Problems
Section 4-2 Node Voltage Analysis of Circuits with Current Sources
P4.2-1
KCL at node 1:
0=
v −v
− 4 −4 − 2
1
1
2
+
+i =
+
+ i = −1.5 + i ⇒ i = 1.5 A
8
6
8
6
v
(checked using LNAP 8/13/02)
P4.2-2
KCL at node 1:
v −v
v
1
2
1
+
+ 1 = 0 ⇒ 5 v − v = −20
1
2
20
5
KCL at node 2:
v −v
v −v
1
2
2
3
+2=
⇒ − v + 3 v − 2 v = 40
1
2
3
20
10
KCL at node 3:
v −v
v
2
3
3
+1 =
⇒ − 3 v + 5 v = 30
2
3
10
15
Solving gives v1 = 2 V, v2 = 30 V and v3 = 24 V.
(checked using LNAP 8/13/02)
P4.2-3
KCL at node 1:
v −v
v
4 − 15 4
1
2
1
+
=i ⇒ i =
+
= −2 A
1
1
5
20
5
20
KCL at node 2:
v −v
v −v
1
2
2
3
+i =
2
5
15
⎛ 4 − 15 ⎞ 15 − 18
⇒ i = −⎜
=2A
⎟+
2
15
⎝ 5 ⎠
(checked using LNAP 8/13/02)
P4.2-4
Node equations:
−.003 +
−
v1 v1 − v2
+
=0
R1
500
v1 − v2 v2
+
− .005 = 0
500
R2
When v1 = 1 V, v2 = 2 V
1 −1
1
+
= 0 ⇒ R1 =
= 200 Ω
1
R1 500
.003 +
500
−1 2
2
−
+ − .005 = 0 ⇒ R2 =
= 667 Ω
1
500 R2
.005 −
500
−.003 +
(checked using LNAP 8/13/02)
P4.2-5
Node equations:
v1
v − v 2 v1 − v3
+ 1
+
=0
500
125
250
v − v3
v − v2
− 1
− .001 + 2
=0
125
250
v − v3 v1 − v3 v3
− 2
−
+
=0
250
250 500
Solving gives:
v1 = 0.261 V, v2 = 0.337 V, v3 = 0.239 V
Finally, v = v1 − v3 = 0.022 V
(checked using LNAP 8/13/02)
P4.2-6
12 Ω + ( 40 Ω & 10 Ω ) = 20 Ω
60 Ω & 120 Ω = 40 Ω
The node equations are
3 × 10−3 =
2 × 10−3 +
v2 − v3
v1 − v 2
20
v1 − v 2
+
=
20
v1 − v 3
v1 − v 3
20
v 2 − v3
10
v3
⇒
0.06 = 2v1 − ( v 2 − v 3 )
⇒
0.04 = −v1 + 3v 2 − 2v 3
⇒
0 = − ( 2v 1 + 4 v 2 ) + 7 v 3
⎡ 2 −1 −1⎤ ⎡ v1 ⎤ ⎡.06⎤
⎢ −1 3 −2 ⎥ ⎢v ⎥ = ⎢.04⎥
⎢
⎥⎢ 2⎥ ⎢ ⎥
⎢⎣ −2 −4 +7 ⎥⎦ ⎢⎣ v 3 ⎥⎦ ⎢⎣ 0 ⎥⎦
⎡ v1 ⎤ ⎡0.244⎤
⎢ ⎥ ⎢
⎥
⎢v 2 ⎥ = ⎢ 0.228⎥
⎢ v 3 ⎥ ⎢⎣0.200⎥⎦
⎣ ⎦
10
+
20
=
40
Solving, e.g. using MATLAB, gives
⇒
(
)
(a) The power supplied by the 3 mA current source is 3 ×10−3 ( 0.244 ) = 0.732 mW. The power
(
)
supplied by the 2 mA source is 2 ×10−3 ( 0.228) = 0.456 mW.
(b) The current in the 12 Ω resistor is equal to the current i =
so the power received by the 12 Ω resistor is ( 0.8 ×10
v1 − v 2
20
=
0.244 − 0.228
= 0.8 mA
20
) (12 ) = 7.68 ×10
−3 2
−b
= 7.68 μ W.
(checked: LNAP and MATLAB 5/31/04)
P4.2-7
Apply KCL at node a to get
2=
va
R
+
va
4
+
va − vb
2
=
7 7 7 − 10 7 1
+ +
= +
⇒ R=4Ω
R 4
2
R 4
Apply KCL at node b to get
is +
va − vb
2
=
vb
8
+
vb
8
= is +
7 − 10 10 10
= +
⇒ is = 4 A
2
8 8
(checked: LNAP 6/21/04)
Section 4-3 Node Voltage Analysis of Circuits with Current and Voltage Sources
P4.3-1
Express the branch voltage of the voltage source in terms of its node voltages:
0 − va = 6 ⇒ va = −6 V
KCL at node b:
va − vb
v −v
+2= b c
6
10
KCL at node c:
Finally:
⇒
−6 − vb
v −v
+2= b c
6
10
vb − vc vc
=
10
8
⇒ −1−
⇒ 4 vb − 4 vc = 5 vc
⎛9 ⎞
30 = 8 ⎜ vc ⎟ − 3 vc
⎝4 ⎠
vb
v −v
+2= b c
6
10
⇒ vb =
⇒ 30 = 8 vb − 3 vc
9
vc
4
⇒ vc = 2 V
(checked using LNAP 8/13/02)
P4.3-2
Express the branch voltage of each voltage source in terms of its node voltages to get:
va = −12 V, vb = vc = vd + 8
KCL at node b:
vb − va
= 0.002 + i ⇒
4000
vb − ( −12 )
= 0.002 + i ⇒ vb + 12 = 8 + 4000 i
4000
KCL at the supernode corresponding to the 8 V source:
v
0.001 = d + i ⇒ 4 = vd + 4000 i
4000
so
vb + 4 = 4 − vd
⇒
( vd + 8) + 4 = 4 − vd
Consequently vb = vc = vd + 8 = 4 V and i =
⇒ vd = −4 V
4 − vd
= 2 mA
4000
(checked using LNAP 8/13/02)
P4.3-3
Apply KCL to the supernode:
va − 10 va va − 8
+
+
− .03 = 0 ⇒ va = 7 V
100
100 100
(checked using LNAP 8/13/02)
P4.3-4
Apply KCL to the supernode:
va + 8 ( va + 8 ) − 12 va − 12 va
+
+
+
=0
500
125
250
500
Solving yields
va = 4 V
(checked using LNAP 8/13/02)
P4.3-5
The power supplied by the voltage source is
⎛ v −v v −v ⎞
⎛ 12 − 9.882 12 − 5.294 ⎞
va ( i1 + i 2 ) = va ⎜ a b + a c ⎟ = 12 ⎜
+
⎟
6 ⎠
4
6
⎝
⎠
⎝ 4
= 12(0.5295 + 1.118) = 12(1.648) = 19.76 W
(checked using LNAP 8/13/02)
P4.3-6
Label the voltage measured by the meter. Notice that this is a node voltage.
Write a node equation at the node at which
the node voltage is measured.
⎛ 12 − v m ⎞ v m
v −8
−⎜
+ 0.002 + m
=0
⎟+
3000
⎝ 6000 ⎠ R
That is
6000 ⎞
6000
⎛
⎜3 +
⎟ v m = 16 ⇒ R = 16
R ⎠
⎝
−3
vm
(a) The voltage measured by the meter will be 4 volts when R = 6 kΩ.
(b) The voltage measured by the meter will be 2 volts when R = 1.2 kΩ.
