Nepal Engineering Council · Electrical Engineering · Chapter 5
Power Electronics and Control
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182 questions in 6 syllabus topics.
5.1 Control system fundamentals
30 questions · AElE0501
1. Which of the following is an example of an open-loop control system?
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Ask which device never checks its own output.
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Answer: B. An electric toaster that browns bread for a preset time
A timer-based toaster does not measure the output (degree of browning) and has no feedback, so it is open loop; the others measure and correct their output.
2. The basic feature that distinguishes a closed-loop control system from an open-loop system is
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Think of what the loop is made of.
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Answer: C. Feedback of the output to compare with the reference input
In a closed-loop system the output is measured and fed back so that the error between reference and output drives the controller.
3. Which is a main advantage of an open-loop control system over a closed-loop one?
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The other three are benefits of feedback.
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Answer: B. It is simple, cheap and cannot become unstable due to feedback
With no feedback path an open-loop system is simple and economical and cannot oscillate because of loop gain, but it does not correct errors.
4. Negative feedback in a control system generally
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Gain-bandwidth trade-off.
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Answer: B. Reduces overall gain but increases bandwidth
Negative feedback divides the gain by (1 + GH) and, as a trade-off, widens the bandwidth.
5. Which of the following is NOT an effect of introducing negative feedback?
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Recall what can happen with excessive loop gain.
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Answer: A. It always guarantees stability of the system
Feedback can make an otherwise stable system unstable if the loop gain is too high, so it does not guarantee stability; the other three are standard effects.
6. The transfer function of a system is defined as the ratio of
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Which way round, and what about initial conditions?
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Answer: C. Laplace transform of output to Laplace transform of input, with all initial conditions zero
By definition T(s) = C(s)/R(s) evaluated with zero initial conditions.
7. The transfer function concept is applicable to
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It needs superposition and constant coefficients.
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Answer: C. Linear time-invariant systems
Transfer functions rely on the Laplace transform and superposition, so they apply to linear time-invariant (LTI) systems.
8. The inverse Laplace transform of a system's transfer function gives its
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What input has a Laplace transform equal to 1?
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Answer: C. Impulse response
For a unit impulse R(s) = 1, so C(s) = T(s) and its inverse Laplace transform is the impulse response.
9. The poles of a transfer function are the values of s at which
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Where does T(s) blow up?
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Answer: C. The denominator becomes zero
Poles are the roots of the denominator (characteristic) polynomial, where T(s) becomes infinite.
10. For a closed-loop system with forward path G(s) and feedback path H(s), the characteristic equation is
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Set the closed-loop denominator to zero.
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Answer: A. 1 + G(s)H(s) = 0
For negative feedback T(s) = G/(1 + GH); setting the denominator to zero gives 1 + G(s)H(s) = 0.
11. In a closed-loop system, the signal obtained by subtracting the feedback signal from the reference input is called the
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It is the comparator's output.
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Answer: B. Actuating (error) signal
The comparator output e = r − b is the actuating or error signal that drives the controller.
12. A feedback control system in which the output is a mechanical position, velocity or acceleration is called a
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Think of a servo motor.
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Answer: A. Servomechanism
Servomechanisms are feedback systems whose controlled output is mechanical position or its derivatives.
13. A system has forward gain G = 10 and feedback gain H = 0.4 with negative feedback. Its closed-loop gain is
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Use G/(1 + GH).
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Answer: D. 2
T = G/(1 + GH) = 10/(1 + 4) = 2.
14. A system with forward gain G = 5 has positive feedback of H = 0.1. The closed-loop gain is
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The sign in the denominator changes for positive feedback.
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Answer: D. 10
For positive feedback T = G/(1 − GH) = 5/(1 − 0.5) = 10.
15. For a closed-loop system with G = 100 and H = 0.09, the sensitivity of the closed-loop gain to changes in G is
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Sensitivity to G is 1/(1 + GH).
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Answer: C. 0.1
S = 1/(1 + GH) = 1/(1 + 9) = 0.1.
16. A unity-feedback amplifier has open-loop gain 99. If the open-loop gain changes by 10 %, the closed-loop gain changes by about
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Multiply the change by the sensitivity.
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Answer: D. 0.1 %
Percentage change in T ≈ (1/(1 + GH)) × 10 % = (1/100) × 10 % = 0.1 %.
17. The DC (steady-state) gain of G(s) = 10/(s + 2) is
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Put s = 0.
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Answer: D. 5
DC gain is G(0) = 10/2 = 5.
18. The transfer function (s + 1)/(s² + 5s + 6) has poles at
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Factor the denominator.
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Answer: B. s = −2 and s = −3
s² + 5s + 6 = (s + 2)(s + 3), so the poles are at s = −2 and s = −3 (the zero is at s = −1).
19. The order of a system with transfer function (s + 1)/(s³ + 2s² + s + 5) is
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Look at the denominator only.
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Answer: A. 3
The order equals the highest power of s in the denominator, which is 3.
20. For an RC low-pass network with R = 10 kΩ and C = 10 µF, the transfer function Vo(s)/Vi(s) is
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Find the time constant RC first.
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Answer: D. 1/(0.1s + 1)
Vo/Vi = 1/(RCs + 1) with RC = 10 000 × 10 × 10⁻⁶ = 0.1 s.
21. Two blocks G1 = 2/(s + 1) and G2 = 3/(s + 2) are connected in cascade. The equivalent transfer function is
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Series blocks multiply, parallel blocks add.
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Answer: A. 6/((s + 1)(s + 2))
Blocks in series multiply: (2 × 3)/((s + 1)(s + 2)).
22. Two blocks G1 = 2/(s + 1) and G2 = 3/(s + 2) are connected in parallel with their outputs added. The equivalent transfer function is
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Add over a common denominator.
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Answer: B. (5s + 7)/((s + 1)(s + 2))
G1 + G2 = [2(s + 2) + 3(s + 1)]/((s + 1)(s + 2)) = (5s + 7)/((s + 1)(s + 2)).
23. A unity-feedback system has G(s) = K/(s(s + 4)). Its closed-loop transfer function is
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Use G/(1 + G) and clear the fraction.
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Answer: D. K/(s² + 4s + K)
T = G/(1 + G) = K/(s(s + 4) + K) = K/(s² + 4s + K).
24. In block-diagram reduction, when a take-off point is moved from after a block G to before it, the branch must be multiplied by
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Keep the signal in the branch unchanged.
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Answer: B. G
Before the move the branch carried G×input; after moving the take-off point ahead of G it carries only the input, so a block G must be inserted in the branch.
25. The 'type' of a feedback control system refers to
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Count the integrators.
