22-Elec-B8 Power Electronics and Drives · Undated paper
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Paper format. National Examinations, May 2019 — 16-Elec-B8 Power Electronics and Drives. Three hours; open book; any non-communicating calculator, whose make and model must be written on the first inside sheet of the work book. The paper is in two parts and the candidate must attempt all parts: Part 1-A is ten five-point short-answer items, Part 1-B is ten five-point multiple-choice items each requiring a written explanation, and Part 2 is four thirty-point problems. Page-1 Note 4 states that the maximum total score is 220 points and that 150 points is a full mark (100 per cent), so there are 70 points of built-in bonus. Note 5 warns that an answer without its working scores nothing. All a.c. voltages and currents below are rms unless stated otherwise, and three-phase voltages are line-to-line. Every part of every question is solved below, in the exam's own order, with the six sections mapping to Part 1-A, Part 1-B and PROBLEMs 1 to 4.
Reference texts.
Two printing anomalies in the paper itself, both worth a line in the answer book. First, Part 1-B Question 9 repeats Question 5 word for word (“AC voltage controllers convert…”) — ten of the fifty Part 1-B points are the same item asked twice. Both are answered below, the second with the quantitative detail the first does not need. Second, the arithmetic of the cover page does not close in the candidate's favour by accident: 50 + 50 + 120 = 220 points are on offer against a full mark of 150, so a candidate should attempt every part and let the surplus absorb the inevitable slips rather than budgeting time as if 100 points were the target.
Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.
The mark scheme gives no credit for a bare letter: each box carries an Explanation line, and Note 5 on page 1 warns that failure to show the reasoning results in a null mark for that part. Each answer below therefore states the choice and then the physics that eliminates the distractors.
In a three-phase half wave rectifier the primary side of the transformer is delta connected because:
Answer: (c) it provides a path for the triplen harmonics.
A three-phase half-wave (three-pulse) rectifier draws one unidirectional current pulse per phase, so each secondary winding carries a d.c. component together with a strong third-harmonic component. The third harmonic is a triplen: it is in phase in all three lines, so it is a zero-sequence quantity and cannot flow in a three-wire star primary. Denied a path, it appears instead as third-harmonic flux in the core, distorting the phase voltages and adding loss. A delta-connected primary provides a closed loop in which the triplen ampere-turns can circulate and balance the secondary triplen m.m.f., so the flux and the terminal voltages stay clean. Option (a) is a description of the delta connection, not a reason; the output voltage $V_{dc}=1.17V_{ph}$ is set by the secondary and by the pulse number, not by the primary connection, so (b) is wrong; and (d) is irrelevant.
Which device can be used in a chopper circuit?
Answer: (d) All the above.
A chopper is a d.c.–d.c. converter, and the only requirement its switch must meet is forced commutation: because the source is d.c. there is no current zero to turn the device off naturally, so the switch must be turnable off by its own control terminal. The power BJT (base current removal), the power MOSFET (gate discharge) and the gate turn-off thyristor (a negative gate pulse) all satisfy this, and each is used in practice at a different point of the voltage–frequency plane: MOSFETs for low-voltage high-frequency choppers, BJTs and IGBTs at intermediate ratings, GTOs and IGCTs in traction and other megawatt drives. Only the ordinary SCR fails the test — it needs an auxiliary commutation circuit — and it is not among the options. Hence (d).
The chopper is a
Answer: (a) Time ratio controller.
The defining feature of a chopper is that it regulates its d.c. output by varying the ratio of on-time to period, $\delta=T_{on}/T$, giving $V_{o}=\delta V_{i}$ — which is precisely what the term time-ratio control names. The two sub-variants are constant-frequency (pulse-width) modulation and variable-frequency (pulse-rate) modulation, and both are time-ratio schemes. Option (b) is simply the wrong conversion: a chopper is d.c. to d.c., and a.c. to d.c. is a rectifier. Option (c) is a loose analogy some texts use for the step-up/step-down behaviour, but a transformer cannot pass d.c. and the analogy is not the operating principle. Option (d) describes the chopper's switch, not the converter. The correct classification is therefore (a).
Static UPS requires
Answer: (c) both inverter and rectifier.
A static uninterruptible power supply has to do two things continuously. It must keep its battery charged from the a.c. mains, which requires a rectifier (in the on-line topology this also feeds the d.c. link directly); and it must produce clean a.c. for the critical load from that d.c., which requires an inverter. In the on-line or double-conversion arrangement the load is permanently fed a.c.→d.c.→a.c., so a mains failure causes no transfer transient at all — the battery simply takes over the d.c. link. A static bypass switch and, usually, an output filter and isolating transformer complete the unit. Neither converter alone is sufficient, so the answer is (c).
