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013213215X ism02.pdf
013213215X_ism02.pdf
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013213215X ism02.pdf-Circuit Elements a Assessment...
013213215X_ism02.pdf-Circuit Elements a Assessment Problems AP
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013213215X_ism02.pdf-Circuit Elements a Assessment Problems AP
Page 28
2–28
CHAPTER 2. Circuit Elements
25
i
1
+
(
-
1
.
25)
20
+(
-
0
.
015625) = 0;
i
1
=3
.
125 mA
v
g
= 60
i
1
+ 260
i
1
= 320
i
1
Therefore,
v
g
=1V
.
P 2.30
[a]
-
50
-
20
i
σ
+ 18
i
Δ
=0
-
18
i
Δ
+5
i
σ
+ 40
i
σ
=0
so
18
i
Δ
= 45
i
σ
Therefore,
-
50
-
20
i
σ
+ 45
i
σ
=0
,
so
i
σ
=2A
18
i
Δ
= 45
i
σ
= 90; so
i
Δ
=5A
v
o
= 40
i
σ
= 80 V
[b]
i
g
= current out of the positive terminal of the 50 V source
v
d
= voltage drop across the 8
i
Δ
source
i
g
=
i
Δ
+
i
σ
+8
i
Δ
=9
i
Δ
+
i
σ
= 47 A
v
d
= 80
-
20 = 60 V
P
gen
=
50
i
g
+ 20
i
σ
i
g
= 50(47) + 20(2)(47) = 4230 W
P
diss
=
18
i
2
Δ
+5
i
σ
(
i
g
-
i
Δ
) + 40
i
2
σ
+8
i
Δ
v
d
+8
i
Δ
(20)
=
(18)(25) + 10(47
-
5) + 4(40) + 40(60) + 40(20)
=
4230 W; Therefore,
P
gen
=
P
diss
= 4230 W
P 2.31
i
E
-
i
B
-
i
C
=0
i
C
=
βi
B
therefore
i
E
= (1 +
β
)
i
B
i
2
=
-
i
B
+
i
1
V
o
+
i
E
R
E
-
(
i
1
-
i
B
)
R
2
=0
-
i
1
R
1
+
V
CC
-
(
i
1
-
i
B
)
R
2
=0
or
i
1
=
V
CC
+
i
B
R
2
R
1
+
R
2
V
o
+
i
E
R
E
+
i
B
R
2
-
V
CC
+
i
B
R
2
R
1
+
R
2
R
2
=0
Now replace
i
E
by (1 +
β
)
i
B
and solve for
i
B
. Thus
i
B
=
[
V
CC
R
2
/
(
R
1
+
R
2
)]
-
V
o
(1 +
β
)
R
E
+
R
1
R
2
/
(
R
1
+
R
2
)
© 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 29
Problems
2–29
P 2.32
Here is Equation 2.25:
i
B
=
(
V
CC
R
2
)
/
(
R
1
+
R
2
)
-
V
0
(
R
1
R
2
)
/
(
R
1
+
R
2
) + (1 +
β
)
R
E
V
CC
R
2
R
1
+
R
2
=
(10)(60
,
000)
100
,
000
= 6V
R
1
R
2
R
1
+
R
2
=
(40
,
000)(60
,
000)
100
,
000
= 24 kΩ
i
B
=
6
-
0
.
6
24
,
000 + 50(120)
=
5
.
4
30
,
000
=0
.
18 mA
i
C
=
βi
B
= (49)(0
.
18) = 8
.
82 mA
i
E
=
i
C
+
i
B
=8
.
82 + 0
.
18 = 9 mA
v
3
d
= (0
.
009)(120) = 1
.
08V
v
bd
=
V
o
+
v
3
d
=1
.
68V
i
2
=
v
bd
R
2
=
1
.
68
60
,
000
= 28
μ
A
i
1
=
i
2
+
i
B
= 28 + 180 = 208
μ
A
v
ab
= 40
,
000(208
×
10
-
6
)=8
.
32 V
i
CC
=
i
C
+
i
1
=8
.
82 + 0
.
208 = 9
.
028 mA
v
13
+ (8
.
82
×
10
-
3
)(750) + 1
.
08 = 10 V
v
13
=2
.
305 V
P 2.33
[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 30
2–30
CHAPTER 2. Circuit Elements
[b]
P 2.34
[a]
From the simplified circuit model, using Ohm’s law and KVL:
400
i
+ 50
i
+ 200
i
-
250 = 0
so
i
= 250
/
650 = 385 mA
This current is nearly enough to stop the heart, according to Table 2.1,
so a warning sign should be posted at the 250 V source.
[b]
The closest value from Appendix H to 400 Ω is 390 Ω; the closest value
from Appendix H to 50 Ω is 47 Ω. There are two possibilites for replacing
the 200 Ω resistor with a value from Appendix H – 180 Ω and 220 Ω. We
calculate the resulting current for each of these possibilities, and
determine which current is closest to 385 mA:
390
i
+ 47
i
+ 180
i
-
250 = 0
so
i
= 250
/
617 = 405
.
2 mA
390
i
+ 47
i
+ 220
i
-
250 = 0
so
i
= 250
/
657 = 380
.
5 mA
Therefore, choose the 220 Ω resistor to replace the 200 Ω resistor in the
model.
P 2.35
P 2.36
[a]
p
=
i
2
R
p
arm
=
250
650
2
(400) = 59
.
17 W
p
leg
=
250
650
2
(200) = 29
.
59 W
p
trunk
=
250
650
2
(50) = 7
.
40 W
© 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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