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Answer: A. The number of open-loop poles at the origin
System type is the number of pure integrators (poles at s = 0) in the open-loop transfer function G(s)H(s).
26. In Mason's gain formula, Δ is equal to
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It is built only from loop gains.
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Answer: A. 1 − (sum of individual loop gains) + (sum of products of gains of all pairs of non-touching loops) − …
Δ = 1 − ΣL1 + ΣL2 − ΣL3 + …, where L2, L3 are products of non-touching loop pairs, triples, etc.
27. A signal flow graph has one forward path of gain 20 and a single loop of gain −4 touching that path. The overall gain is
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Δ = 1 − (loop gain); Δ1 = 1 since the loop touches the path.
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Answer: D. 4
By Mason's rule, T = P1Δ1/Δ = 20 × 1/(1 − (−4)) = 20/5 = 4.
28. A unity-feedback type-0 system has position error constant Kp = 9. The steady-state error for a unit-step input is
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For a step, ess = 1/(1 + Kp).
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Answer: B. 0.1
ess = 1/(1 + Kp) = 1/(1 + 9) = 0.1.
29. A transfer function in which the degree of the numerator is less than the degree of the denominator is called
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Compare the degrees of numerator and denominator.
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Answer: A. Strictly proper
A strictly proper transfer function has numerator degree lower than denominator degree, as in most physical systems.
30. Which statement about the transfer function of an LTI system is correct?
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It is a property of the system, not of the signal.
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Answer: C. It does not depend on the magnitude or type of the input
The transfer function is a property of the system alone; it is independent of input and gives no unique physical structure (different systems can share one TF).
5.2 Time domain analysis
30 questions · AElE0502
31. The time constant of a first-order system is the time taken by its unit-step response to reach
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Evaluate 1 − e⁻¹.
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Answer: B. 63.2 % of its final value
For c(t) = 1 − e^(−t/τ), at t = τ the response is 1 − e⁻¹ = 0.632 of the final value.
32. For the first-order system G(s) = 10/(s + 5), the time constant and the steady-state value of the unit-step response are
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Put it in the form K/(τs + 1).
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Answer: D. 0.2 s and 2
Write G = 2/(0.2s + 1): τ = 0.2 s and the final value is G(0) = 10/5 = 2.
33. Using the 2 % criterion, the settling time of a first-order system with time constant τ is approximately
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When does e^(−t/τ) drop to 0.02?
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Answer: A. 4τ
e^(−t/τ) falls to 2 % at t ≈ 3.9τ, commonly taken as 4τ.
34. The 10 %–90 % rise time of a first-order system with time constant τ is about
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Subtract the times at which the response reaches 10 % and 90 %.
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Answer: B. 2.2τ
The response reaches 10 % at t = τ ln(1/0.9) ≈ 0.105τ and 90 % at t = τ ln 10 ≈ 2.303τ, so tr = τ ln 9 ≈ 2.2τ.
35. For a second-order system with characteristic equation s² + 4s + 16 = 0, the natural frequency and damping ratio are
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Match with s² + 2ζωn s + ωn².
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Answer: D. 4 rad/s and 0.5
Compare with s² + 2ζωn s + ωn²: ωn² = 16 → ωn = 4; 2ζωn = 4 → ζ = 0.5.
36. A second-order system has ζ = 0.5. The peak percentage overshoot of its unit-step response is about
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Overshoot depends only on ζ.
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Answer: D. 16.3 %
Mp = exp(−ζπ/√(1 − ζ²)) = exp(−0.5π/0.866) ≈ 0.163, i.e. 16.3 %.
37. A second-order system has ωn = 4 rad/s and ζ = 0.5. Its peak time is about
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Use the damped frequency, not ωn.
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Answer: A. 0.91 s
tp = π/ωd, with ωd = ωn√(1 − ζ²) = 4 × 0.866 = 3.46 rad/s, so tp ≈ 0.91 s.
38. For a second-order system with ζ = 0.6 and ωn = 5 rad/s, the 2 % settling time is approximately
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Use 4/(ζωn).
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Answer: B. 1.33 s
ts ≈ 4/(ζωn) = 4/(0.6 × 5) = 1.33 s.
39. A second-order system whose damping ratio lies between 0 and 1 is
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Complex poles with negative real part.
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Answer: B. Underdamped, with oscillatory decaying step response
For 0 < ζ < 1 the poles are complex conjugate in the left half plane, giving a decaying oscillation.
40. When the damping ratio of a second-order system is zero, its step response is
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Where are the poles when ζ = 0?
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Answer: A. A sustained oscillation at the natural frequency
With ζ = 0 the poles are at ±jωn on the imaginary axis, so the response oscillates undamped at ωn.
41. Which change increases the percentage overshoot of a standard second-order system?
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Overshoot is a function of one parameter only.
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Answer: B. Decreasing the damping ratio
Overshoot depends only on ζ and rises as ζ decreases; ωn affects speed, not overshoot.
42. A linear time-invariant system is stable if all the closed-loop poles
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Negative real parts mean decaying terms.
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Answer: C. Lie in the left half of the s-plane
Poles with negative real parts give decaying exponential terms, so the response is bounded.
43. A system with a non-repeated pair of closed-loop poles on the imaginary axis and all other poles in the left half plane is
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Neither decaying nor growing.
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Answer: A. Marginally stable
Simple poles on the jω axis produce constant-amplitude oscillation, which is marginal stability.
44. In the Routh-Hurwitz criterion, the number of sign changes in the first column of the Routh array equals the number of
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Sign changes signal instability.
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Answer: D. Roots in the right half of the s-plane
Each sign change in the first column corresponds to one characteristic root with a positive real part.
45. Using the Routh array, the number of roots of s³ + s² + 2s + 8 = 0 in the right half plane is
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Build the first column: 1, 1, ?, 8.
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Answer: A. 2
First column: 1, 1, (1×2 − 1×8)/1 = −6, 8. Two sign changes, so two right-half-plane roots.
46. For the characteristic equation s³ + 2s² + 3s + K = 0, the closed-loop system is stable for
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Require (2×3 − K)/2 > 0 and K > 0.
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Answer: C. 0 < K < 6
Routh: s¹ row = (2×3 − K)/2 must be positive, so K < 6; s⁰ row K > 0. Hence 0 < K < 6.
47. A necessary (but not sufficient) condition for stability is that all coefficients of the characteristic polynomial
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Look at the coefficients before building the array.
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Answer: C. Are present and have the same sign
A missing term or a sign change among coefficients guarantees at least one root not in the open left half plane.
48. In a pole-zero plot, poles and zeros are conventionally marked respectively by
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Standard s-plane notation.
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Answer: A. Crosses (×) and small circles (o)
By convention × marks a pole and o marks a zero on the s-plane.
