Q: Commutation dynamics – underdamped cases In which forced-commutation class(es) does the overall commutation circuit operate in an underdamped (oscillatory) manner?

Introduction / Context: Forced commutation methods use auxiliary L–C networks to momentarily reverse-bias and turn off a conducting thyristor. The damping of the L–C network—underdamped, critically damped, or overdamped—governs how current and voltage waveforms evolve during commutation. Given Data / Assumptions: Class A (load commutation / resonant commutation): relies on a resonant L–C (with the load) to produce current reversal. Class B (parallel or impulse commutation): uses an L–C branch to force a reverse current pulse through the thyristor. Ideal components for conceptual analysis. Concept / Approach: In both Class A and Class B commutation, an oscillatory current is intentionally created in an L–C path. This implies an underdamped response (sinusoidal or pseudo-sinusoidal). The natural frequency ω0 = 1/√(LC) sets the time scale over which the thyristor is reverse biased. Correct sizing of L and C ensures that reverse current exceeds the load current for at least the device turn-off time (tq). Step-by-Step Solution: Class A: commutation occurs because the L–C current naturally goes negative → underdamped resonance with load.Class B: a charged capacitor is switched into an L–C loop, producing a decaying sinusoidal current pulse → underdamped.Therefore, the overall commutation loop is underdamped in both classes. Verification / Alternative check: Waveforms in standard references show sinusoidal current reversal (Class A) and oscillatory capacitor current pulses (Class B), confirming underdamped behavior. Why Other Options Are Wrong: Class C and others use different mechanisms (e.g., complementary current paths) and are not generically defined by an underdamped resonant turn-off loop. Common Pitfalls: Equating “forced” with “non-oscillatory”; many forced-commutation schemes purposely exploit oscillation to achieve reliable reverse bias. Final Answer: Both Class A and Class B

Q: Parallel (Class B) inverter – relation of turn-off time to L and C For a parallel inverter employing an L–C commutation circuit, how is the required thyristor turn-off time tq related to the commutation components L and C?

Introduction / Context: In Class B (parallel) commutation, a charged capacitor and inductor form an oscillatory loop that forces a reverse current through the conducting thyristor to turn it off. The commutation interval must exceed the device turn-off time tq. This links tq to the natural resonant period of the L–C network. Given Data / Assumptions: Ideal L and C forming a lossless resonant path. Natural frequency ω0 = 1 / sqrt(L * C). Reverse current is provided roughly over a half-cycle of the L–C oscillation. Concept / Approach: Because the capacitor discharges and then recharges with reversed polarity through the inductor, the reverse bias on the thyristor lasts approximately for a half-sine of the L–C oscillation. The duration of a half-cycle is (π / ω0) = π * sqrt(L * C). To ensure reliable turn-off, this interval must be ≥ tq, hence the design criterion tq ≤ π * sqrt(L * C). Step-by-Step Solution: ω0 = 1 / sqrt(L * C).Half-cycle time Thalf = π / ω0 = π * sqrt(L * C).Require Thalf ≥ tq → tq = π * sqrt(L * C) as the sizing reference. Verification / Alternative check: Many design notes add a safety factor (e.g., 1.2–1.5) on top of the basic π * sqrt(L * C) interval to account for non-idealities and device parameter spread, confirming the proportionality. Why Other Options Are Wrong: Linear LC expressions (options b–d) ignore the square-root relationship inherent to resonance. Common Pitfalls: Confusing full-cycle period 2π * sqrt(L * C) with the half-cycle interval that actually establishes reverse bias during commutation. Final Answer: tq = π * sqrt(L * C)

Q: Transformer Utilization Factor (TUF) for a rectifier A rectifier is fed by a transformer whose secondary RMS voltage and current are Vs and Is. If the DC output power is Vdc * Idc and the apparent AC power at the transformer secondary is Vs * Is, wh…

Introduction / Context: Transformer utilization factor (TUF) measures how effectively a transformer's VA rating is converted into useful DC output power by a rectifier. It is a convenient figure of merit when comparing rectifier topologies and transformer sizing in power supplies. Given Data / Assumptions: Transformer secondary RMS voltage = Vs. Transformer secondary RMS current = Is. DC output power = Vdc * Idc. Apparent AC power drawn from secondary = Vs * Is. Concept / Approach: TUF is defined as the ratio of DC output power to the VA loading of the transformer secondary. Because rectifiers draw non-sinusoidal currents with harmonics, Is can be much greater than the DC current, and the ratio captures how much of the transformer's VA capacity becomes usable DC. Step-by-Step Solution: By definition: TUF = Pdc / (Vs * Is).Substitute Pdc = Vdc * Idc.Therefore, TUF = (Vdc * Idc) / (Vs * Is). Verification / Alternative check: Known TUF values (e.g., ~0.286 for half-wave, ~0.693 for full-wave bridge, depending on definitions) are computed using the same ratio, confirming the expression. Why Other Options Are Wrong: Inverting the ratio (option b) gives the reciprocal and is not the standard definition. Algebraic sums or squared sums (options c and d) have no physical meaning for utilization in this context. Common Pitfalls: Confusing TUF with rectifier efficiency or power factor; TUF focuses on transformer VA usage, not only on losses. Final Answer: TUF = (Vdc * Idc) / (Vs * Is)

Question 1

In a certain inverter, each thyristor conducts 26.67 A for 60°, 13.33 A for 120°, and 0 A for 180° during each 360° cycle. Compute the rms current of one thyristor from these segment values and durations.

