Confirmed Errata for ‘Study Guide for Fundamentals of Engineering (FE) Electrical & Computer CBT Exam’
Corrections in Problems section:
Problem 1.1 b) Which of the following techniques is not used for corrosion prevention?
(D) Options A, B and C provide corrosion prevention
Problem 1.1 d) Units for activation energy should be J/mol in this question.
Problem 1.2 e) Calculate the magnetic permeability …. at a distance of 50cm perpendicular to the wire.
Problem 1.4 f) Students are measuring heat capacities of three different samples of same liquid. Sample 1 is 1kg, sample 2 is 2kg and sample 3 is 3kg. Which of the following options correctly indicates their heat capacities?
Problem 2.3 a)
(A) 22.6 x 106
Problem 2.4 c)
Problem 3.1 d)
Problem 3.5 c) Capacitive reactance in the problem shall be -j100Ω
(D) 0.14 A phase angle 11.3⁰
Problem 4.2a) The circuit diagram incorrectly shows inductance on resistor and resistance on inductor.
(D) 1000 rad/s
(D) Poles 1st order @ 0.2, 2nd order @ 0, Zero – none, Gain = 20dB
(C) Output function should be 16 for 2<t<6
(D) 1 – cos(t) 0 < t ≤ π, 1- cos(t) π < t ≤ 2π
(D) Ic = 1.13mA, Vce = 1.3V
(C) Ie = 3.2mA, Vec = 5.1V
(D) Ie = 1.5mA, Vec = 2V
Problem 6.4b) Emitter voltage shall be 6V (assume k=0.5mA/V^2 and Vt = 1V)
Problem 6.4e) 10kΩ collector resistor source voltage shall be 10V
Problem 6.6 e) Which of the following devices can be used for resistance measurements?
(B) Strain Gage
(D) Options A, B and C are correct
Problem 7.1 c) Units for options should be ‘VA’
Problem 7.1 g)
(A) 5 – 2j Ω
Problem 7.3 c)
Problem 7.6 c) Missing information – Vrms = 400V and frequency = 60Hz.
Problem 7.6 d) A three phase power utility provides 75 kVA at 0.85 lagging and 35 kVA at 0.75 lagging power factor…..
Problem 7.6 e) A balanced 3-phase ……. Y-connected load that consumes 375kW at 0.85 power factor lagging ….
Problem 9.2 e) H(s) = 100s / (s2 + 150s + 5000)
Problem 9.6 a) y” + 3y’ + 2y = 2u(t)
Problem 10.1 d) A message signal …. using a carrier signal of 50sin2π(4000t). Calculate the modulation index.
Problem 10.3 b) Calculate the minimum bandwidth required to transmit a pulse code modulated message m(t), with M(f) = 0 for f >= 100 Hz using 256 quantization level.
Problem 11.2 a)
Corrections in Solutions section:
1.4 e) Heat capacity is directly proportional to amount of material. Sample # 3 will have the higher heat capacity because it contains the largest amount of substance being tested. Sample # 1 has the least amount of substance therefore it will have the lowest heat capacity.
2.3 a) Since Q = CV this implies that V = Q/C
2.4 c) Therefore energy stored = 500 J
3.5 c)I = 0.14/11.3◦ A
4.2 e) Q = R/(Lω0)
Therefore Q = 500/707=0.707
4.4 e) H(s) = 10 / [s2 (s/0.2 + 1), |gain| = 20log(10) = 20 dB
Therefore Q = 500/707=0.707
5.1 c) Region # 3 f(t) = 1- cos(t) π < t ≤ 2π
6.6 e) Wheatstone bridge, RTD and Strain Gage can be used to measure resistance.
7.1 g) Zload = Zthevenin* = (5 + 2jΩ)* = 5 – 2jΩ
Zline = 1 + jΩ
7.3 c) Vp/a = 427.8V, V.R = (427.8 – 400)/400 = 7%
7.6 c) Solving for Ɵ2 gives 8.337⁰, cos(Ɵ2) = 0.989
7.6 e) C1φ = [375000(tan31.87◦ – tan18.19◦)] / [(2π)(60)(120)2] = 6.7mF ≈ 7mF
10.1 d) Modulation Index = signal amplitude / carrier amplitude
12.5 h) 1st clock cycle column should read Q0 and 2nd column as Q1