49. A pole located at s = −10, far to the left of a dominant pole pair at −1 ± j2, has the effect that
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Compare e^(−10t) with e^(−t).
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Answer: A. Its effect dies out quickly, so the dominant pair governs the response
The term e^(−10t) dies out much faster than e^(−t), so the poles nearest the jω axis dominate.
50. The root locus of a system starts (K = 0) at the
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Set K = 0 in 1 + KG(s)H(s) = 0.
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Answer: B. Open-loop poles
At K = 0 the characteristic equation reduces to the open-loop denominator, so the branches start at the open-loop poles.
51. The number of branches of the root locus equals
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Each branch starts somewhere.
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Answer: D. The number of open-loop poles (when poles ≥ zeros)
One branch starts from each open-loop pole, so the number of branches equals the number of poles when P ≥ Z.
52. A point on the real axis lies on the root locus if the total number of open-loop poles and zeros to its right is
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Angle condition: 180° from real poles and zeros to the right.
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Answer: D. Odd
The angle condition is satisfied on real-axis segments that have an odd number of real poles plus zeros to their right.
53. For G(s)H(s) = K/(s(s + 2)(s + 4)), the root-locus asymptotes meet the real axis (centroid) at
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Sum of poles divided by (P − Z).
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Answer: C. −2
Centroid = (Σpoles − Σzeros)/(P − Z) = (0 − 2 − 4)/3 = −2.
54. For G(s)H(s) = K/(s(s + 2)(s + 4)), the angles of the root-locus asymptotes are
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Use (2q + 1)180°/(P − Z).
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Answer: B. 60°, 180° and 300°
Angles = (2q + 1)180°/(P − Z) with P − Z = 3, giving 60°, 180°, 300°.
55. For the unity-feedback system G(s) = K/(s(s + 2)(s + 4)), the value of K at which the root locus crosses the imaginary axis, and the crossing frequency, are
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Make the s¹ row of the Routh array zero.
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Answer: C. K = 48 and ω ≈ 2.83 rad/s
Characteristic equation s³ + 6s² + 8s + K = 0. Routh s¹ row zero gives K = 6 × 8 = 48; auxiliary 6s² + 48 = 0 → ω = √8 ≈ 2.83 rad/s.
56. For G(s)H(s) = K/(s(s + 2)), the breakaway point of the root locus is at
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Set dK/ds = 0.
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Answer: D. s = −1
K = −s(s + 2) = −s² − 2s; dK/ds = −2s − 2 = 0 gives s = −1.
57. Adding integral action to a proportional controller mainly
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It adds a pole at s = 0.
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Answer: B. Eliminates steady-state error for a step input
Integral action adds a pole at the origin, raising the system type and driving step steady-state error to zero.
58. Derivative action in a PID controller
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Derivative gain grows with frequency.
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Answer: C. Improves damping by reacting to the rate of change of error, but amplifies high-frequency noise
The D term anticipates error changes, adding damping and reducing overshoot, but its gain rises with frequency so it amplifies noise.
59. The transfer function of an ideal PID controller is
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Integration is 1/s; differentiation is s.
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Answer: C. Kp + Ki/s + Kd·s
Proportional, integral (1/s) and derivative (s) terms are added: Gc(s) = Kp + Ki/s + Kd·s.
60. A unity-feedback system has G(s) = 10/(s(s + 2)). The steady-state error for a unit-ramp input is
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Find the velocity error constant Kv.
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Answer: A. 0.2
Type 1 system: Kv = lim s·G(s) = 10/2 = 5, so ess = 1/Kv = 0.2.
5.3 Frequency domain analysis
30 questions · AElE0503
61. A Bode plot consists of
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Two separate graphs on semi-log axes.
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Answer: D. Magnitude in dB and phase angle, each plotted against log of frequency
Bode plots are two separate graphs, 20 log|G(jω)| and ∠G(jω), versus frequency on a logarithmic scale.
62. A gain of 100 expressed in decibels (20 log10) is
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log10(100) = 2.
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Answer: C. 40 dB
20 log10(100) = 20 × 2 = 40 dB.
63. The asymptotic Bode magnitude plot of a simple pole factor 1/(1 + sT) has a slope above its corner frequency of
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One first-order pole.
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Answer: B. −20 dB/decade
Above ω = 1/T the factor behaves like 1/(ωT), falling 20 dB for every tenfold increase in frequency.
64. A slope of −20 dB/decade is equivalent to approximately
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An octave is a doubling of frequency.
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Answer: A. −6 dB/octave
An octave is a factor of 2 in frequency; 20 log10(2) ≈ 6 dB.
65. The corner frequency of the factor 1/(1 + 0.1s) is
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ωc = 1/T.
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Answer: A. 10 rad/s
Corner frequency ωc = 1/T = 1/0.1 = 10 rad/s.
66. At the corner frequency of a simple pole 1/(1 + sT), the error between the actual and asymptotic magnitude, and the phase angle, are
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Put ωT = 1 into the exact expressions.
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Answer: C. −3 dB and −45°
At ωT = 1, |G| = 1/√2 (−3 dB) and ∠G = −tan⁻¹(1) = −45°, while the asymptote gives 0 dB.
67. For the factor 1/s (an integrator), the Bode magnitude plot is a straight line of slope
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Evaluate 1/(jω) at ω = 1.
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Answer: A. −20 dB/decade passing through 0 dB at ω = 1 rad/s, with constant phase −90°
|1/jω| = 1/ω gives −20 dB/decade crossing 0 dB at ω = 1, and ∠(1/jω) = −90° at all frequencies.
68. The Bode magnitude plot of G(s) = 10/s crosses 0 dB at
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Set 10/ω = 1.
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Answer: A. ω = 10 rad/s
|G(jω)| = 10/ω = 1 when ω = 10 rad/s.
69. For G(s) = 100/(s(1 + 0.1s)), the low-frequency asymptote has a magnitude at ω = 1 rad/s and the slope after ω = 10 rad/s of
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Integrator plus one corner.
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Answer: B. 40 dB and −40 dB/decade
The low-frequency asymptote 100/ω gives 20 log 100 = 40 dB at ω = 1; after the pole at 10 rad/s the slope becomes −20 − 20 = −40 dB/decade.
70. The phase angle of G(s) = 1/(s(s + 1)) at ω = 1 rad/s is
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Add the phase of each factor.
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Answer: D. −135°
Phase = −90° (from 1/s) − tan⁻¹(1) = −90° − 45° = −135°.
71. Increasing only the gain K of an open-loop transfer function shifts the Bode plot by
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What is the phase of a positive constant?