P4.3-7
Apply KCL at nodes 1 and 2 to get
10 − v1
1000
10 − v 2
=
v1
+
v1 − v 2
3000 5000
v1 − v 3
v3
+
=
4000
5000 2000
⇒
23v1 − 3v 2 = 150
⇒
-4v1 + 19v 3 = 50
Solving, e.g. using MATLAB, gives
⎡ 23 −3⎤ ⎡ v1 ⎤ ⎡150⎤
⎢ −4 19 ⎥ ⎢v ⎥ = ⎢ 50 ⎥
⎣
⎦⎣ 2⎦ ⎣ ⎦
Then
ib =
v1 − v 2
5000
⇒
v1 = 7.06 V and v1 = 4.12 V
=
7.06 − 4.12
= 0.588 mA
5000
=
7.06 − 10 4.12 − 10
+
= −4.41 mA
1000
4000
Apply KCL at the top node to get
ia =
v 1 − 10
1000
+
v 2 − 10
4000
(checked: LNAP 5/31/04)
P4.3-8
vo
R3
+
v o − v1
+
R1
vo − v2
=0
R2
⇒
vo =
v1
v2
+
R
R
R
R
1+ 1 + 1 1 + 2 + 2
R 2 R3
R1 R 3
(a) When R 1 = 10 Ω, R 2 = 40 Ω and R 3 = 8 Ω
vo =
v1
v2
+
= 0.4v1 + 0.1v 2
1 5 1+ 4 + 5
1+ +
4 4
So a = 0.4 and b = 0.1.
(b) When R 1 = R 2 and R 3 = R 1 & R 2 = R 1 / 2
vo =
v1
v2
+
= 0.25v1 + 0.25v 2
1+1+ 2 1+1+ 2
So a = 0.25 and b = 0.25.
(checked: LNAP 5/31/04)
P4.3-9
Express the voltage source voltages as functions of the node voltages to get
v 2 − v1 = 5 and v 4 = 15
Apply KCL to the supernode corresponding to the 5 V source to get
1.25 =
v1 − v 3
8
+
v 2 − 15
20
=0
⇒
80 = 5v1 + 2v 2 − 5v 3
Apply KCL at node 3 to get
v1 − v 3
8
=
v3
40
+
v 3 − 15
12
⇒
− 15v1 + 28v 3 = 150
Solving, e.g. using MATLAB, gives
⎡ −1 1 0 ⎤ ⎡ v 1 ⎤ ⎡ 5 ⎤
⎢ 5 2 −5⎥ ⎢ v ⎥ = ⎢ 80 ⎥
⎢
⎥⎢ 2⎥ ⎢ ⎥
⎢⎣ −15 0 28 ⎥⎦ ⎢⎣ v 3 ⎥⎦ ⎢⎣150 ⎥⎦
⇒
⎡ v1 ⎤ ⎡ 22.4 ⎤
⎢ ⎥ ⎢
⎥
⎢ v 2 ⎥ = ⎢ 27.4 ⎥
⎢ v 3 ⎥ ⎢⎣17.4 ⎥⎦
⎣ ⎦
So the node voltages are:
v1 = 22.4 V, v 2 = 27.4 V, v 3 = 17.4 V, and v 4 = 15
(checked: LNAP 6/9/04)
P4.3-10
Write a node equation to get
⎛ 12 − 4.5 ⎞ 4.5 4.5 − 6
7.5 4.5 1.5
−⎜
+
+
=0 ⇒ −
+
−
=0
⎜ R1 ⎟⎟ R 3
R2
R1 R 3 R 2
⎝
⎠
Notice that
Similarly,
7.5
is either 0.75 mA or 1.5 mA depending on whether R1 is 10 kΩ or 5 kΩ.
R1
4.5
1.5
is either 0.45 mA or 0.9 mA and
is either 0.15 mA or 0.3 mA. Suppose R1
R3
R2
and R2 are 10 kΩ resistors and R3 is a 5 kΩ resistor. Then
−
7.5 4.5 1.5
+
−
= −0.75 + 0.9 − 0.15 = 0
R1 R 3 R 2
It is possible that two of the resistors are 10 kΩ and the third is 5 kΩ. R3 is the 5 kΩ resistor.
(checked: LNAP 6/9/04)
P4.3-11
Label the node voltages:
Express the voltage source voltages in terms of the node voltages:
v1 − v 2 = 8 and v 5 = −28
Apply KCL to the supernode corresponding to the 8-V source:
2=
v1 − v 6
16
+
v2
12
+
v 2 − v3
3
+
v2 − v3
6
+3
⇒
− 3v1 + 3v 6 − 28v 2 + 24v 3 = 48
Apply KCL at node 6 to get
2+
v 6 − v1
16
+
v6
20
=0
⇒
5v1 − 9v 6 = 160
Apply KCL at node 3 to get
v 2 − v3
6
+
v2 − v3
3
=
v3 − v 4
⇒
10
− 15v 2 + 18v 3 − 3v 4 = 0
Apply KCL at node 4 to get
3+
v3 − v4
10
=
v 4 − v5
7
⇒
210 = −7v 3 + 17v 4 − 10v 5
Solving, e.g. using MATLAB, gives
0
0 ⎤ ⎡ v1 ⎤ ⎡ 8 ⎤
⎡ 1 −1 0 0
⎢ ⎥
⎢0
0
0 0
1
0 ⎥⎥ ⎢v 2 ⎥ ⎢⎢ −28⎥⎥
⎢
⎢ −3 −28 24 0
0
3 ⎥ ⎢ v 3 ⎥ ⎢ 48 ⎥
⎢
⎥⎢ ⎥ = ⎢
⎥
5
0
0
0
0
−
9
⎢
⎥ ⎢ v 4 ⎥ ⎢160 ⎥
⎢ 0 −15 18 −3 0
0 ⎥ ⎢v 5 ⎥ ⎢ 0 ⎥
⎢
⎥⎢ ⎥ ⎢
⎥
0 −7 17 −10 0 ⎦⎥ ⎣⎢ v 6 ⎦⎥ ⎣⎢ 210 ⎦⎥
⎣⎢ 0
⇒
⎡ v1 ⎤ ⎡ −8.5 ⎤
⎢v ⎥ ⎢
⎥
⎢ 2 ⎥ ⎢ −16.5⎥
⎢ v 3 ⎥ ⎢ −15.5⎥
⎢ ⎥=⎢
⎥
⎢ v 4 ⎥ ⎢ −10.5⎥
⎢ v 5 ⎥ ⎢ −28 ⎥
⎢ ⎥ ⎢
⎥
⎣⎢ v 6 ⎦⎥ ⎣⎢ −22.5⎦⎥
(checked: PSpice 6/12/04)
P4.3-12
Express the voltage source voltages in terms of the node voltages:
v 2 − v1 = 8 and v 3 − v 1 = 12
Apply KVL to the supernode to get
v2
10
+
v1
4
+
v3
5
=0
⇒
2v 2 + 5v1 + 4v 3 = 0
So
2 ( 8 + v1 ) + 5v1 + 4 (12 + v1 ) = 0
⇒
v1 = −
64
V
11
The node voltages are
v1 = −5.818 V
v 2 = 2.182 V
v 3 = 6.182 V
(checked: LNAP 6/21/04)
Section 4-4 Node Voltage Analysis with Dependent Sources
P4.4-1
Express the resistor currents in terms of the
node voltages:
va − vc
= 8.667 − 10 = −1.333 A and
1
v −v
2 − 10
i 2= b c =
= −4 A
2
2
i 1=
Apply KCL at node c:
i1 + i 2 = A i1 ⇒ − 1.333 + ( −4 ) = A (−1.333)
⇒
A=
−5.333
=4
−1.333
(checked using LNAP 8/13/02)
P4.4-2
Write and solve a node equation:
va − 6
v
v − 4va
+ a + a
= 0 ⇒ va = 12 V
1000 2000 3000
ib =
va − 4va
= −12 mA
3000
(checked using LNAP 8/13/02)
P4.4-3
First express the controlling current in terms of
the node voltages:
2 − vb
i =
a
4000
Write and solve a node equation:
−
2 − vb
v
⎛ 2 − vb ⎞
+ b − 5⎜
⎟ = 0 ⇒ vb = 1.5 V
4000 2000
⎝ 4000 ⎠
(checked using LNAP 8/14/02)
P4.4-4
Apply KCL to the supernode of the CCVS to get
12 − 10 14 − 10 1
+
− + i b = 0 ⇒ i b = −2 A
4
2
2
Next
10 − 12
1⎫
=− ⎪
−2
V
=4
4
2⎬ ⇒ r =
1
A
−
r i a = 12 − 14 ⎪⎭
2
ia =
(checked using LNAP 8/14/02)
P4.4-5
First, express the controlling current of the CCVS in
v2
terms of the node voltages: i x =
2
Next, express the controlled voltage in terms of the
node voltages:
v2
24
12 − v 2 = 3 i x = 3
⇒ v2 =
V
2
5
so ix = 12/5 A = 2.4 A.
