Transcript Document

Drill Exercise
A linear transformer couples a load consisting of a 360 Ω resistor in series with a 0.25 H
inductor to a sinusoidal voltage source, as shown. The voltage source has an internal
impedance of 184+j0 Ω and a maximum voltage of 245.20 V, and it is operating at 800
rad/s. The transformer parameters are R1 = 100Ω, L1 = 0.5 H, R2 = 40Ω, L2 = 0.125 H,
and k = 0.4. Calculate :
a). The reflected impedance,
b). The primary current, c). The secondary current, and
d). The average power delivered to the primary terminals of the transformer.
Ideal Transformer
Example
4:1
ideal
60Ω
500 A
rm s
40Ω
20Ω
a). Find the average power delivered by the sinusoidal current source in the circuit
shown.
b). Find the average power delivered to the 20 Ω resistor.
60Ω
+
4:1
V1
-
+
30000V
rm s
V2
ideal
40Ω
i1
20Ω
Solution
a).
+
i2
300  60I1  V1  20I1  I 2 
0  20I 2  I1   V2  40I 2
V2  14 V1
I2  4I1
The solutions for V1, V2, I1 and I2 are
V1  260V rm s
V2  65 V rm s
I1  0.25 A rm s
I 2  1.0 A rm s
The voltage across the 5 A current source is
V5 A  V1  20I1  I 2 
 260 200.25   1  285V rms
The average power associated with the current source is
P  2855  1425W
b).
To find the average power delivered to the 20Ω resistor
I 20   I1  I 2  0.25   1  1.25 A rm s
P20   1.25 20  31.25 W
2
Drill Exercise
Find the average power delivered to the 4 kΩ resistor in circuit shown.
10Ω
1:2.5
1:4
4kΩ
10000V
rm s
ideal
ideal
Equivalent Circuits for Magnetically Coupled Coils
di1
di2
v1  L1
M
dt
dt
di1
di2
v2  M
 L2
dt
dt
Rangkaian Ekivalen model T
R1
a
+
v1
b
L1-M
i1
M
L2-M
c
i2
+
v2
d
R2
Rangkaian Ekivalen model 
R1
L1 L2  M 2
M
a
i1
+
v1
b
L1L2  M 2
L1  M
L1L2  M 2
L2  M
c
+
v2
-
d
R2
i2
Example
j100
500
200
100
j1200
+
I1
V1
j 3600
j1600
800
+
I2
V2
 j 2500
0
3000 V
-
a.
-
6H
1H
3H
For the polarity dots shown in this example, M carries a value of
+3 H in the T equivalent circuit.
L1  M  9  3  6 H
L2  M  4  3  1H
M  3H
At an operating frequency of 400 rad/s,
j2400
j400
j1200
V  300
V
V


0
700 j 2500 j1200 900 j 2100
V  136 j8  136,24  3,37 V
o
300 136 j8
I1 
 63,25  71,57o m Arm s
700 j 2500
136 j8
I2 
 59,6363,43o m Arm s
900 j 2100
b). When the polarity dot is moved to the lower terminal of the
secondary coil, M carries a value of -3 H in the T equivalent
circuit.
L1  M  9   3  12H
L2  M  4   3  7 H
j4800
j2800
M  3H
At an operating frequency of
400 rad/s,
-j1200
I1  63,25  71,57 mArms
o
I 2  59,63 116,57 mArms
o
V  300
V
V


0
700 j 4900  j1200 900 j 300
V  8  j 56  56.57  98.130 V rm s
300  8  j 56
I1 
 63.25  71.57 m A rm s
700 j 4900
 8  j 56
I2 
 59.63  116.570 m A rm s
900 j 300
Drill Exercise
A linear transformer couples a load consisting of a 360 Ω resistor in series with a
0.25 H inductor to a sinusoidal voltage source, as shown. The voltage source has
an internal impedance of 184+j0 Ω and a maximum voltage of 245.20 V, and it is
operating at 800 rad/s. The transformer parameters are R1 = 100Ω, L1 = 0.5 H,
R2 = 40Ω, L2 = 0.125 H, and k = 0.4. Calculate :
a). The reflected impedance,
b). The primary current, c). The secondary current, and
d). The average power delivered to the primary terminals of the transformer.
Use the T-equivalent circuit.
15Ω
j50Ω
+
j20Ω
256∠0o V
(rms)
80Ω
j32Ω
V
0
-
Calculate :
a). The rms magnitude of V0
b). The average power dissipated in the 80 Ω resistor.
+
-
20
Ω
j35Ω
40
j50Ω
15
Ω
Ω
j45
j80Ω
Ω
ZL
480∠0
o V
(rms)
The impedance ZL in the circuit shown is adjusted for maximum
average power transfer to ZL. The internal impedance of the
sinusoidal voltage source is 20+j35 Ω.
What is the maximum average power delivered to ZL?
15mH
20mH
25mH
88Ω
200Ω
vg
Find the average power delivered to the 200 Ω resistor in
the circuit shown if
vg= 424 cos 8000t V