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Answer: A. Moving the magnitude curve up, leaving the phase curve unchanged
A positive constant K adds 20 log K dB at all frequencies and contributes 0° phase.
72. The gain crossover frequency is the frequency at which
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Magnitude crosses a particular line.
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Answer: A. |G(jω)H(jω)| = 1 (0 dB)
Gain crossover is where the open-loop magnitude equals unity; phase margin is measured there.
73. The phase margin of a system is defined as
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Measured where |GH| = 1.
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Answer: B. 180° plus the phase angle of G(jω)H(jω) at the gain crossover frequency
PM = 180° + ∠G(jωgc)H(jωgc): the additional phase lag needed to bring the system to the verge of instability.
74. At the gain crossover frequency, the open-loop phase angle of a system is −150°. Its phase margin is
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PM = 180° + ∠GH.
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Answer: C. 30°
PM = 180° + (−150°) = 30°.
75. At the phase crossover frequency, the open-loop magnitude |G(jω)H(jω)| is 0.25. The gain margin is about
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GM = 20 log(1/|GH|) at phase crossover.
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Answer: D. 12 dB
GM = 20 log10(1/0.25) = 20 log10(4) ≈ 12.04 dB.
76. For the unity-feedback system G(s) = 3/(s(s + 1)(s + 2)), the gain margin is approximately
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Phase crossover of this system is at ω = √2 rad/s.
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Answer: C. 6 dB
Phase is −180° at ω = √2 rad/s where |G| = 3/6 = 0.5, so GM = 20 log 2 ≈ 6 dB.
77. For a minimum-phase system, which combination indicates a stable closed loop?
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Both margins measure distance from instability.
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Answer: C. Positive gain margin and positive phase margin
For minimum-phase systems both margins must be positive for closed-loop stability; zero margins mean marginal stability.
78. A minimum-phase transfer function is one that has
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Think about where the zeros sit.
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Answer: B. No poles or zeros in the right half of the s-plane
Minimum-phase systems have all poles and zeros in the left half plane (no time delay), so magnitude uniquely fixes phase.
79. A pure time delay e^(−sT) in the open loop affects the Bode plot by
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What is the magnitude of e^(−jθ)?
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Answer: D. Adding phase lag ωT radians without changing the magnitude
|e^(−jωT)| = 1 for all ω, while its phase is −ωT, which grows linearly with frequency.
80. For a second-order system with ζ = 0.5 and ωn = 10 rad/s, the resonant peak Mr of the closed-loop frequency response is about
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Use Mr = 1/(2ζ√(1 − ζ²)).
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Answer: D. 1.15
Mr = 1/(2ζ√(1 − ζ²)) = 1/(2 × 0.5 × 0.866) ≈ 1.15.
81. For a second-order system with ζ = 0.5 and ωn = 10 rad/s, the resonant frequency is about
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ωr = ωn√(1 − 2ζ²), not the damped frequency.
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Answer: D. 7.07 rad/s
ωr = ωn√(1 − 2ζ²) = 10√0.5 ≈ 7.07 rad/s.
82. The bandwidth of a closed-loop control system is the frequency at which the closed-loop magnitude
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Half-power point.
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Answer: C. Falls to 0.707 (−3 dB) of its zero-frequency value
Bandwidth is the range up to the frequency where |T(jω)| drops 3 dB below |T(0)|.
83. The Nyquist stability criterion is based on
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A theorem from complex variable theory.
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Answer: B. Cauchy's principle of the argument (conformal mapping)
Nyquist maps the s-plane contour through G(s)H(s) and counts encirclements using Cauchy's argument principle.
84. In the Nyquist criterion Z = N + P (N = clockwise encirclements of −1 + j0, P = open-loop right-half-plane poles), Z represents the number of
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Zeros of 1 + G(s)H(s).
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Answer: B. Closed-loop poles in the right half of the s-plane
Z is the number of zeros of 1 + GH in the right half plane, i.e. closed-loop poles there; stability needs Z = 0.
85. An open-loop transfer function has no poles in the right half plane. Its Nyquist plot encircles −1 + j0 twice in the clockwise direction. The closed-loop system has
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Apply Z = N + P.
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Answer: B. Two poles in the right half plane, so it is unstable
Z = N + P = 2 + 0 = 2 closed-loop poles in the right half plane, so the system is unstable.
86. An open-loop system has one pole in the right half plane. Its Nyquist plot encircles −1 + j0 once counter-clockwise. The closed-loop system is
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Counter-clockwise encirclements count as negative N.
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Answer: A. Stable
Counter-clockwise encirclement gives N = −1, so Z = N + P = −1 + 1 = 0: no closed-loop right-half-plane poles.
87. The Nyquist plot of an open-loop transfer function crosses the negative real axis at −0.5. The gain margin is
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How much can the gain rise before the plot passes through −1?
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Answer: A. 2 (about 6 dB)
GM = 1/|crossing| = 1/0.5 = 2, i.e. 20 log 2 ≈ 6 dB.
88. If the Nyquist plot of G(jω)H(jω) passes exactly through the point −1 + j0, the closed-loop system is
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1 + GH = 0 on the jω axis.
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Answer: D. Marginally stable
Passing through −1 means 1 + GH = 0 at some jω, i.e. closed-loop poles on the imaginary axis.
89. The polar plot of G(jω) = 1/(1 + jωT) for ω from 0 to ∞ is
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Check the points ω = 0, 1/T and ∞.
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Answer: B. A semicircle in the fourth quadrant from 1 to the origin
At ω = 0 it is 1∠0°, at ω = 1/T it is 0.707∠−45°, and it approaches 0∠−90° as ω → ∞, tracing a semicircle of diameter 1 below the real axis.
90. As ω → ∞, the polar plot of G(s) = K/(s(s + 1)(s + 2)) approaches the origin tangent to the direction of
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Each excess pole contributes −90° at high frequency.
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Answer: C. −270°
With three more poles than zeros, the phase at high frequency tends to 3 × (−90°) = −270°.
5.4 Power semiconductor switches
32 questions · AElE0504
91. A power diode differs from a low-power signal diode mainly by having
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Where is the high blocking voltage dropped?
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Answer: B. A lightly doped n⁻ drift region between the p and n⁺ layers to block high reverse voltage
The lightly doped drift layer supports a wide depletion region and hence a high reverse breakdown voltage.
92. The reverse recovery time trr of a power diode is the time
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Stored minority carriers must go somewhere.
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Answer: D. During which reverse current flows after the forward current falls to zero, while stored charge is removed
After forward conduction, stored minority charge must be swept out, so the diode conducts in reverse for trr before blocking.
93. A power diode has reverse recovery time 2 µs and peak reverse recovery current 30 A. Assuming a triangular recovery current, the stored (recovered) charge is about
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Area of a triangle.
