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
Apply KVL to mesh 2 to get
2i 2 − 8 − 4 ( −i 3 ) = 0
⇒
⇒
5i1 − 7i 3 = −16
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
With the given values:
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 ⎥⎦
⎡ 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)
P4.7-16
Expressing the dependent source currents in terms of the mesh currents we get:
i1 = 4 i a = 4 ( i 2 + 1) ⇒ 4 = i1 − 4 i 2
Apply KVL to mesh 2 to get
2 i 2 + 2 ( i 2 + 1) − 2 ( i1 − i 2 ) = 0 ⇒ − 2 = −2 i1 + 6 i 2
Solving these equations using MATLAB we
get
i1 = −8 A and i2 = −3 A
P4.7-17
Apply KVL to mesh 1 to get
2 i1 + 4 i a + 2 ( i1 − i 2 ) − 12 = 0 ⇒ 2 i1 + 4 ( i 2 + 1) + 2 ( i1 − i 2 ) − 12 = 0 ⇒ 8 = 4 i1 + 2 i 2
Apply KVL to mesh 2 to get
2 i 2 + 2 ( i 2 + 1) − 2 ( i1 − i 2 ) = 0 ⇒ − 2 = −2 i1 + 6 i 2
Solving these equations using MATLAB we
get
i1 = 1.8571 A and i2 = 0.2857 A