(checked using ELab 9/5/02)
P4.4-6
Pick a reference node and label the unknown node voltages:
Express the controlling current of the dependent source in terms of the node voltages:
i4 = −
va
. Then v b = 2 i 4 = −
6
Apply KCL at node a:
va
3
.
v a − 12
3
+
va
6
+
va − vb
4
=0
So:
⎛ v ⎞
va − ⎜ − a ⎟
v a − 12 v a
⎝ 3 ⎠ =0
+ +
3
6
4
va ⎞
⎛
4 ( v a − 12 ) + 2 v a + 3 ⎜ v a + ⎟ = 0 ⇒ v a = 4.8 V
3⎠
⎝
12 − 4.8 7.2
The current in the 12-V voltage source is i =
=
= 2.4 A
3
3
So the power supplied by the voltage source is 12(2.4) = 28.8 W.
(checked: LNAP 5/18/04)
P4.4-7
Label the node voltages:
First, v2 = 10 V, due to the independent
voltage source. Next, express va and ib, the
controlling voltage and current of the
dependent sources, in terms of the node
voltages:
ib =
v3 − v2
8
=
v 3 − 10
8
and
v a = v 1 − v 2 = v 1 − 10
Next, express ib and 3va, the controlled voltages of the dependent sources, in terms of the node
voltages:
⎛ v 3 − 10 ⎞
8i b = v 1 − v 3
⇒
8⎜
⎟ = v1 − v 3
⎝ 8 ⎠
and
3v a = v1
⇒
3 ( v1 − 10 ) = v1
⇒
v1 = 15 V
So
v 3 − 10 = 15 − v 3
⇒
v 3 = 12.5 V
Next
v a = 15 − 10 = 5 V
ib =
and
12.5 − 10
= 0.3125 A
8
Finally, apply KCL to the top node to get
ic =
va
2
+ ib =
5
+ 0.3125 = 2.8125 A
2
(checked: LNAP 6/3/04)
P4.4-8
Label the node voltages.
First, v2 = 10 V due to the independent voltage
source. Next, express the controlling current
of the dependent source in terms of the node
voltages:
ia =
v3 − v2
16
=
v 3 − 10
16
Now the controlled voltage of the dependent source can be expressed as
⎛ v 3 − 10 ⎞
v1 − v 3 = 8 i a = 8 ⎜
⎟
⎝ 16 ⎠
3
v1 = v 3 − 5
2
⇒
Apply KCL to the supernode corresponding to the dependent source to get
v1 − v 2
4
+
v1
12
+
v3 − v 2
16
+
v3
8
=0
Multiplying by 48 and using v2 = 10 V gives
16v 1 + 9v 3 = 150
Substituting the earlier expression for v1
⎛3
⎞
16 ⎜ v 3 − 5 ⎟ + 9v 3 = 150
⎝2
⎠
⇒
v 3 = 6.970 V
Then v1 = 5.455 V and ia = -0.1894 A. Applying KCL at node 2 gives
v1
12
= ib +
10 − v1
4
⇒
12 i b = −3 + 4 v1 = −30 + 4 ( 5.455 )
So
i b = −0.6817 A.
Finally, the power supplied by the dependent source is
p = ( 8 i a ) i b = 8 ( −0.1894 ) ( −0.6817 ) = 1.033 W
(checked: LNAP 5/24/04)
P4.4-9
Apply KCL at node 2:
i a + bi a = i b =
but
ia =
v3 − v2
20
v 2 − v1
40
=
=
−6 − ( 0 )
= −0.3 A
20
0−4
= −0.1
40
so
(1 + b )( −0.1) = ( −0.3)
⇒
b =2
A
A
Next apply KCL to the supernode corresponding to the voltage source.
v1
10
+ 2 ia +
v3
R
=0
⇒
4
−6
+ 2 ( −0.1) +
=0
10
R
⇒
R=
6
= 30 Ω
.2
(checked: LNAP 6/9/04)
P4.4-10
(a) Express the controlling voltage of the dependent source in terms of the node voltages:
va = 9 − vb
Apply KCL at node b to get
9 − vb
100
= A(9 − v b ) +
vb
200
⇒
A=
18 − 3v b
200 ( 9 − v b )
= 0.02
(b) The power supplied by the dependent source is
− ( Av a ) v b = − ( 0.02 ( 9 − 18 ) ) (18 ) = 3.24 W
(checked: LNAP 6/06/04)
P4.4-11
This circuit contains two ungrounded voltage sources, both incident to node x. In such a circuit
it is necessary to merge the supernodes corresponding to the two ungrounded voltage sources
into a single supernode. That single supernode separates the two voltage sources and their nodes
from the rest of the circuit. It consists of the two resistors and the current source. Apply KCL to
this supernode to get
v x − 20 v x
+ +4=0
⇒
v x = 10 V .
2
10
The power supplied by the dependent source is
( 0.1 v ) ( −30 ) = −30 W .