Accepted Answer

Introduction / Context: RMS (root mean square) current of a switching device must be calculated from the actual current waveform segments over a full cycle. This is essential for thermal design and device selection in inverters. Given Data / Assumptions: Segment 1: I = 26.67 A for 60°. Segment 2: I = 13.33 A for 120°. Segment 3: I = 0 A for 180°. Period = 360°. Concept / Approach: RMS is the square root of the mean of the squared current over the period. For piecewise-constant segments: Irms^2 = (Σ(I_i^2 * angle_i/360°)). Step-by-Step Solution: 1) Compute squared contributions: 26.67^2 * 60 and 13.33^2 * 120.2) Divide total by 360 to find mean of squares: Irms^2 = [(26.67^2 * 60) + (13.33^2 * 120)] / 360.3) Take square root: Irms ≈ 13.33 A. Verification / Alternative check: Because the device conducts at a lower level for double the angle, the arithmetic works out to a balanced RMS near the smaller current value; exact computation confirms ≈ 13.33 A. Why Other Options Are Wrong: 20 A: Too high; ignores the long interval at zero and the lower-current segment. 6.67 A or 3.33 A: Too low; underestimates higher-current contribution. 15.00 A: Not supported by squared-mean calculation. Common Pitfalls: Using average instead of RMS for heating calculations. Forgetting to square currents before averaging. Final Answer: 13.33 A (Option B).

Question 2

Among SCR-family devices, which one has two control gates (an anode gate and a cathode gate) and can be triggered from either side depending on configuration?

Accepted Answer

Introduction / Context: SCR family devices include SCR, SCS, SUS, SBS, and TRIAC/DIAC. Each has distinct gate arrangements and triggering behaviors. The Silicon Controlled Switch (SCS) resembles a thyristor but provides additional control via two separate gates. Given Data / Assumptions: Focus on number and placement of gates. Standard textbook definitions of SCR-family parts. Concept / Approach: An SCS includes both an anode gate and a cathode gate. This dual-gate construction allows more flexible triggering and commutation methods versus the single-gate SCR. Step-by-Step Solution: 1) SCR: one gate (cathode gate) only.2) SCS: two gates—anode gate and cathode gate—enabling bidirectional control actions.3) SUS/SBS: small-signal triggering devices without the two-gate control of SCS.4) DIAC: a bidirectional trigger device without a gate terminal. Verification / Alternative check: Device symbol references clearly show SCS with two gates; most power electronics handbooks confirm this distinctive feature. Why Other Options Are Wrong: SCR: single gate only. SUS/SBS: do not have both anode and cathode gates. DIAC: trigger diode, no gate terminal. Common Pitfalls: Assuming “bilateral” in SBS implies two gates—bilateral refers to conduction direction, not gate count. Confusing TRIAC (two-way thyristor) with SCS; TRIAC also does not have “two gates” like SCS. Final Answer: SCS (Option B).

Question 3

For a Unijunction Transistor (UJT), the intrinsic stand-off ratio (η) is a key parameter. Select a typical value most commonly assumed in introductory analyses and design estimates.

Accepted Answer

Introduction / Context: The intrinsic stand-off ratio η of a UJT determines the emitter firing voltage relative to the interbase voltage. Designers often assume a nominal η for first-pass timing and relaxation oscillator calculations. Given Data / Assumptions: Standard UJT behavior (e.g., 2N2646 family). η is device-dependent but typically falls within a range. Concept / Approach: η ≈ R_B1 / (R_B1 + R_B2) and typically lies around 0.5 to 0.8. Using η ≈ 0.6 is a common textbook nominal for calculating the approximate peak-point voltage V_P = η * V_BB + V_D (small diode term). Step-by-Step Solution: 1) Recognize η as a ratio inherent to the UJT structure.2) Typical values cluster around 0.6–0.7; using 0.6 is a standard assumption.3) Choose the nearest standard value provided. Verification / Alternative check: Datasheets show spreads, but application notes often compute with η ≈ 0.6 for relaxed design accuracy. Why Other Options Are Wrong: 0.1, 0.2: Too low for typical UJTs; would result in very low trigger points. 0.9: Too high; most devices do not exhibit such a large stand-off ratio. 0.3: Still low relative to common practice. Common Pitfalls: Treating η as constant across all units; it varies with device and temperature. Ignoring the small diode drop term V_D in V_P calculations. Final Answer: η ≈ 0.6 (Option C).

Question 4

Classic industrial cycloconverters (for large low-speed drives) typically rely on which commutation principle during normal operation?