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Answer: C. 30 µC
Qrr ≈ ½ × trr × IRR = 0.5 × 2 µs × 30 A = 30 µC.
94. Compared with a p-n power diode, a Schottky power diode has
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It is a majority-carrier device.
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Answer: A. Lower forward voltage drop and negligible reverse recovery, but a lower reverse voltage rating
A Schottky diode is a majority-carrier metal-semiconductor device: low forward drop and almost no recovery, but limited blocking voltage and higher leakage.
95. A thyristor (SCR) has
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Count layers, then junctions between them.
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Answer: B. Four layers (p-n-p-n), three junctions and three terminals
The SCR is a p-n-p-n structure with junctions J1, J2, J3 and terminals anode, cathode and gate.
96. For a thyristor, the latching current is
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Staying on just after turn-on needs more current than staying on later.
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Answer: B. Larger than the holding current
Latching current (minimum anode current to stay on just after the gate pulse is removed) is typically two to three times the holding current.
97. Once a thyristor is conducting, it can be turned off by
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The gate cannot help once it latches.
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Answer: D. Reducing its anode current below the holding current
The gate loses control after latching; the SCR turns off only when anode current falls below the holding current (natural or forced commutation).
98. Which of the following is NOT a method of turning on a thyristor?
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Gate triggering needs current in a particular direction.
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Answer: A. Applying a negative gate current
Thyristors can be turned on by gate current, breakover voltage, dv/dt, temperature or light; a negative gate current does not turn it on.
99. In the two-transistor model of a thyristor, α1 = 0.4, α2 = 0.5 and the gate current is 50 mA. Neglecting leakage currents, the anode current is
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Ia = α2Ig/(1 − α1 − α2).
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Answer: A. 0.25 A
Ia = α2·Ig/(1 − (α1 + α2)) = 0.5 × 0.05/(1 − 0.9) = 0.25 A.
100. For a thyristor to turn off reliably by commutation, the circuit turn-off time must be
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Give the device enough time to recover.
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Answer: A. Greater than the device turn-off time tq
The device must stay reverse biased (or at zero current) longer than tq so the junction J2 recovers its blocking ability.
101. A gate turn-off thyristor (GTO) is turned off by
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It is in the name.
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Answer: D. Applying a large negative gate current
Unlike an SCR, a GTO can be turned off from the gate by extracting a large negative gate current.
102. A GTO carries 600 A and has a turn-off gain of 4. The negative gate current needed to turn it off is
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Turn-off gain = anode current / gate turn-off current.
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Answer: B. 150 A
Turn-off gain = Ia/Ig(off), so Ig(off) = 600/4 = 150 A.
103. A TRIAC is equivalent to
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It conducts in both half-cycles.
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Answer: A. Two SCRs connected in inverse parallel with a common gate
A TRIAC conducts in both directions and can be triggered for either polarity from one gate, like two anti-parallel SCRs.
104. A typical application of a TRIAC is
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It handles both half-cycles of mains AC.
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Answer: D. AC phase-control such as light dimmers and fan regulators
TRIACs are low-cost bidirectional switches for 50 Hz AC power control; their dv/dt and frequency capability are limited.
105. A TRIAC is most sensitive (needs the least gate current) when triggered with
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The mode that behaves most like a normal SCR.
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Answer: C. MT2 positive and gate positive with respect to MT1
Mode I+ (MT2 and gate both positive relative to MT1) has the highest gate sensitivity; mode III+ is the least sensitive.
106. The device commonly used to trigger a TRIAC in a simple AC lamp dimmer is a
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A bidirectional breakover device.
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Answer: D. DIAC
A DIAC breaks over symmetrically in both directions and delivers a trigger pulse to the TRIAC gate in each half-cycle.
107. An IGBT combines
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Input like one device, output like another.
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Answer: B. The high-impedance voltage-controlled gate of a MOSFET with the low conduction drop of a BJT
The IGBT has a MOS gate (voltage drive, low gate power) and bipolar conductivity modulation in the drift region (low on-state drop).
108. The three terminals of an IGBT are
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A mix of MOSFET and BJT terminal names.
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Answer: B. Gate, collector and emitter
IGBT terminals are named gate, collector and emitter, reflecting its MOS input and bipolar output.
109. Latch-up in an IGBT is caused by
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Count the layers in an IGBT.
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Answer: C. Turn-on of the parasitic p-n-p-n thyristor structure inside the device
The four-layer IGBT contains a parasitic thyristor; if it latches, the gate loses control of the collector current.
110. A power MOSFET is a
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No minority carrier storage.
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Answer: A. Majority-carrier, voltage-controlled device with very fast switching
Conduction is by majority carriers only, so there is no stored charge, and the gate is voltage controlled.
111. Power MOSFETs can be paralleled easily because their on-state resistance
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Hotter device should take less current.
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Answer: A. Has a positive temperature coefficient
If one device heats up, its RDS(on) rises and it takes less current, so current sharing is self-balancing.
112. Arranged in order of decreasing maximum switching frequency, typical power devices are
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Majority-carrier devices are fastest.
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Answer: B. MOSFET, IGBT, GTO, SCR
MOSFETs switch at hundreds of kHz, IGBTs at tens of kHz, while GTOs and SCRs are limited to around 1 kHz or line frequency.
113. A power MOSFET with RDS(on) = 0.1 Ω carries an RMS current of 10 A. Its conduction loss is
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A MOSFET behaves as a resistor when on.
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Answer: C. 10 W
Pcond = I²rms × RDS(on) = 10² × 0.1 = 10 W.
114. A switch turns on and off 400 V, 20 A at 20 kHz with rise time 0.1 µs and fall time 0.2 µs. With linear transitions, the switching loss is about
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Psw = ½ V I (tr + tf) f.
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Answer: D. 24 W
Psw = ½ V I (tr + tf) f = 0.5 × 400 × 20 × 0.3 µs × 20 000 = 24 W.
115. A switch dissipates 24 W of switching loss when operated at 20 kHz. If the switching frequency is raised to 50 kHz with the same voltage, current and transition times, the switching loss becomes
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Energy per transition times transitions per second.
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Answer: A. 60 W
Switching energy per cycle is fixed, so loss scales with frequency: 24 × 50/20 = 60 W.
116. A thyristor with an on-state drop of 1.2 V (assumed constant) carries an average current of 50 A. Its average conduction loss is about
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Constant voltage drop times average current.
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Answer: C. 60 W
With a constant on-state drop, Pcond = VT × Iavg = 1.2 × 50 = 60 W.
