013213215X ism02.pdf-Circuit Elements a ...
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 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
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:
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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