for inversion (back emf > DC link, proper firing angles). Semi-converter uses two SCRs and two diodes, reducing component cost. Concept / Approach: A fully controlled bridge can deliver power to the load (Quadrant I) and, when firing angle exceeds 90° with suitable load back-emf, it can operate in inversion to return power (Quadrant IV with positive voltage but negative current direction relative to the source). The reason statement concerns cost, which is true but does not explain why the fully controlled bridge achieves two-quadrant operation (that is due to controllability of all four devices). Step-by-Step Solution: Check Assertion: Fully controlled bridge → bidirectional power flow capability → True.Check Reason: Semi-converter is indeed cheaper due to fewer controllable devices → True.Causality: Cost difference does not explain two-quadrant capability → R is not the correct explanation. Verification / Alternative check: Drive literature shows inversion mode with α > 90° for fully controlled bridges; semi-converters cannot fully reverse power flow due to diodes blocking inversion paths. Why Other Options Are Wrong: Claiming R explains A is incorrect; saying A is wrong contradicts standard converter behavior. Common Pitfalls: Equating cost with capability; assuming semi-converters can invert power without additional hardware. Final Answer: Both A and R are correct but R is not correct explanation of A

Question 13

Auxiliary-commutated (forced) chopper device stress In an auxiliary-commutated chopper, the instantaneous peak current in a thyristor equals:

Accepted Answer

Introduction / Context: Auxiliary (forced) commutation schemes briefly superimpose a resonant current on the load current to reverse-bias or momentarily reduce current through a main thyristor for turn-off. Understanding peak device current is essential for safe ratings. Given Data / Assumptions: Chopper uses LC commutation applying a reverse or additional current pulse. Load current is nonzero at the commutation instant. Idealized short commutation interval with low damping. Concept / Approach: At commutation, device currents combine vectorially along the same conduction path instant by instant. A thyristor may momentarily conduct both the load current and the resonant capacitor current (depending on topology and which device is carrying which component), producing a peak equal to the sum of those peak magnitudes in the worst case. Step-by-Step Solution: Identify currents: I_load(t) and I_c(t) from LC resonance.Path analysis: during commutation, a device may carry I_load + I_c or be relieved, depending on whether it is the outgoing or auxiliary device.Worst-case device stress: I_thyristor,peak = I_load,peak + I_c,peak. Verification / Alternative check: Design guides recommend sizing SCRs for the algebraic sum during overlap, considering tolerances, since the resonant current pulse adds to the steady load current in a given device path. Why Other Options Are Wrong: Considering only one component underestimates device stress; “none of the above” contradicts standard commutation analysis. Common Pitfalls: Ignoring simultaneous current paths; assuming the capacitor current always subtracts rather than adds; overlooking timing that can produce additive peaks. Final Answer: The sum of the peak load current and the peak commutating capacitor current

Question 14

Single-phase full converter with source inductance (overlap) During the overlap interval in a fully controlled bridge supplied by a source with inductance, how many thyristors conduct simultaneously?

Accepted Answer

Introduction / Context: With source inductance, the current cannot change instantaneously when devices commutate. In a single-phase full converter, this creates an overlap period where both the outgoing and incoming devices conduct together. Counting the devices conducting clarifies current paths and voltage drop during overlap. Given Data / Assumptions: Single-phase full-controlled bridge (four SCRs). Finite source inductance causing finite overlap angle μ. Continuous current operation. Concept / Approach: In normal conduction without overlap, one device from the top leg and one from the bottom leg conduct (two devices). During overlap, the incoming pair turns on before the outgoing pair turns off because source inductance resists current transfer, so both pairs conduct at once. Step-by-Step Solution: Before overlap: two devices conduct (one from each leg).During overlap: outgoing pair + incoming pair → four devices.After overlap: only the incoming two devices remain on. Verification / Alternative check: Equivalent circuit during overlap shows shorted DC link (output voltage tends toward zero during μ), consistent with four devices conducting simultaneously. Why Other Options Are Wrong: 1 or 2 devices describe non-overlap conditions; 3 devices is not the standard conduction state in a symmetrical bridge. Common Pitfalls: Forgetting that overlap reduces output voltage because of simultaneous conduction paths and source inductance. Final Answer: 4

Question 15

Load-commutated chopper topology How many thyristors are typically used in a load-commutated chopper configuration?

