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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 7
Problems
2–7
[b]
Draw the circuit model from part (a) and attach a 25Ω resistor:
To ﬁnd the power delivered to the 25Ω resistor we must calculate the
current through the 25 Ω resistor. Do this by ﬁrst using KCL to recognize
that the current in each of the components is
i
t
, ﬂowing in a clockwise
direction. Write a KVL equation in the clockwise direction, starting
below the voltage source, and using Ohm’s law to express the voltage
drop across the resistors in the direction of the current
i
t
ﬂowing through
the resistors:
-
25 V + 100
i
t
+ 25
i
t
=0
so
125
i
t
= 25
so
i
t
=
25
125
=0
.
2A
Thus, the power delivered to the 25Ω resistor is
p
25
= (25)
i
2
t
= (25)(0
.
2)
2
=1W
.
AP 2.8
[a]
From the graph in Assessment Problem 2.7(a), we see that when
v
t
= 0,
i
t
=0
.
25 A. Therefore the current source must be 0
.
25 A. 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 0
.
25 A current
source in parallel with a 100Ω resistor, as shown below:
[b]
Draw the circuit model from part (a) and attach a 25Ω resistor:
Note that by writing a KVL equation around the right loop we see that
the voltage drop across both resistors is
v
t
. Write a KCL equation at the
top center node, summing the currents leaving the node. Use Ohm’s law
to specify the currents through the resistors in terms of the voltage drop
across the resistors and the value of the resistors.
-
0
.
25 +
v
t
100
+
v
t
25
=0
,
so
5
v
t
= 25
,
thus
v
t
=5V
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 8
2–8
CHAPTER 2. Circuit Elements
p
25
=
v
2
t
25
=1W
.
AP 2.9
First note that we know the current through all elements in the circuit except
the 6 kΩ resistor (the current in the three elements to the left of the 6 kΩ
resistor is
i
1
; the current in the three elements to the right of the 6 kΩ resistor
is 30
i
1
). To ﬁnd the current in the 6 kΩ resistor, write a KCL equation at the
top node:
i
1
+ 30
i
1
=
i
6k
= 31
i
1
We can then use Ohm’s law to ﬁnd the voltages across each resistor in terms
of
i
1
. The results are shown in the ﬁgure below:
[a]
To ﬁnd
i
1
, write a KVL equation around the left-hand loop, summing
voltages in a clockwise direction starting below the 5V source:
-
5 V + 54
,
000
i
1
-
1 V + 186
,
000
i
1
=0
Solving for
i
1
54
,
000
i
1
+ 186
,
000
i
1
=6V
so
240
,
000
i
1
=6V
Thus,
i
1
=
6
240
,
000
= 25
μ
A
[b]
Now that we have the value of
i
1
, we can calculate the voltage for each
component except the dependent source. Then we can write a KVL
equation for the right-hand loop to ﬁnd the voltage
v
of the dependent
source. Sum the voltages in the clockwise direction, starting to the left of
the dependent source:
+
v
-
54
,
000
i
1
+8V
-
186
,
000
i
1
=0
Thus,
v
= 240
,
000
i
1
-
8 V = 240
,
000(25
×
10
-
6
)
-
8V=6V
-
8V=
-
2V
We now know the values of voltage and current for every circuit element.
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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