Math 152, Spring 2011
Solutions Assignment #6
Notes:
•
Each question is worth 5 marks.
•
Due in class: Monday, February 21 for MWF sections; Tuesday, Feb
ruary 22 for TTh sections.
•
Solutions will be posted Tuesday, February 22 in the afternoon.
•
No late assignments will be accepted.
(1) Consider the resistor network given in the following picture:
t
V
V\

/\0\
Use the method of loop currents to find the current though each of the
resistors and the change in voltage E across the current source
Solution:
We are going to solve this problem by applying Kirchhoff’s
second law to each fundamental loop. For the loop on the left we get
i
1
+ 2(
i
1

i
2
) + 1(
i
1

i
3
)

10 = 0;
that is,
4
i
1

2
i
2

i
3
= 10
.
For the right upper loop we get
3
i
2
+ 2(
i
2

i
3
) + 2(
i
2

i
1
) = 0;
that is,

2
i
1
+ 7
i
2

2
i
3
= 0
.
For the right lower loop we get
(
i
3

i
1
) + 2(
i
3

i
2
) +
E
= 0;
1
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2
that is,

i
1

2
i
2
+ 3
i
3
+
E
= 0
.
Finally, the current source makes the current through the right lower
loop to be 2
A
in the direction shown in the picture. Thus
i
3
=

2
.
We can use Gaussian elimination to solve this system. The associated
augmented matrix is
4

2

1
0

10

2
7

2
0

0

1

2
3
1

0
0
0
1
0


2
We can permute the rows and change the signs and we obtain the
matrix
1
2

3

1

0
4

2

1
0

10
0
0
1
0


2

2
7

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 Spring '08
 Caddmen
 Math, Linear Algebra, linear transformation, Identity function, right lower loop

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