117. A device dissipates 50 W. The junction-to-ambient thermal resistance including heat sink is 1.5 °C/W and ambient temperature is 40 °C. The junction temperature is
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Temperature rise = power × thermal resistance.
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Answer: C. 115 °C
Tj = Ta + P × Rθja = 40 + 50 × 1.5 = 115 °C.
118. An RC snubber connected across a thyristor is used mainly to
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The capacitor resists sudden voltage change.
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Answer: C. Limit the rate of rise of forward voltage (dv/dt)
The capacitor slows the rise of voltage across the device, preventing false dv/dt turn-on; the resistor limits discharge current at turn-on.
119. A thyristor has a maximum di/dt rating of 50 A/µs and is supplied from 300 V. The minimum series inductance needed for di/dt protection is
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di/dt = V/L at switch-on.
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Answer: C. 6 µH
L ≥ V/(di/dt) = 300/(50 × 10⁶) = 6 µH.
120. For protection of a thyristor by a fast-acting fuse, the fuse should be chosen so that its
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The fuse must blow before the device fails.
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Answer: D. Total clearing I²t is less than the I²t rating of the thyristor
The fuse must clear a fault before the device's sub-cycle surge (I²t) withstand is exceeded.
121. Four thyristors, each rated 3 kV, are connected in series to block a total of 10 kV. The string derating factor is about
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DRF = 1 − string efficiency.
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Answer: B. 0.17
Derating factor = 1 − Vstring/(n × Vdevice) = 1 − 10/(4 × 3) ≈ 0.17.
122. When thyristors are connected in series, static voltage sharing is achieved by connecting
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Unequal leakage currents need a parallel path.
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Answer: D. Equal resistors across each thyristor
Shunt resistors equalize off-state voltages despite unequal leakage currents; RC networks handle dynamic (transient) sharing.
5.5 Power converters
30 questions · AElE0505
123. A single-phase half-wave diode rectifier with a resistive load is fed from 230 V (rms), 50 Hz. The average output voltage is about
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Vdc = Vm/π for half-wave.
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Answer: C. 103.5 V
Vdc = Vm/π = (√2 × 230)/π = 325.3/π ≈ 103.5 V.
124. A single-phase full-wave diode bridge rectifier with a resistive load is fed from 230 V (rms). The average output voltage is about
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Full-wave gives twice the half-wave average.
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Answer: C. 207.1 V
Vdc = 2Vm/π = 2 × 325.3/π ≈ 207.1 V.
125. The ripple factor of a single-phase half-wave rectifier with resistive load is about
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RF = √(FF² − 1).
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Answer: A. 1.21
With Vrms = Vm/2 and Vdc = Vm/π, form factor = π/2 = 1.57 and RF = √(1.57² − 1) ≈ 1.21.
126. The maximum rectification efficiency (DC output power / AC input power to the load) of a single-phase full-wave rectifier with resistive load is about
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Ratio of (Vdc)² to (Vrms)².
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Answer: B. 81.2 %
η = (2Vm/π)²/(Vm/√2)² = 8/π² ≈ 0.812.
127. In a single-phase centre-tapped full-wave rectifier, the peak inverse voltage across each diode is
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The non-conducting diode sees both halves of the winding.
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Answer: D. 2Vm
When one diode conducts, the other sees the full secondary voltage of both halves, i.e. 2Vm.
128. A single-phase fully controlled bridge with a highly inductive (continuous current) load is fed from 230 V (rms). At a firing angle of 60°, the average output voltage is about
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Vdc = (2Vm/π) cos α for continuous conduction.
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Answer: A. 103.5 V
Vdc = (2Vm/π) cos α = 207.1 × cos 60° ≈ 103.5 V.
129. A single-phase half-controlled (semi-converter) bridge is fed from 230 V (rms). At a firing angle of 60°, the average output voltage is about
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Semi-converter: (Vm/π)(1 + cos α).
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Answer: B. 155.3 V
Vdc = (Vm/π)(1 + cos α) = 103.5 × 1.5 ≈ 155.3 V.
130. A single-phase fully controlled bridge with a highly inductive load can operate in the inverting mode (power flow from DC to AC side) when
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When does cos α become negative?
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Answer: A. The firing angle exceeds 90° and a suitable DC source is present on the load side
For α > 90° the average voltage cos α is negative; with a DC source driving current the converter returns power to the AC supply.
131. Connecting a freewheeling diode across the inductive load of a controlled rectifier
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It removes the negative part of the output.
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Answer: D. Prevents negative output voltage and improves input power factor
The freewheeling diode clamps the output at zero during the negative portions, so energy is not returned to the supply and the input power factor improves.
132. A three-phase fully controlled bridge rectifier is fed from 400 V (line, rms). With continuous current and a firing angle of 30°, the average output voltage is about
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Vdc = 1.35 VL cos α.
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Answer: C. 468 V
Vdc = (3√2/π) VL cos α = 1.35 × 400 × cos 30° ≈ 468 V.
133. The ripple frequency in the output of a three-phase fully controlled bridge (six-pulse) rectifier fed from a 50 Hz supply is
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Count pulses per cycle.
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Answer: B. 300 Hz
A six-pulse converter has six output pulses per supply cycle: 6 × 50 = 300 Hz.
134. The characteristic AC-side current harmonics of a six-pulse converter are of order
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Pulse number p gives harmonics pk ± 1.
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Answer: B. 6k ± 1 (5, 7, 11, 13, …)
Six-pulse converters produce harmonics of order 6k ± 1 on the AC side and 6k on the DC side.
135. The input power factor of a rectifier is equal to
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Non-sinusoidal current affects PF in two ways.
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Answer: B. Distortion factor × displacement factor
PF = (I1/Irms) × cos φ1, i.e. distortion factor multiplied by displacement (fundamental) power factor.
136. A single-phase fully controlled bridge with a ripple-free DC load current is operated at α = 30°. Its input power factor is about
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Multiply the square-wave distortion factor by cos α.
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Answer: C. 0.78
The input current is a square wave: distortion factor = 2√2/π ≈ 0.9; displacement factor = cos 30° = 0.866; PF ≈ 0.9 × 0.866 = 0.78.
137. The total harmonic distortion of an ideal square-wave current (equal positive and negative halves of 180°) is about
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THD = √(Irms² − I1²)/I1.
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Answer: A. 48.3 %
I1 = 0.9 Irms, so THD = √(Irms² − I1²)/I1 = √(1 − 0.81)/0.9 ≈ 0.483.
138. Which of the following is NOT a method of improving the input power factor of a phase-controlled converter?
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Displacement factor is roughly cos α.
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Answer: C. Increasing the firing angle
Increasing α increases the displacement angle and worsens power factor; the other three are forced-commutation PF improvement methods.