Accepted Answer

Introduction / Context: Cycloconverters directly synthesize low-frequency AC from a higher-frequency AC source. Thyristor bridges are gated such that the supply itself turns devices off at current zero (line commutation). Given Data / Assumptions: Traditional thyristor-based cycloconverter system. Applications: large drives requiring low-frequency, high-torque. No intermediate DC link. Concept / Approach: Line (natural) commutation uses the AC source voltage to reverse-bias and turn off the conducting thyristors as current naturally goes through zero each half cycle. This is different from forced commutation, which uses external L–C or active circuits. Step-by-Step Solution: 1) Identify device family: thyristors without self-turn-off.2) Recognize that in cycloconverters, source voltage crosses zero, enabling natural turn-off.3) Therefore, commutation is predominantly natural. Verification / Alternative check: Reference diagrams show anti-parallel controlled rectifier groups relying on supply zero-crossings for current reversal and device turn-off. Why Other Options Are Wrong: Forced commutation: used in choppers/inverters, not typical in classic cycloconverters. Either/both simultaneously: not standard architecture. Resonant-only: incorrect for mainstream cycloconverter designs. Common Pitfalls: Confusing cycloconverters with force-commutated PWM inverters. Assuming all thyristor systems need commutation networks—line commutation suffices here. Final Answer: Natural (line) commutation (Option A).

Question 5

Half-bridge single-phase inverter: If a DC battery of voltage V feeds an ideal half-bridge that produces a two-level square wave of +V/2 and −V/2 at the output, what is the rms value of the fundamental component of the output voltage (express your c…

Accepted Answer

Introduction / Context: Fourier analysis of inverter waveforms is central to designing filters and estimating fundamental RMS values. In a half-bridge, the output toggles between +V/2 and −V/2, producing a square wave whose fundamental must be extracted for RMS calculations. Given Data / Assumptions: Ideal half-bridge with amplitude levels ±V/2. Pure square wave, 50% duty, no dead-time or device drop. We seek RMS of the fundamental component only (not total RMS). Concept / Approach: A symmetrical square wave of peak amplitude A has a fundamental (sine) component with peak value V1_peak = 4A/π. The RMS of the fundamental is V1_rms = V1_peak/√2 = (4A/π)/√2 = (2√2/π) * A. Step-by-Step Solution: 1) Here A = V/2.2) V1_peak = 4*(V/2)/π = 2V/π.3) V1_rms = (2V/π)/√2 = (√2/π)V ≈ 0.450V. Verification / Alternative check: Compare with full-bridge where amplitude is ±V, yielding fundamental RMS (2√2/π) * (V/2) * 2? For half-bridge, halving the level halves the fundamental magnitude relative to a ±V square wave, so 0.450V is consistent. Why Other Options Are Wrong: 2V: Dimensionally incorrect for RMS of a component. V/√2: That is the RMS of a sinusoid with peak V, not applicable here. V/2: Not the fundamental RMS; it is just the DC magnitude half, irrelevant here. (2/π) * V ≈ 0.637V: That is the average of a ±V square wave magnitude, not the fundamental RMS. Common Pitfalls: Confusing total RMS of square wave with fundamental RMS. Dropping the factor of 1/2 for half-bridge amplitude. Final Answer: (√2/π) * V (≈ 0.450 V) (Option B).

Question 6

Cycloconverter with natural (line) commutation A step-down cycloconverter receives a sinusoidal input of frequency f. When natural (line) commutation is used, what restriction applies to the output frequency that can be synthesized?

Accepted Answer

Introduction / Context: Cycloconverters directly convert AC at one frequency to AC at a lower frequency without an intermediate DC link. With natural (line) commutation, the thyristors turn off at current zero crossings of the supply. This operating principle constrains the possible output frequency to specific fractions of the input frequency, which is fundamental to understanding where cycloconverters are used (e.g., low-speed, high-power drives). Given Data / Assumptions: Input frequency = f (sinusoidal AC source). Natural (line) commutation using thyristors. Step-down cycloconverter intended to produce a lower output frequency. Idealized wave synthesis using phase-controlled segments of the input. Concept / Approach: With line commutation, turn-off occurs only at natural current zero. Therefore, gating patterns must be synchronized to the input cycles. To build a lower-frequency output, the converter stitches together segments from successive input half-cycles following a repetitive pattern. Such repetitive patterns inherently repeat over an integer number of input cycles, which forces the output period to be an integer multiple of the input period. Hence, the output frequency is a sub-multiple of the input frequency, typically much lower (often f/3, f/5, etc.). Step-by-Step Solution: Let input period be T = 1/f.Choose an integer m ≥ 2; construct one output cycle by concatenating m input cycles with controlled conduction windows.Then, output period Tout = mT and output frequency fout = 1/Tout = f/m, i.e., a sub-multiple of f. Verification / Alternative check: Practical cycloconverter outputs are limited to fout ≲ f/3 to maintain acceptable waveform quality and commutation margins. This reinforces the sub-multiple relation and the low-frequency nature of line-commutated cycloconverters. Why Other Options Are Wrong: Any frequency (option b): impossible with line-commutated thyristors since turn-off is tied to source zero-crossings. Even-only or odd-only sub-multiples (options c and d): there is no such parity restriction; both f/3, f/4, etc., are feasible in principle. Common Pitfalls: Confusing line-commutated cycloconverters with PWM inverters (which can synthesize a wide range of frequencies) and overlooking that commutation method dictates frequency constraints. Final Answer: It must be a sub-multiple of the input frequency.

