The Mesh-Current Method
We can use the mesh-current method in frequency domain circuits as using the same methods in
section 4.5-4.7.
Example:
V3
V1
1Ω
j2Ω
1Ω
12Ω
V2
150∠0° I1
j3Ω
+
Ix
-
I2
39Ix
-j16Ω
For mesh 1:
(1 + 12 + j (2 − 16) )I1 − (12 − j16)I 2 = 150
(13 − j (14) )I1 − (12 − j16)I 2 = 150
(1)
For mesh 2:
(1 + 12 + j (3 − 16) )I 2 − (12 − j16)I1 + 39I x
− (12 − j16)I1 + (13 − j13)I 2 + 39I x = 0
=0
and I x = I1 − I 2
− (12 − j16 )I1 + (13 − j13)I 2 + 39(I1 − I 2 ) = 0
(27 − j16)I1 − (26 + j13)I 2 = 0
(2)
Using Equation (1) and (2), to find I1 and I2
I1 = −26 − j52 A
I 2 = −24 − j58 A
I x = −2 + j 6 A
The voltages at nodes
V1 = (1 + j 2) I1 = 78 − j104 V
V2 = (12 − j16) I x = 72 + j104V
V3 = (1 + j 3) I 2 = 150 − j130 V
39I x = −78 + j 234V
4-1
9.10 The Transformer
Two Topics
• The Sinusoidal steady-state behavior of the linear transformer
• Ideal Transformer
The Analysis of a Linear Transformer Circuit
Zs
a
R1
R2
c
jwM
I1
jωL1
V
jωL2
b
I1
ZL
d
R1 : The resistance of the primary winding
R2 : The resistance of the secondary winding
L1 : The self-inductance of the primary winding
L2 : The self-inductance of the secondary winding
M: The mutual inductance
Vs : The sinusoidal voltage source
Zs : The internal impedance of the source Vs
ZL : The internal impedance of the load connected to the secondary winding
Ii : The primary current
I2: The secondary current
Lets write two mesh-current equation
Vs = ( Z s + R1 + jωL1 )I1 − jωMI 2
0 = − jωMI1 + ( R2 + Z L + jωL2 )I 2
Let say that
Z 11 = Z s + R1 + jωL1
Z 22 = R2 + Z L + jωL2
Therefore we can find
I1 =
Z 22
Vs
Z 11 Z 22 + ω 2 M 2
I2 =
jωM
jω M
Vs =
I1
2
2
Z 22
Z 11 Z 22 + ω M
The impedance between node a and b
4-2
Z ab
Vs
Z 11 Z 22 + ω 2 M 2
ω 2M 2
=
- Zs =
− Z s = Z 11 +
− Zs
I1
Z 22
Z 22
or
Z ab = R1 + jωL1 +
ω 2M 2
R 2 + Z L + jω L 2
The third term in Zab is called reflected impedance
Zr =
ω 2M 2
R 2 + Z L + jω L 2
We can see that if M becomes zero then Zr becomes zero.
The Ideal Transformer
Determining the Voltage and Current Ratios
The primary coil is wound so that it has N1 turns. The secondary coil has N2 turns.
The turn ratio is
N=
N2
N1
The relationship between primary and secondary current in ideal transformer is is
V2 =
N
N
2
V1
and
I2 =
N1
I1
N2
+
V1
-
I1
N1 N
2
I2
ideal
V1 V2
=
, N 1 I1 = − N 2 I 2
N1 N 2
+
+
V2
V1
-
+
N1 N
2
I1
I2 V2
ideal
-
-
V1
V
= − 2 , N 1 I1 = N 2 I 2
N1
N2
4-3
+
V1
I1
N1 N
2
+
I2 V2
ideal
-
V1 V2
=
, N 1 I1 = N 2 I 2
N1 N 2
-
+
N1 N
2
V1
I1
V
I2 2
ideal
-
+
-
V1
V
= − 2 , N 1 I1 = − N 2 I 2
N1
N2
4-4