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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