x
(checked: LNAP 6/6/04)
P4.4-12
Express the voltages of the independent voltage sources in terms of the node voltages
v 1 − v 2 = 16 and v 4 − v 5 = 8
Express the controlling current of the dependent source in terms of the node voltages
ix =
v3
6
Express the controlled voltage of the dependent source in terms of the node voltages
⎛ v3 ⎞
v 2 − v 4 = 4i x = 4 ⎜ ⎟
⎝6⎠
⇒
− 6v 2 + 4v 3 + 6v 4 = 0
Apply KCL to the supernode to get
v1 − v 3
2
+
v 4 − v3
3
+
v5
8
=1
⇒
12v1 − 20v 3 + 8v 4 + 3v 5 = 24
Apply KCL at node 3 to get
v 3 − v1
2
+
v3
6
+
v3 − v4
Solving, e.g. using MATLAB, gives
3
=0
⇒
− 3v1 + 6v 2 − 2v 4 = 0
0 0 ⎤ ⎡ v1 ⎤ ⎡16 ⎤
⎡ 1 −1 0
⎢ ⎥
⎢0 0
0
1 −1⎥⎥ ⎢v 2 ⎥ ⎢⎢ 8 ⎥⎥
⎢
⎢ 0 −6 4
6 0 ⎥ ⎢v 3 ⎥ = ⎢ 0 ⎥
⎢
⎥⎢ ⎥ ⎢ ⎥
⎢12 0 −20 8 3 ⎥ ⎢ v 4 ⎥ ⎢ 24⎥
6 −2 0 ⎦⎥ ⎣⎢ v 5 ⎦⎥ ⎣⎢ 0 ⎦⎥
⎣⎢ −3 0
⇒
⎡ v1 ⎤ ⎡ 24 ⎤
⎢v ⎥ ⎢ ⎥
⎢ 2⎥ ⎢ 8 ⎥
⎢ v 3 ⎥ = ⎢12 ⎥
⎢ ⎥ ⎢ ⎥
⎢v 4 ⎥ ⎢ 0 ⎥
⎢ ⎥
⎣ v 5 ⎦ ⎢⎣ −8⎥⎦
(checked: LNAP 6/13/04)
P4.4-13
Express the voltage source voltages in terms of the node voltages:
v 1 − v 2 = 8 and v 4 − v 3 = 16
Express the controlling current of the dependent source in terms of the node voltages:
ix =
v 2 − v3
10
+
v1 − v 3
5
= 0.2v1 + 0.1v 2 − 0.3v 3
Express the controlled voltage of the dependent source in terms of the node voltages:
v 5 = 4i x = 0.8v1 = 0.4v 2 − 1.2v 3
⇒
0.8v1 + 0.4v 2 − 1.2v 3 − v 5 = 0
Apply KVL to the supernodes
v1 − v 5
+
v2 − v4
2
v4 − v2
4
+
4
v3
8
+
+
v 2 − v3
10
v3 − v2
10
+
+
v1 − v 3
5
v 3 − v1
5
=0
=2
⇒
⇒
14v1 + 7v 2 − 6v 3 − 5v 4 − 10v 5 = 0
− 8v1 − 14v 2 + 17v 3 + 10v 4 = 80
Solving, e.g. using MATLAB, gives
−1
0
0
0 ⎤ ⎡ v1 ⎤ ⎡ 8 ⎤
⎡1
⎢0
⎥ ⎢v ⎥ ⎢16 ⎥
−
0
1
1
0
⎢
⎥⎢ 2⎥ ⎢ ⎥
⎢ 0.8 0.4 −1.2 0 −1 ⎥ ⎢ v 3 ⎥ = ⎢ 0 ⎥
⎢
⎥⎢ ⎥ ⎢ ⎥
−6 −5 −10 ⎥ ⎢v 4 ⎥ ⎢ 0 ⎥
7
⎢ 14
⎢⎣ −8 −14 17 10
0 ⎥⎦ ⎢⎣ v 5 ⎥⎦ ⎢⎣80 ⎥⎦
⇒
⎡ v1 ⎤ ⎡11.32 ⎤
⎢v ⎥ ⎢
⎥
⎢ 2 ⎥ ⎢ 3.32 ⎥
⎢ v 3 ⎥ = ⎢ 2.11 ⎥
⎢ ⎥ ⎢
⎥
⎢v 4 ⎥ ⎢18.11⎥
⎢ v 5 ⎥ ⎢⎣ 7.85 ⎥⎦
⎣ ⎦
(checked: LNAP 6/13/04)
P4.4-14
Express the voltage source voltage in terms of the node voltages:
v 3 = 12
Express the controlling signals of the dependent sources in terms of the node voltages:
v y = v1 − v 3 and i x = −
v2
8
Express the controlled voltage of the CCVS in terms of the node voltages:
⎛ v2 ⎞
v 2 − v 4 = 3i x = 3 ⎜ − ⎟
⎝ 8⎠
⇒
11v 2 − 8v 4 = 0
Apply KCL to the supernode corresponding to the dependent voltage source:
v 2 − v1
5
+
v2
8
+
v4 − v3
2
= 2 ( v1 − v 3 )
⇒
− 88v1 + 13v 2 + 60v 3 + 20v 4 = 0
Apply KCL at node 1
3+
v1 − v 3
4
+
v1 − v 2
5
=0
⇒
− 9v1 + 4v 2 + 5v 3 = 60
Solving, e.g. using MATLAB, gives
0 1 0 ⎤ ⎡ v1 ⎤ ⎡12 ⎤
⎡ 0
⎢ −88 13 60 20 ⎥ ⎢v ⎥ ⎢ 0 ⎥
⎢
⎥⎢ 2⎥ = ⎢ ⎥
⎢ 0 11 0 −8⎥ ⎢ v 3 ⎥ ⎢ 0 ⎥
⎢
⎥⎢ ⎥ ⎢ ⎥
⎣ −9 4 5 0 ⎦ ⎢⎣v 4 ⎥⎦ ⎣60⎦
⇒
⎡ v1 ⎤ ⎡ −230.4⎤
⎢v ⎥ ⎢
⎥
⎢ 2 ⎥ = ⎢ −518.4⎥
⎢ v 3 ⎥ ⎢ 12 ⎥
⎢ ⎥ ⎢
⎥
⎢⎣ v 4 ⎥⎦ ⎣ −712.8⎦
(checked: LNAP 6/13/04)
P4.4-15
Express the controlling voltage and current of
the dependent sources in terms of the node
voltages:
v3 − v4
v a = v 4 and i b =
R2
Express the voltage source voltages in terms of
the node voltages:
v 1 = V s and v 2 − v 3 = A v a = Av 4
Apply KCL to the supernode corresponding to the dependent voltage source
v 2 − v1
+
R1
v3 − v 4
R2
= Is
⇒ − R 2 v1 + R 2 v 2 + R1 v 3 − R1 v 4 = R1 R 2 I s
Apply KCL at node 4:
B
v3 − v4
R2
+
v3 − v4
R2
=
v4
R3
⇒
⎛
( B + 1) v 3 − ⎜⎜ B + 1 +
⎝
R2 ⎞
⎟ v4 = 0
R 3 ⎟⎠
Organizing these equations into matrix form:
⎡ 1
⎢
⎢ 0
⎢−R2
⎢
⎢
⎢ 0
⎢⎣
0
1
R2
0
⎤
⎥ ⎡ v1 ⎤ ⎡ V s ⎤
⎥ ⎢v ⎥ ⎢
⎥
⎥ ⎢ 2⎥ = ⎢ 0 ⎥
⎥ ⎢v 3 ⎥ ⎢ R R I ⎥
⎛
R 2 ⎞⎥ ⎢ ⎥ ⎢ 1 2 s ⎥
B +1 − ⎜ B +1+
⎟⎟ ⎥ ⎣⎢v 4 ⎦⎥ ⎣ 0 ⎦
⎜
R
3
⎝
⎠ ⎥⎦
0
−1
R1
0
−A
− R1
With the given values:
⎡ v1 ⎤ ⎡ 25 ⎤
0 0
0 ⎤ ⎡ v1 ⎤ ⎡ 25 ⎤
⎡ 1
⎢v ⎥ ⎢
⎢v ⎥ ⎢
⎢ 0
⎥
⎥
−5 ⎥ ⎢ 2 ⎥ ⎢ 0 ⎥
1 −1
44.4 ⎥⎥
2⎥
⎢
⎢
⎢
=
⇒
=
⎢ v 3 ⎥ ⎢ 8.4 ⎥
⎢ −20 20 10
−10 ⎥ ⎢ v 3 ⎥ ⎢ 400 ⎥
⎢ ⎥ ⎢
⎥
⎢
⎥⎢ ⎥ ⎢
⎥
0 4 −4.667 ⎦ ⎢⎣v 4 ⎥⎦ ⎣ 0 ⎦
⎢⎣v 4 ⎥⎦ ⎣ 7.2 ⎦
⎣ 0
(Checked using LNAP 9/29/04)
P4.4-16
Express the controlling voltage and current of
the dependent sources in terms of the node
voltages:
v a = v 4 = 22.5 V
and
v 3 − v 4 −15 − 22.5
ib =
=
= −0.75
R2
50
Express the dependent voltage source voltage
in terms of the node voltages:
v 2 − v 3 = Av a = Av 4
so
A=
v2 − v3
v4
=
75 − ( −15 )
= 4 V/V
22.5
Apply KCL to the supernode corresponding to the dependent voltage source
v 2 − v1
R1
+
v3 − v 4
R2
= Is
⇒
75 − 10 −15 − 22.5
+
= 2.5 ⇒ R1 = 20 Ω
R1
50
Apply KCL at node 4:
v3 − v 4
R2
=
v4
R3
+B
v3 − v 4
R2
⇒
−15 − 22.5 22.5
−15 − 22.5
=
+B
⇒ B = 2.5 A/A
50
20
50
(Checked using LNAP 9/29/04)
P4.4-17
v 2 − v1 21 − 12
v2
−3
a. R1 =
=
=4Ω
=
= 6 Ω and R 2 =
2 − 0.5
1.5
1.25 − 2 −0.75
b. The power supplied by the voltage source is 12 ( 0.5 + 1.25 − 2 ) = −3 W . The power supplied
by the 1.25-A current source is 1.25 ( −3 − 12 ) = −18.75 W . The power supplied by the 0.5-A
current source is −0.5 ( 21) = −10.5 W . The power supplied by the 2-A current source is
2 ( 21 − ( −3) ) = 48 W .
