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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 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
)
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]
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
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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