139. A voltage source inverter (VSI) is characterized by
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Stiff voltage versus stiff current.
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Answer: A. A stiff DC voltage source (large capacitor) at its input and an output voltage waveform independent of load
A VSI is fed from a stiff DC voltage, so its output voltage waveform is fixed by switching while the current depends on load; a CSI uses a large series inductor.
140. The output voltage of a single-phase square-wave inverter contains
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Fourier series of a square wave.
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Answer: A. Only odd harmonics, with the nth harmonic amplitude equal to 1/n of the fundamental
A square wave with half-wave symmetry has only odd harmonics with amplitudes (4V/nπ).
141. A single-phase full-bridge inverter operating in square-wave mode is supplied from 220 V DC. The rms value of the fundamental output voltage is about
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Fundamental peak of a square wave is 4/π times its height.
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Answer: D. 198 V
V1(rms) = (4Vdc/π)/√2 = 0.9 Vdc = 0.9 × 220 ≈ 198 V.
142. A three-phase bridge inverter in 180° conduction mode is fed from 600 V DC. The rms line-to-line output voltage is about
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VL = 0.816 Vdc for 180° conduction.
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Answer: B. 490 V
VL(rms) = √(2/3) Vdc = 0.816 × 600 ≈ 490 V.
143. In a single-phase full-bridge inverter, making the output a quasi-square wave with each pulse 120° wide
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Harmonic n is proportional to sin(n × half pulse width).
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Answer: D. Eliminates the third harmonic
For a pulse width 2d, harmonic n ∝ sin(nd); with 2d = 120°, sin(3 × 60°) = 0, so the third and all triplen harmonics vanish.
144. In sinusoidal PWM, the amplitude modulation index is defined as the ratio of
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It compares the two amplitudes.
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Answer: C. Peak of the sinusoidal reference to peak of the triangular carrier
ma = Vref(peak)/Vcarrier(peak); the frequency modulation ratio mf = fcarrier/fref.
145. A single-phase half-bridge inverter with sinusoidal PWM is fed from 400 V DC (total). With an amplitude modulation index of 0.8 in the linear range, the peak fundamental output voltage is
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The half bridge swings ±Vdc/2.
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Answer: B. 160 V
For a half bridge, V1(peak) = ma × Vdc/2 = 0.8 × 200 = 160 V.
146. In a sinusoidal PWM inverter with a 2550 Hz carrier and a 50 Hz reference, the dominant output harmonics are located around the
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Compute mf = fcarrier/freference.
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Answer: B. 51st harmonic (2550 Hz) and its sidebands
mf = 2550/50 = 51; SPWM pushes harmonics to sidebands around mf and its multiples, which are easy to filter.
147. A 'dead time' (blanking time) is provided between turning off one switch and turning on the other switch of the same inverter leg to
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Two switches in one leg must never be on together.
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Answer: D. Prevent shoot-through short circuit of the DC supply
Since switches take finite time to turn off, both switches of a leg could briefly conduct and short the DC link; dead time prevents this.
148. A DC-DC chopper is supplied from 200 V. Its switch is on for 2 ms and off for 3 ms in each period. The average output voltage of this step-down chopper is
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The period is Ton + Toff.
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Answer: D. 80 V
D = Ton/T = 2/5 = 0.4, so Vo = D × Vs = 0.4 × 200 = 80 V.
149. An ideal buck converter has an input of 24 V and a duty ratio of 0.4. Its output voltage is
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Buck: Vo = D·Vin.
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Answer: A. 9.6 V
Vo = D × Vin = 0.4 × 24 = 9.6 V.
150. An ideal boost converter operating in continuous conduction has an input of 24 V and a duty ratio of 0.4. Its output voltage is
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Boost: Vo = Vin/(1 − D).
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Answer: A. 40 V
Vo = Vin/(1 − D) = 24/0.6 = 40 V.
151. An ideal buck-boost converter in continuous conduction has 24 V input and duty ratio 0.4. The magnitude of its output voltage is
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Buck-boost: D/(1 − D).
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Answer: D. 16 V
|Vo| = Vin × D/(1 − D) = 24 × 0.4/0.6 = 16 V (with reversed polarity).
152. A buck converter has Vin = 48 V, D = 0.5, L = 1 mH and switching frequency 20 kHz. The peak-to-peak inductor current ripple is
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During on-time the inductor sees Vin − Vo.
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Answer: C. 0.6 A
ΔI = (Vin − Vo)·D/(L·f) = (48 − 24) × 0.5/(1 × 10⁻³ × 20 000) = 0.6 A.
5.6 Applications of power electronics
30 questions · AElE0506
153. In an on-line (double-conversion) UPS, under normal mains conditions the load is supplied
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Double conversion means the power always takes the same path.
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Answer: B. Continuously through the rectifier and inverter
In an on-line UPS the mains is rectified and inverted all the time, so the load always runs from the inverter and sees no transfer when mains fails.
154. In an off-line (standby) UPS, the inverter
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Standby means waiting.
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Answer: C. Supplies the load only after the mains fails, after a short transfer time
The load normally runs from the mains; on failure a transfer switch connects the inverter, causing a transfer time of a few milliseconds.
155. Which UPS type provides zero transfer time to the load when the mains supply fails?
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Which one already runs from the inverter?
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Answer: D. On-line double-conversion UPS
Since the inverter already feeds the load, a mains failure only changes the source of the DC link (rectifier to battery), with no switching at the output.
156. A line-interactive UPS differs from a simple standby UPS mainly because it
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It interacts with the line voltage level.
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Answer: C. Corrects mains sags and swells by voltage regulation without using the battery
Line-interactive UPSs include automatic voltage regulation, so moderate voltage variations are corrected without using the battery.
157. Which of the following is NOT a usual component of a static on-line UPS?
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Look for the rotating machine.
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Answer: C. Synchronous condenser
A static UPS consists of a rectifier/charger, battery, inverter and static bypass switch; a synchronous condenser is a rotating reactive-power device used in power systems.
158. The static bypass switch of a UPS, typically made of anti-parallel thyristors, is used to
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A fast switch to an alternative source.
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Answer: B. Transfer the load to the mains quickly if the inverter fails or is overloaded
On inverter fault or overload, the static switch connects the load directly to the bypass mains within a fraction of a cycle.
159. A UPS supplies a 3 kW load at 0.8 power factor. The minimum UPS rating needed is
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UPS ratings are in kVA.
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Answer: D. 3.75 kVA
S = P/pf = 3/0.8 = 3.75 kVA.
160. A UPS uses two 12 V, 100 Ah batteries in series. With inverter efficiency 90 % and a 600 W load, the approximate backup time (ignoring battery derating) is
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Battery energy × efficiency ÷ load.
