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7/27/2019 ECE202 Ch13-Fall2014
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Chapter 13Magnetically Coupled Circuits
13.1 What is a transformer?13.2 Mutual Inductance
13.3 Energy in a Coupled Circuit13.4 Linear Transformers13.5 Ideal Transformers
13.6 Ideal Autotransformers13.8 Multisim Analysis13.9 Applications
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13.1 What is a transformer? (1) It is an electrical device designed on the basis of the
concept of magnetic coupling It uses magnetically coupled coils to transfer energy from
one circuit to another It is the key circuit elements for stepping up or stepping
down ac voltages or currents, impedance matching ,isolation , etc.
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Mutual Inductance When two inductors (or coils) are in close proximity
of each other, the magnetic flux caused by currentin one coil links with the other coil, thereby inducing
voltage in the latter.
3
13.2 Mutual Inductance (1)
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First consider a single inductor, a coil with N turns:
According to Faradays law, the voltage v inducedin the coil is proportional to the number of turns Nand the time rate of change of magnetic flux :
4
13.2 Mutual Inductance (2)
When current i flows through thecoil, a magnetic flux isproduced around it.
t d
d N v
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Voltage induced in the coil: But the flux is produced by current i
so that any change in is caused by achange in the current:
Recall the voltage-current relationshipfor an inductor.
The inductance L of the inductor (Self inductance) is given by:
Self-inductance L relates the voltageinduced in a coil by a time-varying
current in the same coil. 5
t d
id Lv
t d
id
id
d N v
13.2 Mutual Inductance (3)
t d
d
N v
id
d N L
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Now consider two coils with self-inductances L 1 and L 2 that are in close proximity of each other:
The total magnetic flux 1
emanating from coil 1has two components:
11 links only coil 1 12 links both coils
6
13.2 Mutual Inductance (4)
Coil 1 has N 1 turns
Coil 2 has N 2 turnsAssume coil 2 carries no current.
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Although the two coils are physically separated,they are magnetically coupled.
The voltage induced in each coil is proportional tothe flux in each coil.
7
13.2 Mutual Inductance (5)
ECE 202 Ch 13
t d
d N v
111
t d
d N v
1222
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The voltage equations can be rewritten as follows:
8
13.2 Mutual Inductance (6)
ECE 202 Ch 13
t d
d N v
1
11
t d
d N v
12
22
t d
id
id
d N v
1
1
111
t d
id Lv
111
t d
id
id
d N v
1
1
1222
t d
id M v
1212
L1 is the self-inductance of coil 1 M21 is the mutual inductance ofcoil 2 with respect to coil 1
v 2 is the open-circuit mutualvoltage (or induced voltage)across coil 2
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Suppose now we let current i 2 flow in coil 2 while
coil 1 carries no current:
The magnetic flux 2 emanating from coil 2comprises flux 22 that links only coil 2 and flux 21 that links both coils:
The resulting symmetry is true:
9
222
13.2 Mutual Inductance (7)
ECE 202 Ch 13
t d
id Lv
222
t d
id M v
2121
2
21112
id
d N M
L2 is the self inductance of coil 2M12 is the mutual inductance ofcoil 1 with respect to coil 2
v 1 is the open-circuit mutualvoltage (or induced voltage)across coil 1
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13.2 Mutual Inductance (8)
10
M12 = M21 = M - Mutual inductance between thecoils
As with self-inductance L, mutual inductance M ismeasured in Henrys
Mutual inductance only exists when inductors orcoils are in close proximity and the circuits aredriven by time-varying sources
Although mutual inductance M is always apositive quantity, the mutual voltage M di/dt maybe negative or positive, just like the self-inducedvoltage L di/dt
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13.2 Mutual Inductance (9)
11
Self-induced voltage polarity is determined by the
reference direction of the current and the referencepolarity of the voltage
The polarity of the mutual voltage is not as easy todetermine (depends on the winding direction of the coils)
We use the dot convention to determine
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13.2 Mutual Inductance (10)Dot Convention
12
If a current enters the dottedterminal of one coil, the referencepolarity of the mutual voltage inthe second coil is positive at thedotted terminal of the second coil.
Alternatively, if a current leaves thedotted terminal of one coil, thereference polarity of the mutualvoltage in the second coil isnegative at the dotted terminal ofthe second coil.
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13.2 Mutual Inductance (11)
13
Replace the dot with a dependent voltage source. I1 induces a voltage in the second coil (j MI1 term) while I 2 induces a
voltage in the first coil (j MI2 term). I1 enters L1 at the dotted end, then the polarity of the mutual voltage
(j MI1) in L2 is positive at the dotted terminal on top and negative at thebottom. Therefore, the controlled voltage source has a plus sign on top andminus sign at the bottom.
I2 leaves L2 at the dotted end, then the polarity of the mutual voltage in L 1 isnegative at the dotted terminal on top and positive at the bottom.Therefore, the controlled voltage source (j MI2) has a minus sign on topand plus sign at the bottom.
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13.2 Mutual Inductance (12)Recommended
14ECE 202 Ch 13
Watch these videos illustrating solving mutualinductance problems:http://www.youtube.com/watch?v=tD35a-uzd34 http://www.youtube.com/watch?v=hzU4XKQYTWw
http://www.youtube.com/watch?v=OqSvesTtnUo
http://www.youtube.com/watch?v=tD35a-uzd34http://www.youtube.com/watch?v=hzU4XKQYTWwhttp://www.youtube.com/watch?v=OqSvesTtnUohttp://www.youtube.com/watch?v=OqSvesTtnUohttp://www.youtube.com/watch?v=hzU4XKQYTWwhttp://www.youtube.com/watch?v=tD35a-uzd34http://www.youtube.com/watch?v=tD35a-uzd34http://www.youtube.com/watch?v=tD35a-uzd347/27/2019 ECE202 Ch13-Fall2014
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13.2 Mutual Inductance (13)
15
)connectionaiding-(series
221 M L L L
)connectionopposing-(series
221 M L L L
Dot convention for coils in series; the sign indicates thepolarity of the mutual voltage; (a) series-aiding connection,(b) series-opposing connection.
