Fundamentals of Electrical Circuits,
7th Edition by Charles K Alexander
Complete Chapter Solutions Manual
are included (Ch 1 to 19)
** Immediate Download
** Swift Response
** All Chapters included
** EOC problems Solutions Files
** Practice Problem Solutions
,Table of Contents are given below
1) Basic Concepts
2) Basic Laws
3) Methods of Analysis
4) Circuit Theorems
5) Operational Amplifiers
6) Capacitors and Inductors
7) First-Order Circuits
8) Second-Order Circuits
9) Sinusoids and Phasors
10) Sinusoidal Steady-State Analysis
11) AC Power Analysis
12) Three-Phase Circuits
13) Magnetically Coupled Circuits
14) Frequency Response
15) Introduction to the Laplace Transform
16) Applications of the Laplace Transform
17) The Fourier Series
18) Fourier Transform
19) Two-Port Networks
, CHAPTER 1
P.P.1.1 A proton has 1.602 x 10-19 C. Hence, 6.667 billion protons have
+1.602 x 10-19 x 6.667 x 109 = 1.6021 x 10–9 C
P.P.1.2 i = dq/dt = d(20–15t–10e–3t)/dt = (–15–10(–3)e–3t) mA
At t = 1.0 sec, i = –15+30e–3 = –15+1.4936 = –13.506 mA
1 2 8 2
P.P.1.3 q = ∫ idt = ∫0 8dt + ∫1 8t 2 dt = 8t|10 + t 3 |
3 1
= 8 + 8(8–1)/3 = 26.67 C
P.P.1.4 (a) Vab = dw/dq = 100/5 = 20 V
(b) Vab = dw/dq = 100/–10 = –10 V
P.P.1.5 (a) v = 6 i = 30 cos (60 t)
p = v i = 150 cos2 (60 t)
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.
7th Edition by Charles K Alexander
Complete Chapter Solutions Manual
are included (Ch 1 to 19)
** Immediate Download
** Swift Response
** All Chapters included
** EOC problems Solutions Files
** Practice Problem Solutions
,Table of Contents are given below
1) Basic Concepts
2) Basic Laws
3) Methods of Analysis
4) Circuit Theorems
5) Operational Amplifiers
6) Capacitors and Inductors
7) First-Order Circuits
8) Second-Order Circuits
9) Sinusoids and Phasors
10) Sinusoidal Steady-State Analysis
11) AC Power Analysis
12) Three-Phase Circuits
13) Magnetically Coupled Circuits
14) Frequency Response
15) Introduction to the Laplace Transform
16) Applications of the Laplace Transform
17) The Fourier Series
18) Fourier Transform
19) Two-Port Networks
, CHAPTER 1
P.P.1.1 A proton has 1.602 x 10-19 C. Hence, 6.667 billion protons have
+1.602 x 10-19 x 6.667 x 109 = 1.6021 x 10–9 C
P.P.1.2 i = dq/dt = d(20–15t–10e–3t)/dt = (–15–10(–3)e–3t) mA
At t = 1.0 sec, i = –15+30e–3 = –15+1.4936 = –13.506 mA
1 2 8 2
P.P.1.3 q = ∫ idt = ∫0 8dt + ∫1 8t 2 dt = 8t|10 + t 3 |
3 1
= 8 + 8(8–1)/3 = 26.67 C
P.P.1.4 (a) Vab = dw/dq = 100/5 = 20 V
(b) Vab = dw/dq = 100/–10 = –10 V
P.P.1.5 (a) v = 6 i = 30 cos (60 t)
p = v i = 150 cos2 (60 t)
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.
