3560 = a) 5×30° (6-j8t ² 1 = 50.845 < -15-560 J.P.T b) (10560°) (355-50°) ( ۶ ) - ( ماز + 2 ) 60·0255-110-964" - a) v (t) = 21 cos (4+ -15°)V v(t) vmcos (wt + 0) V = V₁₂ ²0 |v = 215-15°V b) i(t) = - 8 sin (100 + 70 °) ^ A From -sinA= cus (A+ 90°) i(t) = -8 sin ( 10t 170°) = 8 cos ( 10t +70° +90") = 8 cos (10€ +160°) I=8 <160⁰ MA
3560 = a) 5×30° (6-j8t ² 1 = 50.845 < -15-560 J.P.T b) (10560°) (355-50°) ( ۶ ) - ( ماز + 2 ) 60·0255-110-964" - a) v (t) = 21 cos (4+ -15°)V v(t) vmcos (wt + 0) V = V₁₂ ²0 |v = 215-15°V b) i(t) = - 8 sin (100 + 70 °) ^ A From -sinA= cus (A+ 90°) i(t) = -8 sin ( 10t 170°) = 8 cos ( 10t +70° +90") = 8 cos (10€ +160°) I=8 <160⁰ MA
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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Question
Explain how the answer is obtain using calculator. NOTE: j = i

Transcribed Image Text:3560
ja) 5×30° (6-j8t etj.
1 = 50.845 < -15.560 j.p.f|
b) (10560°) (355-50°)
(24j6)-(tj)
1 = 60·02155-110-964
1 a) v (t) = 21 cos (47-15°)V
v(t) vicos (wt + 0)
V = Um ²0
| v = 215-15°V
b) i(t) = - 8 sin (100 + 70°) ^ A
From -sin A = cos (A + 90°)
i(t) = -as in ( 10t +70°)
= 8 cos ( 10t +70° +90')
= 8 cos [ 10€ +160°)
I=8 <160⁰mA
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