EECE 210 Homework 5, Fall 2014
American University of Beirut · Electric Circuits · 8 pages
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Homework 5
P5.1.5
Determine IO in Figure
P5.1.5.
Solution: From KCL, the
current through
the connection
between the
100 and 300
resistors is zero. The 2 A current flows through the 100 resistor and
establishes a voltage of 200 V across the 100 and 300 resistors. It follows
that IO = 200 V/300 = 2/3 A.
P5.1.8
Determine the power dissipated in
1-5
2 A
50
100
300
25
500
IO
IO
+
–
15 V
Figure P5.1.5
2 A
50
100
300
25
500
IO
IO
+
–
15 V
2 A
I = 0
+
–
200 V
4
4
– +
3VO
the 5 resistor due to each source in Figure P5.1.8, and the total power
dissipated in this resistor. Note that it should be obvious from superposition that
the dependent source does not contribute to current in this resistor.
Solution: If the dependent source is replaced
by an independent source and
applied alone, with the two current
sources set to zero, no current flows
through the 5 resistor. If the 2 A
source is applied alone, with the 4 A
source set to zero, the current
through the 5 resistor is 2 A, and
the power dissipated is 45 = 20 W.
If the 4 A source is applied alone, with
the 2 A source set to zero, the current
through the 5 resistor is 4 A, and the power dissipated is 165 = 80 W. If the
two sources are applied together, the current through the 5 resistor is 6 A, and
the power dissipated is 365 = 180 W.
P5.1.11
Determine the power delivered or absorbed by the
current source in Figure P5.1.11.
2-5
4
4
12
4
Figure P5.1.11
+
–
120 V
a
b
3
+
–
80 V
2 A
4 A
2
4
4
2
5
+
VO
– +
3VO
–
2 A
4 A
6 A
4
8
24
4
Figure P5.1.11
32
+
–
120 V
Solution
When the upper delta is transformed to a star and
the 120 V source is applied alone, V.
When the 6 A source is applied alone, the
source current divides into = 2 A on the RHS and 4
A on the LHS. Hence, V. It follows that Vab = 30 V
and the current source absorbs, 306 = 180 W.
P5.1.20
Determine VO in Figure
P5.1.20.
Solution: When the 10 V source is
applied alone, =, A and A.
3-5
4
4
12
4
+
–
120 V
a
b
3
+
–
80 V
4
4
12
4
6 A
a
b
3
+
–
50 V
4 A 2 A
10 V
–
+
4
10 A
8
Ix
4Ix
+
–
VO
2
2
a
b
Figure P5.1.20
10 V
–
+
4
8
IS
+
–
VO
2
2
a
b
xI
When the 10 A source is applied alone, with the other sources set to zero, the
8 resistor is in series with the parallel combination of 2 and 4 . The total
resistance is 8 + 8/6 = 28/3 . The current in 8 resistor is A and A.
When the dependent source is replaced by an independent source ISRC
applied alone, the 2 resistor is in series with the parallel combination of 2
and 4 . The total resistance is 2 + 8/6 = 10/3 . The current in 2 resistor is A
and A.
Substituting ISCR = 4Ix, it
follows from superposition
that Ix = 25/17 – 20/17 –
(32/17)Ix, which gives Ix =
5/49 A.
In the original circuit, Va =
2Ix = 10/49 A. The current in
the 4 resistor is (10 –
10/49)/4 = 120/49 A. From
KCL, the current in the 2 resistor is 10 + 5/49 – 120/49 = 375/49 A as shown. It
follows that VO = 10/49 + 2375/49 = 760/49 = 15.51 V.
4-5
10 V
–
+
4
10 A
8
4Ix
+
–
VO
2
2
a
b
10/49 V
120/49 A
5/49 A
375/49 A
4
8
+
–
VO
2
2
a
b
xI
ISRC
12/17 A
P5.1.22
Determine IO in Figure P5.1.22.
Solution: When the 10 A source is
applied alone, A and V. When
the dependent source is
replaced by an independent
source VSRC and applied
alone, . By superposition, Vx =
5/4 + VSRC/2. Substituting VSRC
= 5Vx, gives Vx = -5/6 V. It
follows that IO = -45/6 = -10/3 A.
5-5
4 S
10 A
4 S
5Vx
+
–
2 S
2 S
+
–
Vx
IO
Figure P5.1.22
4 S
10 A
4 S
+
–
2 S
2 S
Vx
5 A
5 A
4 S
4 S
+
–
2 S
2 S
VSRC /2
–
+
VSRC
P5.2.3
Determine VSRC in
Figure P5.2.3
assuming all
resistances are 1
. Note that this is
a special ladder
known as the R-2R
ladder, in which the resistance between each of the upper essential nodes and
the lower common node in 2R, and the resistance looking to the right of each of
the upper essential nodes is also 2R. Consequently, the voltages at the upper
nodes are successive multiplied by 2 in going from right to left.
P5.2.3
It is seen that Va = 2
V, Iad = 1 A, Iba = 2 A,
Vb = 4 V. Ibd = 2 A, Icb
= 4 A, Vcd = VSRC = 8
V.
6-5
VSRC
+
–
1 A
R
R
R
R
2R
2R
2R
Figure P5.2.3
c
2 V
2 A
4 V
4 A
a
b
8 V
1 A
2 A
d
VSRC
+
–
1 A
R
R
R
R
2R
2R
2R
Figure P5.2.3
a
b
c
d
20
30
35
–
+
4Ix
IO
15
13
0.75 A
50
50
10
+
–
30 V
Figure P5.2.5
Ix
P5.2.5
Determine
IO in
Figure
P5.2.5.
Solution: With Ix = -3 A, the
dependent source
becomes an
independent source
of 12 V and
reversed polarity.
The 10 resistor in
parallel with the 30
V source and the 20 resistor in series with the 0.75 A source are redundant for
the purpose of calculating IO and could be removed, the circuit becoming as
shown.
The 30 V source in series 50 is transformed to a 0.6 A current source in
parallel with 50 . This, in parallel with 50 becomes 25 . The 0.6 A current
source in parallel with 25 is transformed to a voltage source of 15 V in series
with 25 . This, in series with 35 becomes 60 . The 15 V source in series
with 60 is transformed to a current source of
0.25 A in parallel with 60 . This, in parallel with
30 becomes 20 and the 0.25 A source is
added to the 0.75 A source to give a 1 A source
in parallel with 20 . This is transformed to a 20
V source in series with 20 , the circuit
becoming as shown. It follows that IO = (12 +
20)/(20 + 15 + 13) = 32/48 = 2/3 A.
7-5
12 V
IO
15
13
+
–
20 V
+
–
20
30
35
12 V
IO
15
13
0.75 A
50
50
+
–
30 V
+
–
P5.2.12
Determine ISRC assuming all
resistances are 1 .
Solution: The circuit is symmetrical with respect
to the diagonal and could be split into
two half circuits, as shown in the
figure for one half circuit.
Combining the resistors, the
circuit becomes as shown. It follows
that the current due to one half circuit
is 7 A, so that the current due to the
complete circuit is 14 A.
8-5
+
–
1
21 V
1
6
7 A
1
2
4
1
21 V
ISRC
Figure P5.2.12
+
–
+
–
21 V
1
2
3
5
4
6
1
1
1
1
1
1