P4.4-18
i1 =
12 − ( −1.33)
= 1.666 A
8
and
i2 =
a. R1 =
v 2 − v1
2 − i1
=
9.6
= 2.4 A
4
v3
−1.33
9.6 − 12
= 6 Ω and R 2 =
=
= 3.98 4 Ω
i1 − 2 1.666 − 2
2 − 2.4
b. The power supplied by the voltage source is 12 ( 2.4 + 1.66 − 2 ) = 24.7 W . The power supplied
by the current source is 2 ( 9.6 − ( −1.33) ) = 21.9 W .
(Checked using LNAP 10/2/04)
Section 4-5 Mesh Current Analysis with Independent Voltage Sources
P4.5-1
2 i1 + 9 (i1 − i 3 ) + 3(i1 − i 2 ) = 0
15 − 3 (i1 − i 2 ) + 6 (i 2 − i 3 ) = 0
−6 (i 2 − i 3 ) − 9 (i1 − i 3 ) − 21 = 0
or
14 i1 − 3 i 2 − 9 i 3 = 0
− 3 i 1 + 9 i 2 − 6 i 3 = −15
−9 i1 − 6 i 2 + 15 i 3 = 21
so
i1 = 3 A, i2 = 2 A and i3 = 4 A.
(checked using LNAP 8/14/02)
P4.5-2
Top mesh:
4 (2 − 3) + R (2) + 10 (2 − 4) = 0
so R = 12 Ω.
Bottom, right mesh:
8 (4 − 3) + 10 (4 − 2) + v 2 = 0
so v2 = −28 V.
Bottom left mesh
−v1 + 4 (3 − 2) + 8 (3 − 4) = 0
so v1 = −4 V.
(checked using LNAP 8/14/02)
P4.5-3
Ohm’s Law: i 2 =
−6
= −0.75 A
8
KVL for loop 1:
R i1 + 4 ( i1 − i 2 ) + 3 + 18 = 0
KVL for loop 2
+(−6) − 3 − 4 ( i1 − i 2 ) = 0
⇒ − 9 − 4 ( i1 − ( −0.75 ) ) = 0
⇒ i 1 = −3 A
R ( −3) + 4 ( −3 − ( −0.75) ) + 21 = 0 ⇒ R = 4 Ω
(checked using LNAP 8/14/02)
P4.5-4
KVL loop 1:
25 ia − 2 + 250 ia + 75 ia + 4 + 100 (ia − ib ) = 0
450 ia −100 ib = −2
KVL loop 2:
−100(ia − ib ) − 4 + 100 ib + 100 ib + 8 + 200 ib = 0
−100 ia + 500 ib = − 4
⇒ ia = − 6.5 mA , ib = − 9.3 mA
(checked using LNAP 8/14/02)
P4.5-5
Mesh Equations:
mesh 1 : 2i1 + 2 (i1 − i2 ) + 10 = 0
mesh 2 : 2(i2 − i1 ) + 4 (i2 − i3 ) = 0
mesh 3 : − 10 + 4 (i3 − i2 ) + 6 i3 = 0
Solving:
5
i = i2 ⇒ i = − = −0.294 A
17
(checked using LNAP 8/14/02)
P4.5-6
60 Ω & 300 Ω = 50 Ω
40 Ω + 60 Ω = 100 Ω
and
100 Ω + 30 Ω + ( 80 Ω & 560 Ω ) = 200 Ω
so the simplified circuit is
The mesh equations are
200 i1 + 50 ( i1 − i 2 ) − 12 = 0
100 i 2 + 8 − 50 ( i1 − i 2 ) = 0
or
⎡ 250 −50⎤ ⎡ i1 ⎤ ⎡12 ⎤
⎢ −50 150 ⎥ ⎢i ⎥ = ⎢ −8⎥
⎣
⎦⎣ 2⎦ ⎣ ⎦
⇒
⎡ i1 ⎤ ⎡ 0.04 ⎤
⎢i ⎥ = ⎢
⎥
⎣ 2 ⎦ ⎣ −0.04 ⎦
The power supplied by the 12 V source is 12 i1 = 12 ( 0.04 ) = 0.48 W . The power supplied by the
8 V source is −8i 2 = −8 ( −0.04 ) = 0.32 W . The power absorbed by the 30 Ω resistor is
i12 ( 30 ) = ( 0.04 ) ( 30 ) = 0.048 W .
2
(checked: LNAP 5/31/04)
Section 4-6 Mesh Current Analysis with Voltage and Current Sources
P4.6-1
1
A
2
mesh 2: 75 i2 + 10 + 25 i2 = 0
mesh 1: i1 =
⇒ i2 = − 0.1 A
ib = i1 − i2 = 0.6 A
(checked using LNAP 8/14/02)
P4.6-2
mesh a: ia = − 0.25 A
mesh b: ib = − 0.4 A
vc = 100(ia − ib ) = 100(0.15) =15 V
(checked using LNAP 8/14/02)
P4.6-3
Express the current source current as a function of the mesh currents:
i1 − i2 = − 0.5 ⇒ i1 = i2 − 0.5
Apply KVL to the supermesh:
30 i1 + 20 i2 + 10 = 0 ⇒ 30 (i2 − 0.5) + 20i2 = − 10
50 i2 − 15 = − 10 ⇒ i2 =
5
= .1 A
50
i1 =−.4 A and v2 = 20 i2 = 2 V
(checked using LNAP 8/14/02)
P4.6-4
Express the current source current in terms
of the mesh currents:
ib = ia − 0.02
Apply KVL to the supermesh:
250 ia + 100 (ia − 0.02) + 9 = 0
∴ ia = − .02 A = − 20 mA
vc = 100(ia − 0.02) = −4 V
(checked using LNAP 8/14/02)
P4.6-5
Express the current source current in terms of the mesh currents:
i 3 − i 1 = 2 ⇒ i1 = i 3 − 2
Supermesh: 6 i1 + 3 i 3 − 5 ( i 2 − i 3 ) − 8 = 0 ⇒ 6 i1 − 5 i 2 + 8 i 3 = 8
Lower, left mesh: −12 + 8 + 5 ( i 2 − i 3 ) = 0 ⇒ 5 i 2 = 4 + 5 i 3
Eliminating i1 and i2 from the supermesh equation:
6 ( i 3 − 2 ) − ( 4 + 5 i 3 ) + 8 i 3 = 8 ⇒ 9 i 3 = 24
⎛ 24 ⎞
The voltage measured by the meter is: 3 i 3 = 3 ⎜ ⎟ = 8 V
⎝ 9 ⎠
(checked using LNAP 8/14/02)
P4.6-6
Mesh equation for right mesh:
4 ( i − 2 ) + 2 i + 6 ( i + 3) = 0 ⇒ 12 i − 8 + 18 = 0 ⇒ i = −
10
5
A=− A
12
6
(checked using LNAP 8/14/02)
P4.6-7
i2 = −3 A
i1 − i2 = 5 ⇒ i1 − ( −3) = 5
⇒ i1 = 2 A
2 ( i3 − i1 ) + 4 i3 + R ( i3 − i2 ) = 0
⇒ 2 ( −1 − 2 ) + 4 ( −1) + R ( −1 − ( −3) ) = 0
⇒ R=5 Ω
(checked using LNAP 8/14/02)
P4.6-8
Use units of V, mA and kΩ.