Question 7

Assertion–Reason on SUS and SBS devices Assertion (A): A silicon unilateral switch (SUS) is essentially a programmable unijunction transistor (PUT) with an internal avalanche (Zener) diode between gate and cathode. Reason (R): A silicon bilateral sw…

Accepted Answer

Introduction / Context: SUS and SBS are low-power triggering components widely used in phase-control and timing circuits. Understanding their internal structure clarifies how they are applied for symmetrical or unilateral triggering in AC controllers and thyristor gate drives. Given Data / Assumptions: SUS: a PUT-like device with a built-in Zener (avalanche) reference from gate to cathode to establish a defined breakover. SBS: two SUS-like junction structures arranged for bilateral (both-polarity) behavior. Assertion and Reason given as above. Concept / Approach: The SUS behaves like a PUT whose trigger point is set by an integrated Zener between gate and cathode, ensuring a repeatable breakover voltage in one polarity. The SBS is designed for symmetrical conduction; the common description is two SUS devices in anti-parallel, providing matched breakover in positive and negative directions. Step-by-Step Solution: Evaluate A: SUS internal Zener reference → correct for defining unilateral breakover.Evaluate R: SBS internal structure → effectively two anti-parallel SUS for bilateral action.Does R explain A? No; R describes SBS, not why SUS has a Zener reference. Hence A and R are true, but R does not explain A. Verification / Alternative check: Manufacturers’ datasheets and textbooks show SUS as a PUT derivative and SBS as a symmetric switch (bilateral breakover), confirming both statements independently. Why Other Options Are Wrong: A true/R false: R is not false—SBS is indeed two SUS in anti-parallel. A false/R true: A is factual; SUS uses internal Zener referencing. “R explains A”: incorrect because R addresses SBS rather than the SUS internal Zener rationale. Common Pitfalls: Assuming SBS is just two ordinary diodes; it is a bilateral trigger built from SUS-like structures with matched characteristics. Final Answer: Both A and R are true but R is not correct explanation of A

Question 8

Control nature of BJT versus MOSFET Which statement correctly classifies the controlling quantity of a bipolar junction transistor (BJT) and a metal–oxide–semiconductor field-effect transistor (MOSFET)?

Accepted Answer

Introduction / Context: Knowing whether a device is current- or voltage-controlled is crucial for gate/base drive design, driver power dissipation, and dynamic performance in power electronics. BJTs and MOSFETs represent two fundamentally different control paradigms. Given Data / Assumptions: BJT: requires base current proportional to collector current (via beta) to maintain conduction. MOSFET: gate forms an insulated capacitor; conduction is governed by gate-to-source voltage exceeding threshold. Steady-state viewpoints; transient gate charging for MOSFET acknowledged but does not change classification. Concept / Approach: A BJT is a transconductance device where collector current is controlled by base-emitter current and voltage; practically, the base must be driven with current (current-controlled). A MOSFET channel is formed by an electric field; the gate is insulated and ideally draws negligible steady current, so the device is voltage-controlled, with drive power mainly due to charging/discharging gate capacitances during switching. Step-by-Step Solution: BJT: I_C ≈ β * I_B → base current required → current-controlled.MOSFET: channel forms for V_GS > V_th → gate current ≈ 0 in steady state → voltage-controlled.Therefore, the correct classification: BJT current-controlled, MOSFET voltage-controlled. Verification / Alternative check: Datasheets specify base drive current for BJTs, while MOSFET drivers are rated by gate charge (Q_G) and allowable V_GS—consistent with current- vs voltage-control distinctions. Why Other Options Are Wrong: Both current-controlled or both voltage-controlled: contradict device physics and practical drive requirements. BJT voltage-controlled and MOSFET current-controlled: reversed from reality. Common Pitfalls: Confusing instantaneous charging current of MOSFET gates (during switching) with steady-state control nature; the control variable remains V_GS. Final Answer: BJT is current-controlled and MOSFET is voltage-controlled.

Question 9

Chopper quadrant classification According to the standard chopper classes (A–E), which statement about a Class E chopper is correct?

Accepted Answer

Introduction / Context: Chopper classes categorize the possible voltage–current polarity combinations at the load (i–v plane). This determines whether the converter can handle motoring and regenerative braking in forward and reverse directions, which is critical for electric drives. Given Data / Assumptions: Class A: first quadrant (V > 0, I > 0). Class B: second quadrant (V > 0, I < 0). Class C: first and second quadrants (two-quadrant, non-reversing voltage, reversing current). Class D: first and fourth quadrants (reversing voltage, non-reversing current). Class E: four-quadrant operation (reversing voltage and current). Concept / Approach: A Class E topology combines the features that allow both voltage and current to reverse under control. This enables all combinations of motoring and regenerative braking in both directions, hence fully four-quadrant behavior. Step-by-Step Solution: Identify quadrant capability per class.Class E supports V and I reversal → quadrants I, II, III, IV.Therefore, the correct description is four-quadrant operation. Verification / Alternative check: Standard power-electronics texts depict Class E as combination chopper circuits allowing bidirectional power flow and voltage reversal. Why Other Options Are Wrong: Single- or two-quadrant statements limit operation and refer to other classes (A, B, C, D). Common Pitfalls: Confusing regenerative capability (negative power flow) with mere current reversal; true four-quadrant control requires both voltage and current control. Final Answer: It can operate in all four quadrants.

Question 10

Commutation dynamics – underdamped cases In which forced-commutation class(es) does the overall commutation circuit operate in an underdamped (oscillatory) manner?

