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013213215X_ism02.pdf-Circuit Elements a Assessment Problems AP
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013213215X ism02.pdf-Circuit Elemen...
013213215X_ism02.pdf-Circuit Elements a Assessment Problems AP
##### Page 21
Problems
2–21
P 2.19
[a]
20
i
a
=
80
i
b
i
g
=
i
a
+
i
b
=5
i
b
i
a
=
4
i
b
50
=
4
i
g
+ 80
i
b
= 20
i
b
+ 80
i
b
= 100
i
b
i
b
=
0
.
5 A, therefore,
i
a
=2A
and
i
g
=2
.
5A
[b]
i
b
=0
.
5A
[c]
v
o
= 80
i
b
= 40 V
[d]
p
=
i
2
g
(4) = 6
.
25(4) = 25 W
p
20Ω
=
i
2
a
(20) = (4)(20) = 80 W
p
80Ω
=
i
2
b
(80) = 0
.
25(80) = 20 W
[e]
p
50V
(delivered) = 50
i
g
= 125 W
Check:
P
dis
= 25 + 80 + 20 = 125 W
P
del
= 125 W
P 2.20
[a]
Use KVL for the right loop to calculate the voltage drop across the
right-hand branch
v
o
. This is also the voltage drop across the middle
branch, so once
v
o
is known, use Ohm’s law to calculate
i
o
:
v
o
=
1000
i
a
+ 4000
i
a
+ 3000
i
a
= 8000
i
a
= 8000(0
.
002) = 16 V
16
=
2000
i
o
i
o
=
16
2000
= 8 mA
[b]
KCL at the top node:
i
g
=
i
a
+
i
o
=0
.
002 + 0
.
008 = 0
.
010 A = 10 mA.
[c]
The voltage drop across the source is
v
0
, seen by writing a KVL equation
for the left loop. Thus,
p
g
=
-
v
o
i
g
=
-
(16)(0
.
01) =
-
0
.
160 W =
-
160 mW.
Thus the source delivers 160 mW.
© 2010 Pearson Education, Inc., Upper Saddle River, NJ.
All rights reserved. This publication is protected by Copyright and written permission should be
obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic,
mechanical, photocopying, recording, or likewise. For information
regarding permission(s), write to: Rights and Permissions Department,
Pearson Education, Inc., Upper Saddle River, NJ 07458.

##### Page 22
2–22
CHAPTER 2. Circuit Elements
P 2.21
[a]
v
2
= 150
-
50(1) = 100V
i
2
=
v
2
25
= 4A
i
3
+1=
i
2
,
i
3
=4
-
1 = 3A
v
1
= 10
i
3
+ 25
i
2
= 10(3) + 25(4) = 130V
i
1
=
v
1
65
=
130
65
= 2A
Note also that
i
4
=
i
1
+
i
3
=2+3=5A
i
g
=
i
4
+
i
o
=5+1=6A
[b]
p
=
5
2
(4) = 100 W
p
50Ω
=
1
2
(50) = 50 W
p
65Ω
=
2
2
(65) = 260 W
p
10Ω
=
3
2
(10) = 90 W
p
25Ω
=
4
2
(25) = 400 W
[c]
P
dis
= 100 + 50 + 260 + 90 + 400 = 900 W
P
dev
= 150
i
g
= 150(6) = 900 W
P 2.22
[a]
i
cd
= 80
/
16 = 5 A
© 2010 Pearson Education, Inc., Upper Saddle River, NJ.
All rights reserved. This publication is protected by Copyright and written permission should be
obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic,
mechanical, photocopying, recording, or likewise. For information
regarding permission(s), write to: Rights and Permissions Department,
Pearson Education, Inc., Upper Saddle River, NJ 07458.

##### Page 23
Problems
2–23
v
ac
= 125
-
80 = 45
so
i
ac
= 45
/
15 = 3 A
i
ac
+
i
bc
=
i
cd
so
i
bc
=5
-
3=2A
v
ab
= 15
i
ac
-
5
i
bc
= 15(3)
-
5(2) = 35 V
so
i
ab
= 35
/
7=5A
i
bd
=
i
ab
-
i
bc
=5
-
2=3A
Calculate the power dissipated by the resistors using the equation
p
R
=
Ri
2
R
:
p
= (7)(5)
2
= 175 W
p
30Ω
= (30)(3)
2
= 270 W
p
15Ω
= (15)(3)
2
= 135 W
p
16Ω
= (16)(5)
2
= 400 W
p
= (5)(2)
2
= 20 W
[b]
Calculate the current through the voltage source:
i
=
-
i
ab
-
i
ac
=
-
5
-
3=
-
8A
Now that we have both the voltage and the current for the source, we can
calculate the power supplied by the source:
p
g
= 125(
-
8) =
-
1000 W
thus
p
g
(supplied) = 1000 W
[c]
P
dis
= 175 + 270 + 135 + 400 + 20 = 1000 W
Therefore,
P
supp
=
P
dis
P 2.23
[a]
v
a
= (5 + 10)(4) = 60 V
-
240 +
v
a
+
v
b
=0
so
v
b
= 240
-
v
a
= 240
-
60 = 180 V
i
e
=
v
b
/
(14 + 6) = 180
/
20 = 9 A
i
d
=
i
e
-
4=9
-
4=5A
v
c
=4
i
d
+
v
b
= 4(5) + 180 = 200 V
i
c
=
v
c
/
10 = 200
/
10 = 20 A
v
d
= 240
-
v
c
= 240
-
200 = 40 V
i
a
=
i
d
+
i
c
= 5 + 20 = 25 A
R
=
v
d
/i
a
= 40
/
25 = 1
.
© 2010 Pearson Education, Inc., Upper Saddle River, NJ.
All rights reserved. This publication is protected by Copyright and written permission should be
obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic,
mechanical, photocopying, recording, or likewise. For information
regarding permission(s), write to: Rights and Permissions Department,
Pearson Education, Inc., Upper Saddle River, NJ 07458.

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