Accepted Answer

Introduction / Context: Load-commutated choppers use the energy stored in the load (often an R–L or motor with back-emf) to assist in commutation. The standard arrangement employs a bridge allowing current to be redirected so that a device naturally turns off when current transfers to another path. Given Data / Assumptions: Bidirectional current capability through the load for commutation. Typical implementation uses an H-bridge with thyristors and diodes as needed. Basic topology, not multi-pulse expansions. Concept / Approach: An H-bridge comprises four controlled devices. By gating diagonally opposite devices, current is steered in either direction through the load. Commutation occurs as the load energy assists in transferring current from one device pair to another, turning devices off without auxiliary LC networks. Step-by-Step Solution: Identify bridge: four devices arranged in an H configuration.Steer current: alternate diagonal pairs conduct to reverse current.Commutation: load induces current reversal or decay enabling device turn-off → load commutation. Verification / Alternative check: Canonical references depict load-commutated choppers with four SCRs; more complex versions add diodes or snubbers but retain four primary controlled devices. Why Other Options Are Wrong: One or two thyristors cannot steer bidirectional current; six devices would be characteristic of three-phase bridges, not a basic DC chopper. Common Pitfalls: Confusing current-commutated (Class C/D) with load-commutated (bridge-type) topologies. Final Answer: Four thyristors

Question 16

Center-tapped single-phase full-wave controlled rectifier If the voltage across each half of the transformer secondary is v_s(t) = Vm sin(ωt), what is the peak inverse voltage (PIV) across each thyristor?

Accepted Answer

Introduction / Context: Peak inverse voltage (PIV) is the maximum reverse voltage a device must withstand when it is off. In center-tapped full-wave rectifiers (using two controlled devices on each half of the secondary), knowing the PIV is essential for selecting properly rated SCRs or diodes. Given Data / Assumptions: Center-tapped transformer secondary, each half produces Vm sin(ωt) (Vm is the peak of half-winding). Only one SCR conducts at a time (ideal operation, no overlap). Negligible leakage and device drops for PIV estimation. Concept / Approach: When one SCR is forward-biased and conducts, the other is reverse-biased. The non-conducting device sees its own half-winding voltage (Vm sin(ωt)) plus the instantaneous load potential referred through the conducting half which effectively places approximately the opposite half-winding peak across it. The worst case occurs near the peak of the opposite half cycle, leading to a reverse stress of approximately 2 Vm. Step-by-Step Solution: During conduction of top device: bottom device sees its half-secondary voltage reversed plus the drop across the load referred to the other half.At peak conditions, this sums to about 2 Vm reverse on the off device.Hence, PIV ≈ 2 Vm for each SCR. Verification / Alternative check: Textbook results for center-tapped full-wave rectifiers list PIV of each device as twice the maximum half-secondary peak. Bridge rectifiers (without center tap) reduce individual device PIV requirements compared to center-tap circuits. Why Other Options Are Wrong: Vm underestimates the stress; 0.5 Vm and 0.25 Vm are far too low for the off device in this topology. Common Pitfalls: Confusing bridge and center-tap PIV values; forgetting that the off device sees the sum of two half-winding peaks under worst case. Final Answer: 2 Vm

Question 17

Average DC output voltage for controlled rectifiers (single-phase) Given a peak input voltage Vm and firing angle α (in radians or degrees, as standard in power electronics), determine the expressions for the average output voltage for:  a single-ph…

Accepted Answer

Introduction / Context: In controlled rectifiers, the average (DC) output voltage depends on how long the sinusoidal source is allowed to conduct each half-cycle. The firing angle α delays conduction and therefore reduces the average output. Understanding these relationships is fundamental for designing DC supplies and drives. Given Data / Assumptions: Single-phase sinusoidal input v = Vm sin(ωt). Purely resistive load to avoid phase-shift complications. Gate firing angle α measured from the start of each relevant half-cycle. Concept / Approach: The average value of a periodic waveform equals the integral of the instantaneous voltage over one period divided by the period. Conduction begins at ωt = α and ends at ωt = π for half-wave, while in a full-wave bridge the rectified conduction repeats twice per cycle. Step-by-Step Solution: Half-wave: Vdc = (1 / 2π) * ∫ from α to π of Vm sin θ dθ.Compute integral: ∫ sin θ dθ = −cos θ; evaluate → (Vm / (2π)) * (1 + cos α).Full-wave: Two identical conduction intervals per cycle yield Vdc = (1 / π) * ∫ from α to π of Vm sin θ dθ = (Vm / π) * cos α. Verification / Alternative check: At α = 0°, half-wave gives Vdc = Vm / (2π) * (1 + 1) = Vm / π, and full-wave gives Vdc = Vm / π, which matches standard results. Why Other Options Are Wrong: They misplace the 1/(2π) versus 1/π factor or ignore α, leading to incorrect scaling of the average value. Common Pitfalls: Confusing half-wave and full-wave constants or using RMS instead of average when sizing DC loads. Final Answer: Half-wave: Vdc = (Vm / (2π)) * (1 + cos α); Full-wave: Vdc = (Vm / π) * cos α

Question 18

SCR package styles and typical current ratings Lead-mounted SCR packages are typically used for which approximate current rating range?