Express the currents to the supermesh to get
i1 − i 3 = 2
Apply KVL to the supermesh to get
4 ( i1 − i 3 ) + (1) i 3 − 3 + (1) ( i1 − i 2 ) = 0
⇒
i1 − 5 i 2 + 5 i 3 = 3
Apply KVL to mesh 2 to get
2i 2 + 4 ( i 2 − i 3 ) + (1) ( i 2 − i1 ) = 0
( −1) i1 + 7i 2 − 4i 3 = 0
⇒
Solving, e.g. using MATLAB, gives
⎡ 1 0 −1⎤ ⎡ i1 ⎤ ⎡ 2 ⎤
⎢ 1 −5 5 ⎥ ⎢i ⎥ = ⎢ 3 ⎥
⎢
⎥⎢ 2⎥ ⎢ ⎥
⎣⎢ −1 7 −4 ⎥⎦ ⎢⎣ i 3 ⎥⎦ ⎣⎢ 0 ⎦⎥
⎡ i 1 ⎤ ⎡ 3⎤
⎢ ⎥ ⎢ ⎥
⎢ i 2 ⎥ = ⎢1 ⎥
⎢ i 3 ⎥ ⎢⎣1⎥⎦
⎣ ⎦
⇒
(checked: LNAP 6/21/04)
P4.6-9
Label the mesh currents:
Express the current source currents in terms of the mesh currents
i x = i1 and i y = i 3 − i 2
Apply KVL to the supermesh corresponding to the current source with current iy to get
4 ( i 3 − i1 ) + v z + 12 ( i 3 − i 4 ) + 2i 2 = 0
⇒
4i1 − 2i 2 − 16i 3 + 12i 4 = v z
Substituting i1 = i x and i 2 = i 3 − i y gives
4i x − 2 ( i 3 − i y ) − 16i 3 + 12i 4 = v z
⇒
− 18i 3 + 12i 4 = v z − 2i y − 4i x
Apply KVL to mesh 4 to get
6i 4 + 12 ( i 4 − i 3 ) = 0
So
⇒
3
i3 = i 4
2
⎡
⎤
⎛3⎞
⎢ −18 ⎜ 2 ⎟ + 12 ⎥ i 4 = v z + 2i y − 4i x
⎝ ⎠
⎣
⎦
⇒
i4 =
v z + 2i y − 4i x
−15
Express the output in terms of the mesh currents to get
1
1
1
2
io = i3 − i4 = i 4 = − v z + i y + i x
2
30
15
15
So
a=
2
1
1
, b=
and c = −
15
15
30
(checked: LNAP 6/14/04)
P4.6-10
(a)
50 ( i 3 − i 2 ) + R 3i 3 + 32 = 0 ⇒ 50 ( 0.0770 − 0.7787 ) + R 3 ( 0.0770 ) + 32 = 0
⇒ R 3 = 40 Ω
i1 R1 + 20i 2 + 50 ( i 2 − i 3 ) − 24 = 0 ⇒
R1 ( −2.2213) + 20 ( 0.7787 ) + 50 ( 0.7787 − 0.0770 ) = 24
⇒
(b)
R1 = 12 Ω
I s = i 2 − i1 = 0.7787 − ( −2.2213) = 3 A
The power supplied by the current source is
p = I s ( 24 − R1 i1 ) = 3 ( 24 − 12 ( −2.2213) ) = 152 W
(checked: LNAP 6/19/04)
P4.6-11
3
3
= i1 − i 2 ⇒ i1 = + i 2 .
4
4
3
⎛
⎞
Apply KVL to the supermesh: −9 + 4i1 + 3 i 2 + 2 i 2 = 0 ⇒ 4 ⎜ + i 2 ⎟ + 5 i 2 = 9 ⇒ 9 i 2 = 6
⎝4
⎠
2
4
so i 2 = A and the voltmeter reading is 2 i 2 = V
3
3
Express the current source current in terms of the mesh currents:
P4.6-12
Express the current source current in terms of the mesh currents: 3 = i1 − i 2
⇒ i1 = 3 + i 2 .
Apply KVL to the supermesh: −15 + 6 i1 + 3 i 2 = 0 ⇒ 6 ( 3 + i 2 ) + 3 i 2 = 15 ⇒ 9 i 2 = −3
Finally, i 2 = −
1
A is the current measured by the ammeter.
3
Section 4-7 Mesh Current Analysis with Dependent Sources
P4.7-1
Express the controlling voltage of the
dependent source as a function of the
mesh current
v2 = 50 i1
Apply KVL to the right mesh:
−100 (0.04(50i1 ) − i1 ) + 50i1 + 10 = 0 ⇒ i1 = 0.2 A
v2 = 50 i1 = 10 V
(checked using LNAP 8/14/02)
P4.7-2
ib = 4ib − ia ⇒ ib =
1
ia
3
⎛1 ⎞
−100 ⎜ ia ⎟ + 200ia + 8 = 0
⎝3 ⎠
⇒ ia = − 0.048 A
(checked using LNAP 8/14/02)
P4.7-3
Express the controlling current of
the dependent source as a function
of the mesh current:
ib = .06 − ia
Apply KVL to the right mesh:
−100 (0.06 − i a ) + 50 (0.06 − i a ) + 250 i a = 0 ⇒
ia = 10 mA
Finally:
vo = 50 i b = 50 (0.06 − 0.01) = 2.5 V
(checked using LNAP 8/14/02)
P4.7-4
Express the controlling voltage of
the dependent source as a function
of the mesh current:
vb = 100 (.006 − ia )
Apply KVL to the right mesh:
−100 (.006 − ia ) + 3 [100(.006 − ia ) ] + 250 ia = 0 ⇒ ia = −24 mA
(checked using LNAP 8/14/02)
P4.7-5
apply KVL to left mesh : − 3 + 10 × 103 i1 + 20 × 103 ( i1 − i2 ) = 0 ⇒ 30 × 103 i1 − 20 × 103 i2 = 3
apply KVL to right mesh : 5 × 103 i1 + 100 × 103 i2 + 20 × 103 ( i2 − i1 ) = 0 ⇒ i1 = 8i2
Solving (1) & ( 2 ) simultaneously
Power delivered to cathode =
⇒ i1 =
6
3
mA, i2 =
mA
55
220
( 5 i1 ) ( i2 ) + 100 ( i2 )2
( 55)( 3 220) + 100 ( 3 220)
= 5 6
∴ Energy in 24 hr. =
2
= 0.026 mW
( 2.6 ×10−5 W ) ( 24 hr ) (3600 s hr ) = 2.25 J
( 2)
(1)
P4.7-6
(a)
(b)
vo = − g R L v and v =
∴
vo
= −g
vi
R2
R1 + R 2
vi ⇒
(5 ×103 )(103 ) = −170
1.1×103
RL R2
vo
= −g
vi
R1 + R 2
⇒ g = 0.0374 S
P4.7-7
Express va and ib, the controlling voltage and current of the dependent sources, in terms of the
mesh currents
v a = 5 ( i 2 − i 3 ) and i b = −i 2
Next express 20 ib and 3 va, the controlled voltages of the dependent sources, in terms of the
mesh currents
20 i b = −20 i 2 and 3 v a = 15 ( i 2 − i 3 )
Apply KVL to the meshes
−15 ( i 2 − i 3 ) + ( −20 i 2 ) + 10 i1 = 0
− ( −20 i 2 ) + 5 ( i 2 − i 3 ) + 20 i 2 = 0
10 − 5 ( i 2 − i 3 ) + 15 ( i 2 − i 3 ) = 0
These equations can be written in matrix form
⎡10 −35 15 ⎤ ⎡ i1 ⎤ ⎡ 0 ⎤
⎢ 0 45 −5 ⎥ ⎢i ⎥ = ⎢ 0 ⎥
⎥
⎢
⎥⎢ 2⎥ ⎢
⎢⎣ 0 10 −10 ⎥⎦ ⎢⎣ i 3 ⎥⎦ ⎢⎣ −10 ⎥⎦
Solving, e.g. using MATLAB, gives
i1 = −1.25 A, i 2 = +0.125 A, and i 3 = +1.125 A
(checked: MATLAB & LNAP 5/19/04)
P4.7-8
Label the mesh currents:
Express ia, the controlling current of the CCCS,
in terms of the mesh currents
i a = i 3 − i1
Express 2 ia, the controlled current of the
CCCS, in terms of the mesh currents:
i 1 − i 2 = 2 i a = 2 ( i 3 − i 1 ) ⇒ 3 i1 − i 2 − 2 i 3 = 0
Apply KVL to the supermesh corresponding to the CCCS:
80 ( i1 − i 3 ) + 40 ( i 2 − i 3 ) + 60 i 2 + 20 i1 = 0
⇒
100i1 + 100i 2 − 120i 3 = 0
Apply KVL to mesh 3
10 + 40 ( i 3 − i 2 ) + 80 ( i 3 − i1 ) = 0
⇒
-80 i1 − 40 i 2 + 120 i 3 = −10
These three equations can be written in matrix form
−1
−2 ⎤ ⎡ i 1 ⎤ ⎡ 0 ⎤
⎡ 3
⎢100 100 −120 ⎥ ⎢i ⎥ = ⎢ 0 ⎥
⎥
⎢
⎥⎢ 2⎥ ⎢
⎢⎣ −80 −40 120 ⎦⎥ ⎣⎢ i 3 ⎦⎥ ⎢⎣ −10 ⎥⎦
Solving, e.g. using MATLAB, gives
i1 = −0.2 A, i 2 = −0.1 A and i 3 = −0.25 A
Apply KVL to mesh 2 to get
v b + 40 ( i 2 − i 3 ) + 60 i 2 = 0 ⇒ v b = −40 ( −0.1 − ( −0.25 ) ) − 60 ( −0.1) = 0 V
So the power supplied by the dependent source is p = v b ( 2i a ) = 0 W .