Accepted Answer

Introduction / Context: Forced commutation methods use auxiliary L–C networks to momentarily reverse-bias and turn off a conducting thyristor. The damping of the L–C network—underdamped, critically damped, or overdamped—governs how current and voltage waveforms evolve during commutation. Given Data / Assumptions: Class A (load commutation / resonant commutation): relies on a resonant L–C (with the load) to produce current reversal. Class B (parallel or impulse commutation): uses an L–C branch to force a reverse current pulse through the thyristor. Ideal components for conceptual analysis. Concept / Approach: In both Class A and Class B commutation, an oscillatory current is intentionally created in an L–C path. This implies an underdamped response (sinusoidal or pseudo-sinusoidal). The natural frequency ω0 = 1/√(LC) sets the time scale over which the thyristor is reverse biased. Correct sizing of L and C ensures that reverse current exceeds the load current for at least the device turn-off time (tq). Step-by-Step Solution: Class A: commutation occurs because the L–C current naturally goes negative → underdamped resonance with load.Class B: a charged capacitor is switched into an L–C loop, producing a decaying sinusoidal current pulse → underdamped.Therefore, the overall commutation loop is underdamped in both classes. Verification / Alternative check: Waveforms in standard references show sinusoidal current reversal (Class A) and oscillatory capacitor current pulses (Class B), confirming underdamped behavior. Why Other Options Are Wrong: Class C and others use different mechanisms (e.g., complementary current paths) and are not generically defined by an underdamped resonant turn-off loop. Common Pitfalls: Equating “forced” with “non-oscillatory”; many forced-commutation schemes purposely exploit oscillation to achieve reliable reverse bias. Final Answer: Both Class A and Class B

Question 11

Single-phase full-bridge inverter feeding an RLC load Considering a full-bridge single-phase inverter with antiparallel diodes and an RLC load (allowing current to continue when devices turn off), which devices conduct simultaneously at any instant?

Accepted Answer

Introduction / Context: In voltage-source inverters with antiparallel diodes, device conduction patterns depend on the load current polarity and whether forced commutation has just occurred. Understanding which pair conducts at any instant is essential for analyzing output waveforms and device current ratings. Given Data / Assumptions: Full-bridge with four controlled devices (e.g., thyristors) and their antiparallel diodes. Load is RLC, so current can lead or lag, and can persist after device turn-off. Ideal switches and diodes. Concept / Approach: For a full bridge, power is delivered through a diagonal pair of devices. If the commanded polarity matches the load current direction, two thyristors in opposite legs conduct. When current must flow opposite to the commanded device polarity (e.g., due to energy fed back from the reactive load), the antiparallel diodes of the opposite devices conduct, clamping the load and circulating current until the next gating interval. Thus, conduction alternates between a pair of thyristors or a pair of diodes, never all four simultaneously. Step-by-Step Solution: Positive output polarity interval → one upper and the opposite lower thyristor conduct.If current reverses (reactive load) during this interval → the two corresponding diodes conduct (freewheeling or reactive current path).On the next half-cycle, the complementary diagonal pair takes over with the same logic. Verification / Alternative check: Textbook current paths show either device–device conduction or diode–diode conduction; mixed thyristor–diode conduction happens transiently at commutation but the steady pairing is as described. Why Other Options Are Wrong: “One thyristor and one diode” (option a): not the dominant simultaneous pair in steady conduction intervals. “Two thyristors only” (option b) ignores diode conduction during reactive intervals. “All four thyristors” (option d) is impossible for a voltage-source inverter. Common Pitfalls: Forgetting that antiparallel diodes provide a path for reactive energy return, which is vital with inductive loads. Final Answer: Either two thyristors or two diodes

Question 12

Parallel (Class B) inverter – relation of turn-off time to L and C For a parallel inverter employing an L–C commutation circuit, how is the required thyristor turn-off time tq related to the commutation components L and C?

Accepted Answer

Introduction / Context: In Class B (parallel) commutation, a charged capacitor and inductor form an oscillatory loop that forces a reverse current through the conducting thyristor to turn it off. The commutation interval must exceed the device turn-off time tq. This links tq to the natural resonant period of the L–C network. Given Data / Assumptions: Ideal L and C forming a lossless resonant path. Natural frequency ω0 = 1 / sqrt(L * C). Reverse current is provided roughly over a half-cycle of the L–C oscillation. Concept / Approach: Because the capacitor discharges and then recharges with reversed polarity through the inductor, the reverse bias on the thyristor lasts approximately for a half-sine of the L–C oscillation. The duration of a half-cycle is (π / ω0) = π * sqrt(L * C). To ensure reliable turn-off, this interval must be ≥ tq, hence the design criterion tq ≤ π * sqrt(L * C). Step-by-Step Solution: ω0 = 1 / sqrt(L * C).Half-cycle time Thalf = π / ω0 = π * sqrt(L * C).Require Thalf ≥ tq → tq = π * sqrt(L * C) as the sizing reference. Verification / Alternative check: Many design notes add a safety factor (e.g., 1.2–1.5) on top of the basic π * sqrt(L * C) interval to account for non-idealities and device parameter spread, confirming the proportionality. Why Other Options Are Wrong: Linear LC expressions (options b–d) ignore the square-root relationship inherent to resonance. Common Pitfalls: Confusing full-cycle period 2π * sqrt(L * C) with the half-cycle interval that actually establishes reverse bias during commutation. Final Answer: tq = π * sqrt(L * C)

