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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 16
2–16
CHAPTER 2. Circuit Elements
Using the values in the table of Fig. P2.13(b)
R
=
(
-
10)
2
17
.
86
×
10
-
3
=
(
-
5)
2
4
.
46
×
10
-
3
=
(5)
2
4
.
46
×
10
-
3
=
(10)
2
17
.
86
×
10
-
3
=
(15)
2
40
.
18
×
10
-
3
=
(20)
2
71
.
43
×
10
-
3
5
.
6 kΩ
Note that this value is found in Appendix H.
P 2.14
[a]
Plot the
v
i
characteristic:
From the plot:
R
=
Δ
v
Δ
i
=
(180
-
100)
(16
-
0)
=5Ω
When
i
t
= 0,
v
t
= 100 V; therefore the ideal current source must have a
current of 100
/
5 = 20 A
[b]
We attach a 20Ω resistor to the device model developed in part (a):
Write a KCL equation at the top node:
20 +
i
t
=
i
1
Write a KVL equation for the right loop, in the direction of the two
currents, using Ohm’s law:
5
i
1
+ 20
i
t
=0
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 17
Problems
2–17
Combining the two equations and solving,
5(20 +
i
t
) + 20
i
t
=0
so
25
i
t
=
-
100;
thus
i
t
=
-
4A
Now calculate the power dissipated by the resistor:
p
20 Ω
= 20
i
2
t
= 20(
-
4)
2
= 320 W
P 2.15
[a]
Plot the
v
-
i
characteristic
From the plot:
R
=
Δ
v
Δ
i
=
(130
-
50)
(10
-
0)
=8Ω
When
i
t
= 0,
v
t
= 50 V; therefore the ideal voltage source has a voltage
of 50 V.
[b]
When
v
t
=0
,
i
t
=
-
50
8
=
-
6
.
25A
Note that this result can also be obtained by extrapolating the
v
-
i
characteristic to
v
t
= 0.
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 18
2–18
CHAPTER 2. Circuit Elements
P 2.16
[a]
[b]
Δ
v
= 25V;
Δ
i
=2
.
5 mA;
R
=
Δ
v
Δ
i
= 10 kΩ
[c]
10
,
000
i
1
= 2500
i
s
,
i
1
=0
.
25
i
s
0
.
02 =
i
1
+
i
s
=1
.
25
i
s
,
i
s
= 16 mA
[d]
v
s
(open circuit) = (20
×
10
-
3
)(10
×
10
3
) = 200 V
[e]
The open circuit voltage can be found in the table of values (or from the
plot) as the value of the voltage
v
s
when the current
i
s
= 0. Thus,
v
s
(open circuit) = 140 V (from the table)
[f]
Linear model cannot predict the nonlinear behavior of the practical
current source.
P 2.17
[a]
Begin by constructing a plot of voltage versus current:
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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