


013213215X ism02.pdf
013213215X_ism02.pdf
Showing 56 out of 31
013213215X ism02.pdfCircuit Elements a Assessment...
013213215X_ism02.pdfCircuit Elements a Assessment Problems AP
013213215X ism02.pdfCircuit Elemen...
013213215X_ism02.pdfCircuit Elements a Assessment Problems AP
Page 5
Problems
2–5
A KCL equation at the upper right node gives
i
2
=
i
5
; a KCL equation at
the bottom right node gives
i
5
=

i
1
; a KCL equation at the upper left
node gives
i
s
=

i
2
. Now replace the currents
i
1
and
i
2
in the Ohm’s law
equations with
i
5
:
v
2
=3
i
2
=3
i
5
;
v
5
=7
i
5
;
v
1
=2
i
1
=

2
i
5
Now substitute these expressions for the three voltages into the ﬁrst
equation:
24 =
v
2
+
v
5

v
1
=3
i
5
+7
i
5

(

2
i
5
) = 12
i
5
Therefore
i
5
= 24
/
12 = 2 A
[b]
v
1
=

2
i
5
=

2(2) =

4V
[c]
v
2
=3
i
5
= 3(2) = 6 V
[d]
v
5
=7
i
5
= 7(2) = 14 V
[e]
A KCL equation at the lower left node gives
i
s
=
i
1
. Since
i
1
=

i
5
,
i
s
=

2A. We can now compute the power associated with the voltage
source:
p
24
= (24)
i
s
= (24)(

2) =

48 W
Therefore 24 V source is delivering 48 W.
AP 2.6
Redraw the circuit labeling all voltages and currents:
We can ﬁnd the value of the unknown resistor if we can ﬁnd the value of its
voltage and its current. To start, write a KVL equation clockwise around the
right loop, starting below the 24 Ω resistor:

120 V +
v
3
=0
Use Ohm’s law to calculate the voltage across the 8Ω resistor in terms of its
current:
v
3
=8
i
3
Substitute the expression for
v
3
into the ﬁrst equation:

120 V + 8
i
3
=0
so
i
3
=
120
8
= 15 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 6
2–6
CHAPTER 2. Circuit Elements
Also use Ohm’s law to calculate the value of the current through the 24Ω
resistor:
i
2
=
120 V
24 Ω
=5A
Now write a KCL equation at the top middle node, summing the currents
leaving:

i
1
+
i
2
+
i
3
=0
so
i
1
=
i
2
+
i
3
= 5 + 15 = 20 A
Write a KVL equation clockwise around the left loop, starting below the
voltage source:

200 V +
v
1
+ 120 V = 0
so
v
1
= 200

120 = 80 V
Now that we know the values of both the voltage and the current for the
unknown resistor, we can use Ohm’s law to calculate the resistance:
R=
v
1
i
1
=
80
20
=4Ω
AP 2.7
[a]
Plotting a graph of
v
t
versus
i
t
gives
Note that when
i
t
= 0,
v
t
= 25 V; therefore the voltage source must be 25
V. Since the plot is a straight line, its slope can be used to calculate the
value of resistance:
R
=
Δ
v
Δ
i
=
25

0
0
.
25

0
=
25
0
.
25
= 100 Ω
A circuit model having the same
v

i
characteristic is a 25 V source in
series with a 100Ω resistor, as shown below:
© 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.
Ace your assessments! Get Better Grades
Browse thousands of Study Materials & Solutions from your Favorite Schools
New Jersey Institute of T...
New_Jersey_Institute_of_Technology
School:
Circuits_and_Systems_2
Course:
Introducing Study Plan
Using AI Tools to Help you understand and remember your course concepts better and faster than any other resource.
Find the best videos to learn every concept in that course from Youtube and Tiktok without searching.
Save All Relavent Videos & Materials and access anytime and anywhere
Prepare Smart and Guarantee better grades
Students also viewed documents
lab 18.docx
lab_18.docx
Course
Course
3
Module5QuizSTA2023.d...
Module5QuizSTA2023.docx.docx
Course
Course
10
Week 7 Test Math302....
Week_7_Test_Math302.docx.docx
Course
Course
30
Chapter 1 Assigment ...
Chapter_1_Assigment_Questions.docx.docx
Course
Course
5
Week 4 tests.docx.do...
Week_4_tests.docx.docx
Course
Course
23
Week 6 tests.docx.do...
Week_6_tests.docx.docx
Course
Course
106