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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 13
Problems
2–13
The interconnection is invalid. The voltage drop between the top terminal and
the bottom terminal on the left hand side is due to the 6 V and 8 V sources,
giving a total voltage drop between these terminals of 14 V. But the voltage
drop between the top terminal and the bottom terminal on the right hand side
is due to the 4 V and 12 V sources, giving a total voltage drop between these
two terminals of 16 V. The voltage drop between any two terminals in a valid
circuit must be the same, so the interconnection is invalid.
P 2.6
The interconnection is valid, since the voltage sources can carry the 20 mA
current supplied by the current source, and the current sources can carry
whatever voltage drop is required by the interconnection. In particular, note
the the voltage drop across the three sources in the right hand branch must be
the same as the voltage drop across the 15 mA current source in the middle
branch, since the middle and right hand branches are connected between the
same two terminals. In particular, this means that
v
1
(the voltage drop across the middle branch)
=

20V + 60V

v
2
Hence any combination of
v
1
and
v
2
such that
v
1
+
v
2
= 40V is a valid
solution.
P 2.7
The interconnection is invalid. In the middle branch, the value of the current
i
Δ
must be

25 A, since the 25 A current source supplies current in this
branch in the direction opposite the direction of the current
i
Δ
. Therefore, the
voltage supplied by the dependent voltage source in the left hand branch is
6(

25) =

150 V. This gives a voltage drop from the top terminal to the
bottom terminal in the left hand branch of 50

(

150) = 200 V. But the
voltage drop between these same terminals in the right hand branch is 250 V,
due to the voltage source in that branch. Therefore, the interconnection is
invalid.
P 2.8
First, 10
v
a
= 5 V, so
v
a
=0
.
5 V. Then recognize that each of the three
branches is connected between the same two nodes, so each of these branches
must have the same voltage drop. The voltage drop across the middle branch
is 5 V, and since
v
a
=0
.
5 V,
v
g
=0
.
5

5=

4
.
5 V. Also, the voltage drop
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