AC voltage controllers convert
Answer: (c) fixed ac to variable ac.
An a.c. voltage controller consists of antiparallel thyristors (or a TRIAC) in series with the load; the devices are phase-controlled at a delay angle $\alpha$ and commutate naturally at the current zero. Because no energy storage or frequency changing element is present, the output is at the same frequency as the supply but with a reduced rms value, $V_{o,rms}$ decreasing monotonically with $\alpha$. It is therefore a fixed-a.c. to variable-a.c. converter, option (c). Options (a) and (b) describe rectifiers, and (d) would require a converter that regulates against supply variation, which a simple phase-controlled a.c. controller does not do by itself.
In the principle of phase control
Answer: (b) control is achieved by adjusting the firing angle of the devices.
Phase control works within each half cycle: the gate pulse is delayed by an angle $\alpha$ after the natural turn-on instant, so the device conducts only over the remainder of the half cycle and the mean (rectifier) or rms (a.c. controller) output falls as $\alpha$ increases — for a fully controlled bridge, $V_{dc}=V_{do}\cos\alpha$. That is exactly option (b). Options (a) and (c) both describe the other control family, integral-cycle or on–off control, in which whole cycles are passed or blocked and the resolution is one cycle; that method avoids harmonics of non-integer order but produces flicker, so it is confined to thermally slow loads. Option (d) is plainly false.
HVDC transmission lines are __________ as compared to HVAC lines.
Answer: (c) more expensive for short distances.
The economics of HVDC are dominated by two cost components that scale differently. The terminal cost — converter valves, transformers, filters, reactive support — is large and essentially independent of line length, and it is several times the cost of an equivalent a.c. substation. The line cost per kilometre, by contrast, is lower for d.c.: two conductors instead of three, no charging current, no stability limit, narrower right-of-way and lower losses. The total-cost curves therefore cross at a break-even distance, conventionally about 600–800 km for overhead lines and only 40–60 km for cables. Below that distance HVDC is more expensive, which is option (c); above it, HVDC wins, so option (b) states the reverse of the truth. HVDC lines are mechanically simpler to erect, not harder, so (a) is wrong, and (d) contradicts (c).
A TRIAC is used in
Answer: (c) speed control of universal motor.
A TRIAC is a bidirectional, naturally commutated a.c. switch, so it can only be used where the current passes through zero every half cycle. That immediately excludes the chopper of option (a), which operates from a d.c. source and needs a forced-commutated switch. The classic TRIAC application is phase-angle control of a series (universal) motor in hand tools, food mixers and portable drills: reducing the rms terminal voltage reduces the speed of a universal motor over a wide, genuinely useful range, and the whole controller is a TRIAC, a diac and an R–C phase-shift network. Option (b) is only partially true — a TRIAC fan regulator does control a small single-phase induction motor, but voltage control on an induction machine changes speed only by increasing slip, works for fan-type loads alone, and a three-phase machine would need a three-device controller rather than one TRIAC. The intended answer is therefore (c).
AC voltage controllers convert
Answer: (c) fixed ac to variable ac.
This item is printed a second time, word for word, as Question 9 — see the callout at the head of this solution. The answer is unchanged: option (c). It is worth adding the quantitative side here rather than merely repeating Question 5. For a purely resistive load the rms output is $V_{o,rms}=V_{s}\sqrt{1-\alpha/\pi+\sin 2\alpha/(2\pi)}$, which gives the full supply voltage at $\alpha=0$, about $0.707V_{s}$ at $\alpha=90^\circ$, and zero at $\alpha=180^\circ$; for the inductive load of PROBLEM 2 the useful control range is compressed to $\phi\le\alpha\le180^\circ$, because firing earlier than the load angle $\phi$ simply returns the controller to full conduction. The frequency, however, is never altered — which is the whole content of the question.
In inverters, to make the supply voltage constant
Answer: (c) capacitor is connected in parallel to the supply side.
A voltage-source inverter must see a stiff d.c. bus: the switching action draws a pulsed, discontinuous current from the source, and any series impedance in the supply would turn that pulsed current into bus-voltage ripple that appears directly in the output. The remedy is a large electrolytic capacitor connected across the supply (d.c. link) terminals, which presents a low impedance to the switching harmonics, supplies the pulsed current locally, and holds the link voltage sensibly constant — option (c). A capacitor across the load, option (b), would filter the output, not the supply, and would be short-circuited by the switches. An inductor in series with the load, option (a), belongs to the current-source inverter, where a series reactor on the d.c. side makes the current constant instead.
