(a) 8 cos(5t - 30°) + 6 cos 5t(b) 20 cos(120t + 45°) – 30 sin(1207t + 20)(c) 4 sin 8t + 3 sin(8t - 10°)nswer:(a)= 13.533 cos(5t + 342.81°), (b)= 42.51 cos(120rt + 84.76°), (c)= 6.974 cos(8t + 265.72°)-Qar University/Department of Electrical and Electronic Engineering-Lectures are prepared by M.Sc. Ali Kareem Page 132015-2016ge: Fundamentals of Electrical Engineering Chapter 14 – Sinusoids and Phasorsnework: Obtain the sinusoids corresponding to each of the following phasors:4.12(a) Vi = 60/15°, w = 1(b) V2 = 6+ j8, w = 40(c) I = 2.8e"j7/3, w = 377(d) I2 =-0.5- j1.2, w = 10(a)= 60 cos(t+ 15°), (b)= 10 cos(40t + 53.13°), (c)= 2.8 cos(377t – Tt/3),|(d)= 1.3 cos(10³t + 247.4°)ver:nework: Find v(t) in the following integrodifferential equations using the phasor approach:_4.13(a) v(t) +v dt = 10 cos tdv(b)+ 5v(t) +4v dt = 20 sin(4t + 10°)dtwer:(a)= 7.071 cos(t + 45°), (b)3.43 cos(4t - 110.96°)mework: Using phasors, determine i(t) in the following equations::14.14di(а) 2+ 3i (1) = 4 cos(2t – 45°)dtSidt+di+ 6i (t) = 5 cos(5t + 22°)(b) 10(a)= 0.8 cos(2t –- 98.13°), (b)0.745 cos(5t - 4.56°)wer:mework: The loop equation for a series RLC circuit gives:di+ 2i +dtAssuming that the value of the integral att =-00 is zero, find i(t) using the phasor method.14.15i dt := cos 2tswer:0.4 cos(2t- 36.87°)-mework: A parallel RLC circuit has the node equation:dv+ 50v + 100dt14.16v dt = 110 cos(377t – 10°)Determine v(t) using the phasor method. You may assume that the value of the integral at t =-00 iszero.V110Z-10°, 0= 377jolution:joV + 50V +100-jl00V j377+50-377=110Z-10°V (380.6282.45°) = 1102-10°V = 0.289Z-92.45°Therefore, v(t)0.289 cos(377t- 92.45°).-Qar University/Department of Electrical and Electronic Engineering-Lectures are prepared by M.Sc. Ali Kareem Page 14 Evaluate these complex numbers:(a) [(5+ j2)(-1+ j4) – 5/60°]*10 + j5 + 3/40(b)+ 10/30°-3+ j4(a) -15.5 j13.67, (b) 8.293 + j2.2.

Phasor: Phasor is a physical quantity which has some magnitude with some phase angle. A phasor can…

ework: Obtain the sinusoids corresponding to each of the following phasors: 4.12 (a) Vi = 60/15°, w = ! (b) V2 = 6+ j8, w = 40 %3D (c) I = 2.8e-ja/3_ (d) I2 = -0.5 - j1.2, o = 10 °, w = 377 %3D %3D %3D ver: |(a)= 60 cos(t + 15°), (b)= 10 cos(40t + 53.13°), (c)= 2.8 cos(377t – T/3), (d)= 1.3 cos(10³t + 247.4°)

ework: Obtain the sinusoids corresponding to each of the following phasors: 4.12 (a) Vi = 60/15°, w = ! (b) V2 = 6+ j8, w = 40 %3D (c) I = 2.8e-ja/3_ (d) I2 = -0.5 - j1.2, o = 10 °, w = 377 %3D %3D %3D ver: |(a)= 60 cos(t + 15°), (b)= 10 cos(40t + 53.13°), (c)= 2.8 cos(377t – T/3), (d)= 1.3 cos(10³t + 247.4°)

Introductory Circuit Analysis (13th Edition)

13th Edition

ISBN:9780133923605

Author:Robert L. Boylestad

Publisher:Robert L. Boylestad

Chapter1: Introduction

Section: Chapter Questions

Problem 1P: Visit your local library (at school or home) and describe the extent to which it provides literature...

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