Confirmed Errata – Study Guide for FE Electrical and Computer CBT Exam – Third Edition

Corrections – Problems Section:

Problems 2.2 e) to 2.2 i) scenario statement shall have

A = {1, 3, 5, 7, 10}

Problem 8.4 a) Second graph represents y(t) not h(t).

Problem 9.1 c) Electron and hole concentrations of a semi-conductor wafer at equilibrium are found to be n = p = 2 x 10¹⁰ m¯³….

Problem 10.2 e) Option B) 23.5∠4.5°Ω

Problem 10.4 g) A single-phase 240V power source is feeding a 20A, unity power factor load with a #10AWG copper conductor in a steel conduit such that R = 1.2Ω/1000ft (circuit reactance can be ignored). One-way length of this circuit is 500 feet. Calculate the approximate total voltage drop of given circuit.

Option D) 24V

Problem 11.1 i) Q₂ = 15 x 10¯⁹C

Problem 13.6 e)

Option A) 5µs      Option B) 6.66µs 

Option C) 10µs     Option D) 20µs 

Problem 17.4 j) and Problem 17.4 k)  

Starting point for graph traversal is ‘O’ 

 

Corrections – Solutions Section:

Solution 1.3 e) …. According to the definition, {(1, b), (1,c), (1,d)} is not a function because input 1 is related to multiple outputs.

Solution 1.1 c) x = -2 is an invalid solution because logarithm of negative real number is undefined. 

Solution 1.4 l) …. Eccentricity of a hyperbola is given by….

It is a long established fact that a reader will be

Solution 5.2 b) 𝜀 required = 6 x 10¯³

Solution 8.1 f) CORRECT ANSWER – C 

Reconstructed signal frequency will be 50Hz with a positive phase angle (due to positive sign inside mod operator) as shown below:

x'(t) = cos(2𝜋(50)t + θ)

x'(t) = cos(2𝜋(50)t + θ)

Solution 8.3 l)

Y(z) (1 + z¯¹ + 2z¯²) = X(z) (1 + z¯¹ + 2z¯²)

Solution 9.2 a)

iD1 = (5V – 1V) / 1kΩ = 4mA

Solution 9.5 g)

R₁ = (6V – 1V) / 1mA = 4.8kΩ ≅ 5kΩ

Solution 10.4 g)

Total voltage drop (2-way) in a single-phase system with unity power factor and negligible circuit reactance can be calculated using following formula:

VD = (2 x L x R x I) / (k x 1000)

VD = Voltage Drop – ?

L = 1-way circuit length = 500 ft

R = Conductor resistance = 1.2 Ω/ft

I = Load current = 20A

k = 1.0 for single-phase AC circuit

VD = (2 x 500 x 1.2 x 20) / (1 x 1000) = 24V

Solution 10.6 l) CORRECT ANSWER: 297 Nm

Pm = T𝜔 = T x (2𝜋/60) x n

T = Pm /  [(2𝜋/60) x n] = 297 Nm

Solution 12.1 c) CORRECT ANSWER: (s + 2)² / [s(s + 4)]

T = [(s + 4) + (s + 3)] / [s(s + 4)] = (s + 2)² / [s(s + 4)]

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