Which power semiconductor device is best suited for high-frequency switching applications above 100 kHz?

The correct answer is B: MOSFET. MOSFETs have switching frequencies up to several MHz, making them ideal for high-frequency applications. Thyristors are limited to lower frequencies (few kHz).

What is the minimum number of NAND gates required to implement a NOT gate?

The correct answer is A: 1 NAND gate. A single NAND gate with both inputs tied together acts as a NOT gate (inverter).

For the state-space system: ẋ = Ax + Bu, y = Cx + Du. The controllability matrix rank must equal n for the system to be completely state controllable. What is n?

The correct answer is A: Number of states. For complete state controllability, rank of [B AB A²B ... Aⁿ⁻¹B] must equal n, the order/number of states of the system.

The back EMF of a DC motor is given by:

The correct answer is A: Eb = PΦZn/60A. Back EMF in DC motor Eb = PΦZn/(60A), where P = poles, Φ = flux, Z = conductors, n = rpm, A = parallel paths

In mesh current analysis for a two-mesh circuit with self-resistances R₁ = 10Ω, R₂ = 15Ω and mutual resistance Rm = 5Ω, and voltages V₁ = 20V, V₂ = 0V, what would be the mesh impedance matrix determinant?

The correct answer is B: 125. The determinant of mesh impedance matrix = (R₁+Rm)(R₂+Rm) - Rm² = (10+5)(15+5) - 5² = 15×20 - 25 = 300-25 = 275... Correction: Det = R₁R₂ - Rm² = 10×15 - 5² = 150 - 25 = 125

Home

Home › Subjects › Electrical Engg (EEE)

Electrical Engg (EEE)

Electrical machines, power systems, circuits

321 Q 7 Topics Take Mock Test

Difficulty: All Easy Medium Hard 131–140 of 321

Topics in Electrical Engg (EEE)

All Circuit Analysis 100 Electrical Machines 100 Power Systems 100 Control Systems 100 Power Electronics 100 Electrical Measurements 40 Digital Electronics 100

Q.131 Medium Control Systems

In compensator design, if both transient and steady-state improvements are needed, which approach is preferred?

A Lead compensator only

B Lag compensator only

C Lead-lag compensator

D PID controller without compensation

Correct Answer: C. Lead-lag compensator

EXPLANATION

Lead-lag compensator combines lead (improves transient) and lag (improves steady-state) characteristics for overall system improvement.

Test

Q.132 Medium Control Systems

Peak overshoot of underdamped second-order system depends on:

A Damping ratio only

B Natural frequency only

C Both damping ratio and natural frequency

D Loop gain only

Correct Answer: A. Damping ratio only

EXPLANATION

Peak overshoot Mp = e^(-πζ/√(1-ζ²)). It depends only on damping ratio ζ, not on ωₙ.

Test

Q.133 Medium Control Systems

A system with transfer function H(s) = 100/(s² + 10s + 100) has natural frequency ωₙ and damping ratio ζ respectively as:

A ωₙ = 10 rad/s, ζ = 0.5

B ωₙ = 100 rad/s, ζ = 0.1

C ωₙ = 10 rad/s, ζ = 0.707

D ωₙ = 5 rad/s, ζ = 1

Correct Answer: A. ωₙ = 10 rad/s, ζ = 0.5

EXPLANATION

Standard form: ωₙ² = 100, so ωₙ = 10 rad/s. 2ζωₙ = 10, so ζ = 10/(2×10) = 0.5

Test

Q.134 Medium Control Systems

A lead compensator with transfer function Gc(s) = (1 + aTs)/(1 + Ts) where a > 1 provides:

A Phase lag and reduced bandwidth

B Phase lead and improved transient response

C Reduced steady-state error only

D Increased settling time

Correct Answer: B. Phase lead and improved transient response

EXPLANATION

Lead compensator (a > 1) provides phase lead in mid-frequency range, improving transient response and system speed, hence used for transient improvement.

Test

Q.135 Medium Control Systems

In state-space representation, observability matrix rank should be equal to:

A Number of states

B Number of inputs

C Number of outputs

D Number of poles

Correct Answer: A. Number of states

EXPLANATION

For a system to be completely observable, the observability matrix [C; CA; CA²; ...] must have rank equal to n (number of states).

Test

Q.136 Medium Control Systems

The settling time of a second-order underdamped system is approximately given by:

A ts ≈ 4/(ζωₙ)

B ts ≈ 2/(ζωₙ)

C ts ≈ 1/(ζωₙ)

D ts ≈ π/(ωₙ)

Correct Answer: A. ts ≈ 4/(ζωₙ)

EXPLANATION

For 2% criteria, settling time ts ≈ 4/(ζωₙ). This is the standard formula for second-order systems.

Test

Q.137 Medium Control Systems

A control system with gain margin of 6 dB means:

A Gain can be increased by factor of 2

B Gain can be increased by factor of 4

C Gain can be reduced by factor of 2

D Gain can be reduced by factor of 6

Correct Answer: A. Gain can be increased by factor of 2

EXPLANATION

Gain margin in dB = 20log₁₀(GM). Therefore, 6 = 20log₁₀(GM), GM = 10^(6/20) ≈ 2

Test

Q.138 Medium Control Systems

The Nyquist stability criterion states that for stability, the Nyquist plot should not encircle:

A Origin (0,0)

B Point (-1, 0)

C Point (1, 0)

D Point (0, 1)

Correct Answer: B. Point (-1, 0)

EXPLANATION

The Nyquist criterion checks encirclements of the critical point (-1, 0) in the complex plane. No encirclement indicates stability for minimum phase systems.

Test

Q.139 Medium Control Systems

A system exhibits steady-state error of 0.2 for unit ramp input with loop gain K = 50. The static velocity error constant is:

A 5 sec⁻¹

B 10 sec⁻¹

C 2.5 sec⁻¹

D 20 sec⁻¹

Correct Answer: A. 5 sec⁻¹

EXPLANATION

For ramp input, ess = 1/Kv. Given ess = 0.2, therefore Kv = 1/0.2 = 5 sec⁻¹

Test

Q.140 Medium Control Systems

For a unity feedback control system with G(s) = 10/[s(s+5)], the static velocity error constant Kv is:

A 2 sec⁻¹

B 10 sec⁻¹

C 50 sec⁻¹

D 5 sec⁻¹

Correct Answer: A. 2 sec⁻¹

EXPLANATION

Kv = lim(s→0) s·G(s) = lim(s→0) s·10/[s(s+5)] = 10/5 = 2 sec⁻¹

Test

1…12 13 14 15 16…33

All Subjects & Topics

Govt. Exams

Entrance Exams

Teacher Exams

Subjects

Quick Links

Electrical Engg (EEE)