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Answer: A. 3.6 h
Energy = 24 V × 100 Ah = 2400 Wh; usable output = 2400 × 0.9 = 2160 Wh; time = 2160/600 = 3.6 h.
161. A UPS with a 48 V battery bank must supply a 1 kW load for 2 hours with an inverter efficiency of 80 %. The minimum battery capacity required is about
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Divide by efficiency first, then by battery voltage.
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Answer: C. 52 Ah
Battery energy = 1000 × 2/0.8 = 2500 Wh; Ah = 2500/48 ≈ 52 Ah.
162. The battery bank of a UPS operates on a 240 V DC bus using 12 V batteries. The number of batteries in series required is
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Series voltages add.
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Answer: A. 20
n = 240/12 = 20 batteries in series.
163. The output of a modern on-line UPS inverter is made nearly sinusoidal mainly by
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Push harmonics high, then filter.
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Answer: A. High-frequency PWM switching with an output LC filter
PWM moves harmonics to high frequencies, where a small LC filter removes them, giving a low-THD sinusoidal output.
164. In a line-commutated (LCC) HVDC link, the main switching devices in the converter valves are
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High-power, line-commutated device.
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Answer: A. Thyristors
Classical LCC HVDC uses series strings of high-power thyristors, commutated by the AC network voltage.
165. The smoothing reactor on the DC side of an HVDC converter station is mainly used to
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A large inductor opposes changes in current.
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Answer: D. Reduce ripple in the DC current and limit the rate of rise of fault current
The large series inductor smooths the DC current, limits di/dt during DC faults and helps reduce commutation failures.
166. AC filters at an HVDC converter station serve to
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They are capacitive at 50 Hz.
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Answer: A. Absorb converter harmonic currents and supply part of the reactive power
Tuned AC filters provide low-impedance paths for characteristic harmonics and, being capacitive at fundamental frequency, supply reactive power.
167. A line-commutated HVDC converter, whether rectifying or inverting,
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Think about the phase of the AC line current.
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Answer: B. Consumes reactive power from the AC system
Because of firing delay and commutation overlap, the AC current lags the voltage in both modes, so LCC converters absorb reactive power (roughly 50–60 % of the active power).
168. A monopolar HVDC link uses
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Mono means one.
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Answer: B. One conductor, usually of negative polarity, with return through earth or sea
A monopolar link has a single (usually negative) conductor; the return current flows through earth/sea electrodes or a metallic return.
169. The most widely used HVDC link configuration for long-distance bulk overhead transmission, which can continue at reduced power if one pole fails, is the
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Two poles of opposite polarity.
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Answer: A. Bipolar link
A bipolar link has +Vd and −Vd conductors; if one pole is out, the other can operate as a monopole with earth return.
170. A homopolar HVDC link has
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Homo means same.
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Answer: A. Two or more conductors of the same polarity, usually negative, with earth return
In a homopolar link all conductors share one polarity (normally negative, for less corona) and the return is through ground.
171. A back-to-back HVDC station is mainly used to
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No line between rectifier and inverter.
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Answer: D. Interconnect two asynchronous AC systems with no DC transmission line
Rectifier and inverter are in the same station, allowing power exchange between systems of different frequency or phase without a DC line.
172. A 12-pulse HVDC converter is formed from two six-pulse bridges fed by converter transformers connected
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The two bridges must be phase shifted.
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Answer: C. Star-star and star-delta, giving a 30° phase shift
The 30° shift between the Y-Y and Y-Δ secondaries cancels the 5th and 7th harmonics, leaving 11th, 13th, … on the AC side.
173. The lowest-order characteristic AC-side current harmonic of a 12-pulse converter on a 50 Hz system is at
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Harmonic order 12k ± 1.
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Answer: C. 550 Hz
Characteristic harmonics are 12k ± 1; the lowest is the 11th: 11 × 50 = 550 Hz.
174. Which of the following is an advantage of HVDC over HVAC transmission?
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Which limit depends on line reactance?
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Answer: B. No stability limit on transmissible power over long distances
DC transmission has no reactance-based angle stability limit; the other options are actually advantages of AC.
175. Which of the following is a disadvantage of HVDC transmission?
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Three of these are problems of AC.
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Answer: B. Converters generate harmonics and need reactive power support
Converter harmonics and reactive demand (plus costly terminals and difficult DC breaking) are HVDC drawbacks; the others are AC problems.
176. For underground or submarine cables, HVDC is preferred over AC for long lengths mainly because
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Think of cable capacitance at 50 Hz.
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Answer: D. DC cables have no continuous capacitive charging current
AC cable charging current grows with length and quickly uses up cable capacity; with DC there is no steady charging current.
177. In the usual control scheme of a two-terminal LCC HVDC link under normal operation, the rectifier and inverter respectively operate on
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The inverter must protect against commutation failure.
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Answer: B. Constant current control and constant extinction angle control
The rectifier regulates the DC current, while the inverter holds its extinction angle at a minimum safe value to avoid commutation failure.
178. Compared with line-commutated HVDC, a voltage-source-converter (VSC) HVDC system using IGBTs
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Self-commutated devices do not need the AC network to turn off.
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Answer: A. Can control active and reactive power independently and can supply a weak or dead AC network
Self-commutated VSCs set the magnitude and phase of their AC voltage, giving independent P and Q control and black-start capability.
179. A bipolar HVDC link operates at ±500 kV with a DC current of 2 kA. The power transmitted is
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Each pole carries Vd × Id.
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Answer: B. 2000 MW
P = 2 × Vd × Id = 2 × 500 kV × 2 kA = 2000 MW.
180. A bipolar ±400 kV HVDC line transmits 1000 MW. Each pole conductor has a resistance of 10 Ω. The total line loss is about
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Find the pole current first; two poles carry it.
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Answer: D. 31.25 MW
Id = 1000 MW/(2 × 400 kV) = 1250 A; loss = 2 × Id²R = 2 × 1250² × 10 ≈ 31.25 MW.
181. An LCC HVDC rectifier delivers 1000 MW while operating at an effective AC-side power factor of 0.94. The reactive power it absorbs is about
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Q = P × tan φ, not P × sin φ.
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Answer: D. 363 MVAr
Q = P tan φ; with cos φ = 0.94, tan φ = 0.363, so Q ≈ 1000 × 0.363 ≈ 363 MVAr.
182. The ripple frequency on the DC side of a 12-pulse HVDC converter connected to a 50 Hz AC system is
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Pulse number × supply frequency.
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Answer: C. 600 Hz
A 12-pulse converter produces 12 pulses per cycle: 12 × 50 = 600 Hz (DC-side harmonics of order 12k).