I
out
14.4
19.2i
−
A
=
I
out
24 A
=
arg I
out
()
53.13
−
deg
=
Complex power to the load
S
out
V
out
I
out
⋅
:=
S
out
17.28
23.04i
+
kV A
⋅
=
S
out
28.8 kV A
⋅
=
arg S
out
53.13deg
=
Use the gain to find the input current.
I
in
NI
out
⋅
:=
I
in
144
192i
−
A
=
I
in
240A
=
arg I
in
53.13
−
deg
=
Complex power in
S
in
V
in
I
in
⋅
:=
S
in
17.28
23.04i
+
kV A
⋅
=
S
in
28.8 kV A
⋅
=
arg S
in
53.13deg
=
Problem 1.
Problem 5.1
An ideal single phase transformer has a voltage gain of 10. The secondary is
terminated in a load impedance of ZL=30+j40ohms.
a.
find the primary driving point impedance, i.e., the input impedance of the transformerload pair as
seen from the source terminals.
b. If the primary voltage is 120V, find the primary current, secondary current, complex power into
the load, and complex power into the transformer primary.
j1
−
:=
Restate the given
gain
10
:=
Z
L
30
j 40
⋅
+
(
) ohm
⋅
:=
Because the transformer is ideal, the driving point impedance is the load impedance reflected to
the primary.
Reflecting the load impedance across the transformer means multiplying it by the
square of the turns ratio of the transformer.
The wording of the problem infers that the turns ratio
is the reciprocal of the gain.
N
gain
:=
N1
0
=
Z
Lp
Z
L
NN
⋅
:=
Z
Lp
0.3
0.4i
+Ω
=
This is the driving point impedance.
Restate the given for part b
V
in
120 V
⋅
:=
V
in
120V
=
Find the secondary voltage first.
It is the primary voltage multiplied by the turns ratio.
V
out
V
in
N
⋅
:=
V
out
1.2kV
=
Use Ohm's law to find the secondary current.
I
out
V
out
Z
L
:=
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View Full DocumentLooking back into the circuit from the output terminals, the impedance
that appears from the output terminals is reflected parallel combination
of the leakage and magnetizing reactances.
X
sc
V
2sc
I
2sc
=
n
2
X
1
⋅
X
m
⋅
X
l
X
m
+
=
The input voltage divides across the leakage and magnetizing reactances;
the portion reflected from the magnetizing reactance is the output voltage.
V
1oc
X
l
X
m
+
V
2oc
nX
m
⋅
=
The sum of the leakage and magnetizing reactance is the ratio of the input
voltage to input current under open circuit conditions.
X
l
X
m
+
V
1oc
I
1oc
=
Using the equivalent circuit model of Figure 5.4,
These give reasonable approximations of the items that we seek.
A more detailed solution follows:
X
l
282.353
Ω
=
X
l
V
1oc
I
1oc
:=
The magnetizing reactance is approximately the ratio between open circuit voltage and open circuit
current.
This measurement is taken on the low voltage side, according to the problem statement.
Z
2sc
29.04
Ω
=
Z
2sc
V
2sc
I
2sc
:=
The leakage reactance is approximately the ratio between short circuit voltage and short circuit
current, assuming that the magnetizing reactance is large enough to be insignificant in the
calculations.
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 Spring '11
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 Electrical impedance, Impedance matching, Impedance bridging, deg, LPU

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