(checked: LNAP 6/7/04)
P4.7-9
Notice that i b and 0.5 mA are the mesh currents.
Apply KCL at the top node of the dependent
source to get
1
i b + 0.5 × 10−3 = 4 i b ⇒ i b = mA
6
Apply KVL to the supermesh corresponding to
the dependent source to get
(
)
−5000 i b + (10000 + R ) 0.5 ×10−3 − 25 = 0
(
)
⎛1
⎞
−5000 ⎜ × 10−3 ⎟ + (10000 + R ) 0.5 × 10−3 = 25
⎝6
⎠
125
6
= 41.67 kΩ
R=
0.5 × 10−3
(checked: LNAP 6/21/04)
P4.7-10
The controlling and controlled currents of the CCCS, i b and 40 i b, are the mesh currents. Apply
KVL to the left mesh to get
1000 i b + 2000 i b + 300 ( i b + 40i b ) − v s = 0
The output is given by
⇒
15300i b = v s
v o = −3000 ( 40 i b ) = −120000 i b
(a) The gain is
vo
vs
=−
120000
= −7.84 V/V
15300
(b) The input resistance is
vs
ib
= 15300 Ω
(checked: LNAP 5/24/04)
P4.7-11
Express the current source current in terms of the mesh currents:
i 4 − i3 = 1
Express the controlling current of the dependent source in terms of the mesh currents:
i x = −i 3
Apply KVL to the supermesh corresponding to the current source to get
3 ( i 3 − i1 ) + 8 + 8i 4 + 6i 3 = 0
⇒
− 3i1 + 9i 3 + 8i 4 = −8
Apply KVL to mesh 1 to get
16 + 4 ( −i 3 ) + 3 ( i1 − i 3 ) + 2i1 = 0
⇒
5i1 − 7i 3 = −16
Apply KVL to mesh 2 to get
2i 2 − 8 − 4 ( −i 3 ) = 0
⇒
2i 2 + 4i 3 = 8
Solving, e.g using MATLAB, gives
⎡0
⎢ −3
⎢
⎢5
⎢
⎣0
0 −1 1 ⎤ ⎡ i 1 ⎤ ⎡ 1 ⎤
⎢ ⎥
0 9 8 ⎥⎥ ⎢i 2 ⎥ ⎢⎢ −8 ⎥⎥
=
0 −7 0 ⎥ ⎢i 3 ⎥ ⎢ −16 ⎥
⎥⎢ ⎥ ⎢
⎥
2 4 0 ⎦ ⎢⎣i 4 ⎥⎦ ⎣ 8 ⎦
⇒
⎡ i 1 ⎤ ⎡ −6 ⎤
⎢i ⎥ ⎢ ⎥
⎢ 2⎥ = ⎢ 8 ⎥
⎢ i 3 ⎥ ⎢ −2 ⎥
⎢ ⎥ ⎢ ⎥
⎢⎣i 4 ⎥⎦ ⎣ −1⎦
(checked: LNAP 6/13/04)
P4.7-12
Label the mesh currents.
Express ix in terms of the mesh currents:
i x = i1
Express 4ix in terms of the mesh currents:
4 i x = i3
Express the current source current in terms of the mesh currents to get:
0.5 = i1 − i 2
⇒
i 2 = i x − 0.5
Apply KVL to supermesh corresponding to the current source to get
5i1 + 20 ( i1 − i 3 ) + 10 ( i 2 − i 3 ) + 25i 2 = 0
Substituting gives
5i x + 20 ( −3i x ) + 10 ( i x − 0.5 − 4i x ) + 25 ( i x − 0.5 ) = 0
⇒
ix = −
35
= −0.29167
120
So the mesh currents are
i1 = i x = −0.29167 A
i 2 = i x − 0.5 = −0.79167 A
i 3 = 4i x = −1.1667 A
(checked: LNAP 6/21/04)
P4.7-13
Express the controlling voltage and current of
the dependent sources in terms of the mesh
currents:
v a = R 3 ( i1 − i 2 ) and i b = i 3 − i 2
Express the current source currents in terms of
the mesh currents:
i 2 = − I s and i1 − i 3 = B i b = B ( i 3 − i 2 )
Consequently
i1 − ( B + 1) i 3 = B I s
Apply KVL to the supermesh corresponding to the dependent current source
R1 i 3 + A R 3 ( i 1 − i 2 ) + R 2 ( i 3 − i 2 ) + R 3 ( i 1 − i 2 ) − V s = 0
or
( A + 1) R 3 i1 − ( R 2 + ( A + 1) R 3 ) i 2 + ( R1 + R 2 ) i 3 = V s
Organizing these equations into matrix form:
⎡
0
⎢
1
⎢
⎢
⎣⎢( A + 1) R 3
1
0
− ( R 2 + ( A + 1) R 3 )
0 ⎤ ⎡ i1 ⎤ ⎡ − I s ⎤
⎥⎢ ⎥ ⎢
⎥
− ( B + 1) ⎥ ⎢i 2 ⎥ = ⎢ B I s ⎥
⎥
R1 + R 2 ⎦⎥ ⎢⎣ i 3 ⎥⎦ ⎢⎣ V s ⎥⎦
With the given values:
⎡ i1 ⎤ ⎡ −0.8276 ⎤
1
0 ⎤ ⎡ i1 ⎤ ⎡ −2 ⎤
⎡0
⎢ ⎥ ⎢ ⎥
⎢ ⎥
⎢1
⎥
0 −4 ⎥ ⎢i 2 ⎥ = ⎢ 6 ⎥ ⇒ ⎢i 2 ⎥ = ⎢⎢ −2 ⎥⎥ A
⎢
⎢ i 3 ⎥ ⎢⎣ −1.7069 ⎥⎦
⎢⎣ 60 −80 50 ⎥⎦ ⎢⎣ i 3 ⎥⎦ ⎢⎣ 25 ⎥⎦
⎣ ⎦
(Checked using LNAP 9/29/04)
P4.7-14
Express the controlling voltage and current of the dependent sources in terms of the mesh
currents:
v a = 20 ( i1 − i 2 ) = 20 ( −1.375 − ( −2.5 ) ) = 22.5
and
i b = i 3 − i 2 = −3.25 − ( −2.5) = −0.75 A
Express the current source currents in terms of the mesh currents:
i 2 = −2.5 A
and
i 3 − i1 = B i b
⇒ − 1.375 − ( −2.5) = B ( −0.75) ⇒ B = 2.5 A/A
Apply KVL to the supermesh corresponding to the dependent current source
0 = 20 i 3 + Av a + 50 i b + v a − 10 = 20 ( −3.25) + A ( 22.5) + 50 ( −0.75) + 22.5 − 10 ⇒
A = 4 V/V
(Checked using LNAP 9/29/04)
P4.7-15
Label the node voltages as shown. The controlling
va
.