Question 13

AC voltage controller with resistive load at α = 90° A single-phase full-wave AC regulator (antiparallel thyristors) supplies a purely resistive load with firing angle α = 90°. What is the input power factor (considering distortion and displacement …

Accepted Answer

Introduction / Context: For AC voltage controllers feeding resistive loads, firing angle control chops the input voltage within each half-cycle. The current remains in phase with the instantaneous voltage during conduction but is discontinuous, producing harmonic distortion. The practical input power factor therefore reflects both displacement (≈ unity for R load during conduction) and distortion due to waveform chopping. Given Data / Assumptions: Single-phase, full-wave controller with antiparallel thyristors. Purely resistive load, so within conduction intervals current and voltage are in phase. Firing angle α = 90°; conduction from ωt = π/2 to π in each half-cycle. Power factor defined as real power over apparent power, including distortion effects. Concept / Approach: The input power factor PF can be expressed as PF = (Fundamental active power)/(Apparent power). For a resistive load the displacement factor is ≈ 1, and the PF reduction is dominated by distortion. At α = 90°, the fundamental component of the chopped sine is reduced such that the effective PF is about 0.707. This commonly cited result arises from Fourier analysis of the conduction window and is a standard benchmark for AC regulators at quarter-cycle conduction per half-cycle. Step-by-Step Solution: Conduction per half-cycle: from π/2 to π (and 3π/2 to 2π).Fundamental component magnitude is reduced relative to the RMS of the chopped waveform.Computing PF via Fourier series for the chopped sine at α = 90° yields PF ≈ 0.707. Verification / Alternative check: Design examples in AC voltage controller texts often quote PF ≈ 0.707 for α = 90° with R load, consistent with the intuitive “half-power-like” reduction linked to conduction over a quarter cycle per half-cycle. Why Other Options Are Wrong: Unity PF: ignores distortion due to chopping. 0.5 or 0.2 lagging: too low for R load at α = 90°; these would imply significantly poorer utilization than observed. Common Pitfalls: Confusing displacement factor with total power factor; for R loads the displacement is near 1 but distortion can be significant at large firing angles. Final Answer: 0.707 lagging

Question 14

Transformer Utilization Factor (TUF) for a rectifier A rectifier is fed by a transformer whose secondary RMS voltage and current are Vs and Is. If the DC output power is Vdc * Idc and the apparent AC power at the transformer secondary is Vs * Is, wh…

Accepted Answer

Introduction / Context: Transformer utilization factor (TUF) measures how effectively a transformer's VA rating is converted into useful DC output power by a rectifier. It is a convenient figure of merit when comparing rectifier topologies and transformer sizing in power supplies. Given Data / Assumptions: Transformer secondary RMS voltage = Vs. Transformer secondary RMS current = Is. DC output power = Vdc * Idc. Apparent AC power drawn from secondary = Vs * Is. Concept / Approach: TUF is defined as the ratio of DC output power to the VA loading of the transformer secondary. Because rectifiers draw non-sinusoidal currents with harmonics, Is can be much greater than the DC current, and the ratio captures how much of the transformer's VA capacity becomes usable DC. Step-by-Step Solution: By definition: TUF = Pdc / (Vs * Is).Substitute Pdc = Vdc * Idc.Therefore, TUF = (Vdc * Idc) / (Vs * Is). Verification / Alternative check: Known TUF values (e.g., ~0.286 for half-wave, ~0.693 for full-wave bridge, depending on definitions) are computed using the same ratio, confirming the expression. Why Other Options Are Wrong: Inverting the ratio (option b) gives the reciprocal and is not the standard definition. Algebraic sums or squared sums (options c and d) have no physical meaning for utilization in this context. Common Pitfalls: Confusing TUF with rectifier efficiency or power factor; TUF focuses on transformer VA usage, not only on losses. Final Answer: TUF = (Vdc * Idc) / (Vs * Is)

Question 15

Highly inductive load with single-phase full control For the single-phase full-controlled bridge shown (two devices conducting per interval) supplying a highly inductive load, each thyristor is fired at angle α. What are the conduction intervals of …

Accepted Answer

Introduction / Context: With a highly inductive (large L) load, current is nearly constant and cannot change direction instantaneously. In a full-controlled single-phase bridge, each thyristor therefore conducts for more than a half-cycle once triggered, overlapping into the next half-cycle by approximately the firing angle. Determining conduction intervals is a staple skill in rectifier analysis. Given Data / Assumptions: Single-phase full bridge, firing angle α for each device pair. Load highly inductive → continuous current; commutation at device firing instants. Ideal devices, negligible overlap due to source inductance (focus on load inductance effect). Concept / Approach: For continuous current, when a new pair is fired at angle α, current transfers from the outgoing pair to the incoming pair; the outgoing thyristor pair then turns off. Each thyristor thus conducts for a duration of π radians. Counting from its own firing instant, the conduction ends one half-cycle later at α + π. Therefore, the first device pair conducts from α to α + π; the complementary pair conducts from π + α to 2π + α, and so on. Step-by-Step Solution: Fire Th1 at ωt = α → Th1 conducts until ωt = α + π.Next pair fires at ωt = π + α → Th2 (complementary) conducts until ωt = 2π + α.Hence intervals are [α, π + α] for Th1 and [π + α, 2π + α] for Th2. Verification / Alternative check: Sketching the current flow for a large L load shows rectangle-like current with phase-shifted commutations at every α beyond the standard π boundaries, matching the intervals above. Why Other Options Are Wrong: 0 to π / π to 2π (option a) ignore the firing delay α. α to π; π + α to 2π (option b) truncate conduction to less than π radians. α to π − α; π + α to 2π − α (option d) are characteristic of discontinuous or resistive cases, not highly inductive continuous current. Common Pitfalls: Confusing effects of load inductance (extends conduction) with source inductance (causes overlap but not the same interval extension). For “highly inductive” analyses we assume continuous current transfer at firing instants. Final Answer: α to π + α and π + α to 2π + α

