Errata – Study Guide

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)

(D) 500J

Problem 3.1 d)

(D) 1.7A

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.

Problem 4.2e)

(B) 0.707

(C) 500

Problem 4.2f)

(D) 1000 rad/s

Problem 4.4e)

(D) Poles 1st order @ 0.2, 2nd order @ 0, Zero – none, Gain = 20dB 

Problem 5.1a)

(C) Output function should be 16 for 2<t<6 

Problem 5.1c)

(D)  1 – cos(t) 0 < t ≤ π, 1- cos(t) π < t ≤ 2π

Problem 6.3b)

(D) Ic = 1.13mA, Vce = 1.3V

Problem 6.3e)

(C) Ie = 3.2mA, Vec = 5.1V

Problem 6.3f)

(D) Ie = 1.5mA, Vec = 2V

Problem 6.4b) Emitter voltage shall be 6V (assume k=0.5mA/V^2 and Vt = 1V)

(B) Saturation

(C) Triode

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)

(A) 7%

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)

(C) Star

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

7.2 g) 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) C = [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

CONTACT US





© www.StudyForFE.com ~ 2016 All Rights Reserved. Design, Developed & SEO by Sansoft Web Technologies Pvt. Ltd.