Accepted Answer

Introduction / Context: Silicon controlled rectifiers (SCRs) are available in different package styles suited to different current ranges. Choosing the right package is essential for thermal management and mounting. Given Data / Assumptions: Lead-mounted (axial/through-hole) small-outline SCRs are intended for low currents. Stud-mount and press-pack (capsule) suit higher currents due to better heat sinking. Concept / Approach: Package capability is largely set by thermal resistance and allowable junction temperature rise. Lead-mounted devices have limited heat dissipation relative to stud or press-pack types. Step-by-Step Solution: Identify typical lead-mounted capability → around hundreds of milliamperes up to roughly 1 A.Compare to stud-mount (often 10 A to hundreds of amperes) and press-pack (high tens to kiloampere-class).Therefore, the closest correct band is about 1 A or less. Verification / Alternative check: Check SCR datasheets: small axial devices are commonly rated sub-ampere to about 1 A continuous, aligning with the above. Why Other Options Are Wrong: Less than a few mA: Unrealistically low for SCR usefulness. 20 A/100 A/500 A: These ratings require robust thermal paths beyond simple lead mounting. Common Pitfalls: Assuming physical size alone dictates rating; thermal path and package mounting method are decisive. Final Answer: about 1 A or less

Question 19

TRIAC triggering polarity A TRIAC can be triggered into conduction by what gate signal polarity?

Accepted Answer

Introduction / Context: A TRIAC is a bidirectional thyristor used in AC control applications. Understanding its gate triggering quadrants is crucial for reliable firing in all voltage polarities. Given Data / Assumptions: TRIAC supports conduction in both polarities of main terminal voltage. Gate triggering can occur with gate current of either polarity depending on the operating quadrant. Concept / Approach: The TRIAC has four triggering quadrants: MT2 and gate can be positive or negative relative to MT1. Practical designs often prefer specific quadrants for sensitivity and noise immunity, but both positive and negative gate currents are possible. Step-by-Step Solution: Recognize bidirectional conduction property.Note that gate current of either polarity can initiate conduction when appropriately referenced.Hence, selection: either positive or negative gate signal. Verification / Alternative check: Application notes show triggering in quadrants I and III (and sometimes II/IV) depending on gate drive circuitry. Why Other Options Are Wrong: Positive only / Negative only: Incomplete understanding of TRIAC quadrants. No gate signal / Only AC: Not generally correct; a DC pulse is commonly used. Common Pitfalls: Ignoring quadrant sensitivity differences (some TRIACs are more sensitive in certain quadrants). Final Answer: either positive or negative gate signal

Question 20

Firing frequency in a three-phase series inverter In a three-phase series inverter, the relationship between the firing frequency of the switches and the output (fundamental) frequency is:

Accepted Answer

Introduction / Context: Series inverters commutate current through resonant elements. In three-phase versions, multiple switching events per electrical cycle are required to synthesize the three-phase waveform. Given Data / Assumptions: Three-phase series inverter with six switch operations per cycle to create line-to-line steps. Idealized commutation; focus on event counting per fundamental period. Concept / Approach: A three-phase bridge involves six sequential switching intervals in one electrical period (each 60°), analogous to six-pulse operation. Thus, the gating (firing) events happen six times during each output fundamental cycle. Step-by-Step Solution: Three-phase synthesis → six equal segments per 360°.Each segment requires a firing event → six firings per cycle.Therefore, firing frequency = 6 * f_out. Verification / Alternative check: Waveform diagrams for three-phase inverters show six commutations per period, matching the six-step concept. Why Other Options Are Wrong: Equal/Twice/Thrice: Underestimates the number of commutations needed for three-phase stepped synthesis. Common Pitfalls: Confusing PWM carrier frequency with fundamental commutation count; here we refer to fundamental firing events. Final Answer: six times the output frequency

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