Question 16

Single-phase semiconverter feeding a highly inductive load with a freewheeling diode across the load What is the typical input (source) current waveform drawn from the AC supply under continuous-current conditions?

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Introduction / Context: Power electronic rectifiers seldom draw sinusoidal current unless special conditioning is used. A single-phase semiconverter (half-controlled bridge) supplying a highly inductive load with a freewheeling diode is a classic case where the load current is nearly constant, yet the source current is discontinuous and distinctly non-sinusoidal. Understanding the input current shape is vital for estimating harmonics, power factor, and line-current distortion. Given Data / Assumptions: Single-phase semiconverter (two SCRs and two diodes) on an AC source. Load is highly inductive, so load current is approximately constant (continuous conduction). A freewheeling diode is connected across the load. Commutation overlap is neglected for conceptual clarity. Concept / Approach: With high inductance, the load current changes slowly and can be approximated as constant over a half cycle. When the controlled devices conduct (from firing angle α to near the line-current zero), the source must supply approximately that constant current. When the freewheeling diode conducts, the load current circulates locally and the source current drops to zero, creating a pulsed input current. The approximate constancy of current during conduction makes the pulses nearly rectangular in magnitude. Step-by-Step Solution: Identify conduction intervals: from ωt = α to ωt ≈ π, SCR/diode pair conducts; from π to next firing, the freewheeling diode carries current.During SCR conduction: source current ≈ constant (equal to the nearly constant load current) → flat-topped pulse.During freewheeling interval: source current ≈ 0.Over one cycle: the supply sees discontinuous, rectangular-like current pulses. Verification / Alternative check: Textbook waveforms for semiconverters with freewheeling diodes show source current as rectangular pulses whose width is determined by firing angle and load current continuity. Harmonic analysis confirms high harmonic content typical of such pulsed currents. Why Other Options Are Wrong: Sinusoidal is incorrect because the current does not follow the voltage due to controlled conduction and freewheeling. Triangular would require a linearly changing current, not present with large inductance. Saying “sinusoidal or rectangular depending only on firing angle” ignores the dominant role of load inductance and freewheeling in fixing the pulse shape. Common Pitfalls: Confusing rectifier source current with load current; assuming any AC-fed circuit draws sinusoidal current; forgetting that freewheeling eliminates source current during portions of the cycle. Final Answer: Rectangular (pulsed, near-constant magnitude during conduction)

Question 17

Crowbar protection in power electronics Is a crowbar circuit generally employed to protect equipment under fault or overvoltage conditions?

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Introduction / Context: Crowbar circuits are widely used overvoltage/overcurrent protection schemes in power supplies, converters, and motor drives. They act like an emergency short across the supply (via an SCR or similar device) to quickly clamp dangerous voltages and blow a fuse or trip a breaker, safeguarding sensitive loads such as logic circuits or power semiconductors. Given Data / Assumptions: The protected system is a DC or rectified supply feeding sensitive electronics. Faults may include regulator failure, transient surges, or component short/open that results in dangerous overvoltage at the output. The crowbar uses a fast trigger (e.g., zener reference, comparator) and a latching switch (e.g., SCR or triac). Concept / Approach: When the monitored voltage exceeds a reference threshold, the crowbar triggers an SCR/triac that creates a low-impedance path across the supply lines. This clamps the voltage to a safe level (near zero) and forces an upstream fuse/circuit breaker to open, isolating the fault. Because the SCR latches, the system remains shut down until power is removed and the fault is corrected. Step-by-Step Solution: Detect overvoltage using a reference (zener/IC).Trigger the crowbar device (SCR/triac) upon threshold crossing.Clamp and divert fault energy to ground/return line.Upstream protection clears (fuse blows or breaker trips), preventing damage. Verification / Alternative check: Bench tests show the crowbar rapidly pulling the output low during regulator runaway. The protection is robust, fast, and cost-effective compared to series “foldback” limiters for catastrophic faults. Why Other Options Are Wrong: It is not exclusive to undervoltage, nor limited to only brief transients; it is specifically intended for damaging faults/overvoltages. Calling it false ignores standard power-supply protection practice. Common Pitfalls: Confusing crowbar (shunt/shorting protection) with series pass limiters; forgetting that the crowbar requires proper fuse sizing and safe energy paths. Final Answer: True

Question 18

Single-phase fully controlled bridge converter supplied by v_s(t) = Vm sin(ωt) What are the expressions for the average and RMS output voltages (continuous current, no overlap)?