currents of the CCCS is expressed as i =
28
The node equations are
va va − vb va
12 =
+
+
28
4
14
and
va − vb va vb
+ =
4
14 8
Solving the node equations gives v a = 84 V and v b = 72 V . Then i =
va
28
=
84
=3A .
28
(checked using LNAP 6/16/05)
Section 4.8 The Node Voltage Method and Mesh Current Method Compared
*P4.8-1
Analysis of this circuit requires 1
node equations or 3 mesh equations,
so we will use node equations.
Apply KCL at the top node to R3 to
get
v a − v1
R1
+
va − v2
R2
+
va
R3
+
v a − Av a
R4
=0
Solving gives
1
1
⎡
⎤
⎡
⎤
⎢
⎥
⎢
⎥
R1
R2
⎥ v1 + ⎢
⎥ v2
va = ⎢
⎢ 1 + 1 + 1 + 1− A ⎥
⎢ 1 + 1 + 1 + 1− A ⎥
⎢R R
⎢R R
R3
R 4 ⎥⎦
R3
R 4 ⎥⎦
2
2
⎣ 1
⎣ 1
Suppose we choose A so that
1
1
1
1 1− A
=
+
+
+
⇒
R R1 R 2 R 3
R4
A = 1+
R4
R1
+
where R is a resistance to be determined later. Then
va =
R
R
v1 +
v2
R1
R2
and
v o = Av a =
AR
AR
v1 +
v2
R1
R2
we require
AR
AR
= 2 and
= 0.5
R1
R2
so
R 2 = 4 R1
To simplify matters, choose R3 = R 4 = R 2 . Then
A = 1+ 4 +1+1−
Now
4 R1
R
= 7−
4 R1
R
R4
R2
+
R4
R3
−
R4
R
4 R1 ⎞
⎛
⎜7 −
⎟ R 7R − 4R
R ⎠
AR ⎝
1
=
=
2=
R1
R1
R1
⇒ 2 R 1 = 7 R − 4 R1 ⇒
R=
6
R1
7
Then
A=7−
4 R1 7
=
6
R1 3
7
Any value of R1 will do. For example, pick R1 = 10 Ω. Then R 2 = R3 = R 4 = 40 Ω.
(checked: LNAP 6/22/04)
P4.8-2
(a)
Apply KVL to meshes 1 and 2:
32i1 − v s + 96 ( i1 − i s ) = 0
v s + 30i 2 + 120 ( i 2 − i s ) = 0
150i 2 = +120i s − v s
vs
4
i 2 = is −
5
150
1
v o = 30i 2 = 24i s − v s
5
So a = 24 and b = -.02.
(b)
Apply KCL to the supernode corresponding to
the voltage source to get
va − (vs + vo )
96
So
is =
+
vs + vo
120
va − vo
32
+
vo
30
=
=
vs + vo
120
vs
120
+
+
vo
30
vo
24
Then
1
v o = 24i s − v s
5
So a = 24 and b = -0.2.
(checked: LNAP 5/24/04)
P4.8-3
(a) Label the reference node and node voltages.
v b = 120 V
due to the voltage source.
Apply KCL at the node between the resistors to get
vb − va
50
=
va
10
⇒
v a = 20 V
Then
i a = 0.2 ( 20 ) = 4 A
and the power supplied by the dependent source is
p = v b i a = (120 )( 4 ) = 480 W
(b) Label the mesh currents. Express the controlling
voltage of the dependent source in terms of the mesh
current to get
v a = 10 i 2 − i1
(
)
Express the controlled current of the dependent
source in terms of the mesh currents to get
−i1 = i a = 0.2 ⎡⎣10 ( i 2 − i1 ) ⎤⎦ = 2i 2 − 2i1
⇒
i1 = 2i 2
Apply KVL to the bottom mesh to get
50 ( i 2 − i1 ) + 10 ( i 2 − i1 ) − 120 = 0
⇒
i 2 − i1 = 2
i 2 − 2i 2 = 2
⇒
i1 = −4 A
So
Then
⇒
i 2 = −2 A
v a = 10 ( −2 − ( −4 ) ) = 20 V and i a = 0.2 ( 20 ) = 4 A
The power supplied by the dependent source is
p = 120 ( i a ) = 120 ( 4 ) 480 W
(checked: LNAP 6/21/04)
Section 4.10 How Can We Check … ?
P4.10-1
Apply KCL at node b:
vb − va
v −v
1
−
+ b c = 0
4
2
5
−4.8 − 5.2
1 − 4.8 − 3.0
− +
≠0
4
2
5
The given voltages do not satisfy the KCL
equation at node b. They are not correct.
P4.10-2
Apply KCL at node a:
v
⎛v −v ⎞
−⎜ b a ⎟ − 2 + a = 0
2
⎝ 4 ⎠
4
⎛ 20 − 4 ⎞
−⎜
= −4≠ 0
⎟−2+
2
⎝ 4 ⎠
The given voltages do not satisfy the KCL
equation at node a. They are not correct.
P4.10-3
Writing a node equation:
⎛ 12 − 7.5 ⎞ 7.5 7.5 − 6
−⎜
+
=0
⎟+
R
R
R
⎝
⎠
1
3
2
so
4.5 7.5 1.5
+
+
=0
R1 R3 R2
There are only three cases to consider. Suppose
R1 = 5 kΩ and R 2 = R 3 = 10 kΩ. Then
−
−
4.5 7.5 1.5 −0.9 + 0.75 + 0.15
+
+
=
= 0
R1 R3 R2
1000
This choice of resistance values corresponds to branch
currents that satisfy KCL. Therefore, it is indeed possible
that two of the resistances are 10 kΩ and the other
resistance is 5 kΩ. The 5 kΩ is R1.
P4.10-4
KCL at node 1:
0=
v1 − v 2
20
+
v1
5
+1 ⇒
−8 − ( −20 ) −8
+
+1 = 0
20
5
KCL at node 2:
v1 − v 2
20
= 2+
v 2 − v3
KCL at node 3:
10
−8 − ( −20 )
−20 − ( −6 )
= 2+
20
10
12 6
⇒
=
20 10
⇒
v2 − v3
10
+1 =
v3
15
⇒
−20 − ( −6 )
−6
−4 −6
+1 =
⇒
=
10
15
10 15
KCL is satisfied at all of the nodes so the computer analysis is correct.
P4.10-5
Top mesh: 10 (2 − 4) + 12(2) + 4 (2 − 3) = 0
Bottom right mesh 8 (3 − 4) + 4 (3 − 2) + 4 = 0
Bottom, left mesh: 28 + 10 (4 − 2) + 8 (4 − 3) ≠ 0 (Perhaps the polarity of the 28 V source was
entered incorrectly.)
KVL is not satified for the bottom, left mesh so the computer analysis is not correct.