Solutions for Fundamentals of Electrical Circuits, 7th Edition by Alexander (All Chapters included)
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Module
Fundamentals of Electrical Circuits 7e Alexander
Institution
Fundamentals Of Electrical Circuits 7e Alexander
Complete Solutions Manual for Fundamentals of Electrical Circuits, 7th Edition by Charles K Alexander ; ISBN13: 9781260226409.....(Full Chapters included and organized in reverse order from Chapter 19 to 1)...1) Basic Concepts
2) Basic Laws
3) Methods of Analysis
4) Circuit Theorems
5) Operatio...
At t = 5 ms, p = 150 cos2 (60 5x10-3) = 150 cos2 (0.3 )
= 51.82 watts
t t 50
(b) v = 6 + 10 idt = 6 + ∫0 50cos (60 t) dt = 6 + 60𝜋 sin (60 t)
0
p = vi = 5 cos (60 t)[6 + (50/(60 )) sin (60 t)]
At t = 5 ms, p = 5 cos (0.3){6 + (50/(60 )) sin (0.3 )}
=5(0.58779)(6+(0.26526)(0.80902)) = 18.264 watts
P.P.1.6 p = v i = 115 x 12 = 1380 watts; w = p x t
W = 1380x24 = 33.12 k watt-hours
, P.P.1.7 p1 = 5(–45) = –225 w
p2 = 2(45) = 90 w
p3 = 0.12xI(20) = 0.6(25)(20) = 60 w
p4 = 3(25) = 75 w
Note that all the absorbed power adds up to zero as expected.
dn
P.P.1.8 i = dq/dt = e = –1.6 x 10-19 x 1013 = –1.6 x 10-6 A
dt
p = v0 i = 25 x 103 x (1.6 x 10-6) = 40 mW
P.P.1.9 Minimum monthly charge = $12.00
First 100 kWh @ $0.16/kWh = $16.00
Next 160 kWh @ $0.10/kWh = $16.00
Remaining 0 kWh @ $0.06/kWh = $0.00
Total Charge = $44.00
Average cost = $44/[100+160+0] = 16.923 cents/kWh
P.P.1.10 This assigned practice problem is to apply the detailed problem solving
technique to some of the more difficult problems of Chapter 1.
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