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Introduction / Context: The fully controlled single-phase bridge (four SCRs) allows control of the average DC output voltage by adjusting the firing angle α. Under continuous current and neglecting source inductance (no overlap), the RMS output voltage equals the source RMS voltage because the squared waveform energy over one period is unchanged by phase control, while the average value depends on α. Given Data / Assumptions: Input: v_s(t) = Vm sin(ωt). Fully controlled bridge, continuous load current. No overlap and ideal devices (no drops). Output voltage segments are sinusoidal portions selected by gating (shifted by α). Concept / Approach: For continuous current, the bridge transfers the instantaneous source voltage (with polarity control) to the load from ωt = α to ωt = α + π each cycle. The average over one period is proportional to cos α. The RMS value, being energy-based, remains equal to the source RMS without overlap since the waveform is a full sine segment over π radians each half-cycle. Step-by-Step Solution: Average: V_avg = (1/π) ∫{α}^{α+π} Vm sin θ dθ = (2Vm/π) * cos α.RMS: V_rms^2 = (1/2π) ∫{0}^{2π} v_o^2(θ) dθ = (1/2π) ∫ Vm^2 sin^2 θ dθ = (Vm^2/2) → V_rms = Vm/√2.Therefore: V_avg = (2Vm/π) cos α and V_rms = Vm/√2. Verification / Alternative check: As α increases from 0° to 90°, V_avg decreases from 2Vm/π to 0, while V_rms remains Vm/√2, matching standard converter results for continuous-current operation without overlap. Why Other Options Are Wrong: Options with Vm/π or Vm scaling are dimensionally or conceptually incorrect. Swapping cos α between average and RMS is wrong since RMS is independent of α in this idealized case. Common Pitfalls: Confusing average with RMS; forgetting that overlap and device drops would reduce RMS slightly; misinterpreting Vm as RMS instead of peak. Final Answer: V_avg = (2Vm/π) * cos α and V_rms = Vm/√2

Question 19

Bridge inverter usage of output transformer In a full-bridge DC–AC inverter, is an output transformer essential for operation?

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Introduction / Context: Modern full-bridge inverters (using IGBTs/MOSFETs or SCRs) synthesize AC directly from a DC link by switching legs in an H-bridge. Whether an output transformer is required depends on isolation, voltage matching, and application constraints, not on the fundamental operation of the bridge topology itself. Given Data / Assumptions: H-bridge output can be filtered to produce sinusoidal AC. Load voltage may be matched via PWM modulation without a transformer. Isolation/step-up may be optional depending on safety or grid requirements. Concept / Approach: A bridge inverter can deliver AC directly to the load when voltage levels are appropriate. An output transformer may be added for galvanic isolation or to step the voltage up/down. It is not inherently necessary for the inverter to function. Efficiency may decrease with a transformer due to core and copper losses; using one does not guarantee higher efficiency. Step-by-Step Solution: Assess voltage match: if DC bus and modulation can meet required AC output, no transformer is needed.Assess isolation: regulatory/safety needs may dictate a transformer, otherwise it is optional.Conclude: operation does not require a transformer; it is an application-driven choice. Verification / Alternative check: UPS inverters and motor drives frequently use transformerless full bridges with line filters, proving a transformer is optional, not essential. Why Other Options Are Wrong: “Essential” is incorrect; “cannot be connected” is false since many designs intentionally use transformers; “improves efficiency” is not universally true. Common Pitfalls: Equating isolation needs with operational necessity; ignoring PWM and LC filtering capabilities that obviate transformers. Final Answer: An output transformer is not essential

Question 20

Modified series inverter (two inductors and two capacitors) How many parallel current paths exist for the load current in this topology?

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Introduction / Context: The modified series inverter (e.g., McMurray/McMurray-Bedford variants) uses split capacitors and inductors to realize efficient commutation and improved control of current. Recognizing the number of parallel paths for load current helps in understanding commutation intervals, current sharing, and device stress. Given Data / Assumptions: Two inductors and two capacitors are included as commutating elements. H-type arrangement with series resonance used for device turn-off. Load current is continuous or quasi-continuous due to resonant paths. Concept / Approach: The split L–C network provides alternative current routes during commutation. When one device pair conducts, a resonant path through one L–C branch develops to aid in turning off the conducting device and transferring current. Thus, the load current can flow through either of two parallel resonant paths depending on the switching state. Step-by-Step Solution: During a half-cycle, current flows via the principal path associated with the active device pair.At commutation, the auxiliary resonant branch provides an alternative loop—effectively a second parallel path—allowing current reversal or energy exchange for turn-off.Therefore the load current can utilize two distinct parallel paths over the switching cycle. Verification / Alternative check: Standard circuit diagrams show two L–C branches arranged so that each half-cycle has its own resonant commutation path, confirming two effective parallels for current. Why Other Options Are Wrong: One path ignores the commutation branch; three or four paths overstate the available distinct resonant routes in the canonical two-L, two-C configuration. Common Pitfalls: Confusing conduction paths of switching devices with unique current paths through the resonant network; overlooking the role of the auxiliary branch during commutation. Final Answer: Two parallel paths

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