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13.2 Mutual Inductance (14)
16
Example 13.1 (Textbook)
Calculate the phasor currents I 1 and I 2 in the circuit shownbelow.
A04.1491.2I A;39.4901.13I 21 Ans:ECE 202 Ch 13
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13.3 Energy in a Coupled Circuit (1)
17
The instantaneous energy w stored in the circuit is:
21
2
22
2
11
2
1
2
1i Mii Li Lw
The positive sign is selected for the mutual term if bothcurrents enter or leave the dotted terminals; the negativesign is selected otherwise.
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13.3 Energy in a Coupled Circuit (2)
18
The coupling coefficient, k, is a measure of the magneticcoupling between two coils; 0 k 1.
21 L Lk M K = 1 coils perfectly coupledK < 0.5 coils loosely coupledK > 0.5 coils tightly coupled
1211
12
K
2221
21
K
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13.3 Energy in a Coupled Circuit (3)
19
Example 13.3 (Textbook)Consider the circuit below. Determine the couplingcoefficient. Calculate the energy stored in the coupledinductors at time t = 1s if v=60cos(4t +30) V.
Ans: k =0.56; w(1) =20.73J
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13.4 Linear Transformers (1)
20
A transformer is generally a four-terminal device comprising two (ormore) magnetically coupled coils
ECE 202 Ch 13
The coil directly connected to the voltage source is called the primarywinding.
The coil directly connected to the load is called the secondary winding. Resistances R1 and R2 are included to account for the losses in the coils. A transformer is said to be linear if the coils are wound on a magnetically
linear material for which the permeability is constant. Air, plastic, Bakelite, wood, etc. Most materials are magnetically linear.
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13.4 Linear Transformers (2)
21
Obtain input impedance Z in as seen from the source:
ECE 202 Ch 13
Mesh 1:
Mesh 2:
2111 )( MI j I L j RV
2221 )(0 I Z L j R MI j L
1
22
2 IZ
I L L j R
M j
1
22
22
111 Z
)( I L j R
M I L j RV
L
From Mesh 2: Substituting into Mesh 1 gives:
11 I Z I Z V reflected primary
ZZ
22
22
L
R
L j R
M
Reflected Impedance
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13.4 Linear Transformers (3)
22
Obtain input impedance Z in as seen from the source:
ECE 202 Ch 13
Mesh 1:
Mesh 2:
21 MI j I Z V primary
21 )(0 I Z MI j Secondary
12II
Secondary Z
M j 1
22
1 I
Z
M I Z V
Secondary
primary
From Mesh 2: Substituting into Mesh 1 gives:
11 I Z I Z V reflected primary Z
22
Secondary
R
Z
M
Reflected Impedance
ZSecondary = the total seriesimpedance in the secondary loop
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13.4 Linear Transformers (4)
23
The input impedance Z in can be broken into two parts as
follows:
ECE 202 Ch 13
reflected primaryin Z Z Z
L
in L j Z
M L j Z Z Z
22
22
11
Note: Z in will be the same if the dot on L2 is switched
Secondary
primaryin Z
M Z Z
22
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13.4 Linear Transformers (5)
24
The input impedance Z in can be broken into two parts as
follows:
ECE 202 Ch 13
reflected primaryin Z Z Z
Note: Z in
will be the same if the dot on L2 is switched
Secondary
primaryin Z
M Z Z
22
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13.4 Linear Transformers (6)
25
Example 13.4 (textbook)In the circuit below, calculate the input impedance and current I 1.
Take Z 1=60- j 100, Z 2=30+ j 40, and Z L=80+j60.
A1.1135.0I ;1.5314.100Z 1in Ans:
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13.4 Linear Transformers (7)
26
Analyzing magnetically coupled circuits issomewhat challenging, so it is sometimesconvenient to replace a magneticallycouple circuit with an equivalent circuitwith no magnetic coupling.
ECE 202 Ch 13
We want to replace the linear transformer with anequivalent T or circuit that has no mutual inductance.
Circuit: T Circuit:
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13.4 Linear Transformers (8)Equivalent T Circuit:
27ECE 202 Ch 13
2
1
2
1
2
1
I
I
jwL M j
M j L j
V
V
Linear Transformer Circuit
Equivalent T Network
2
1
2
1
)(
)(
I
I
L L jw jwL
jwL L L j
V
V
cbc
cca
Equating the terms gives the following relationships:
La = L1 - M; Lb = L2 - M; Lc = M
Mesh Analysis of this transformer circuitresults in the following:
Mesh Analysis of the equivalent T networkresults in the following:
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13.4 Linear Transformers (9)Equivalent T Circuit (Swapped Dots):
28ECE 202 Ch 13
2
1
2
1
2
1
I
I
jwL M j
M j L j
V
V
Linear Transformer Circuit
Equivalent T Network
2
1
2
1
)(
)(
I
I
L L jw jwL
jwL L L j
V
V
cbc
cca
Equating the terms gives the following relationships:
La = L1 + M; Lb = L2 + M; Lc = -M
Mesh Analysis of this transformer circuitresults in the following:
Mesh Analysis of the equivalent T networkresults in the following:
13 4 Li T f (10)
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13.4 Linear Transformers (10)Equivalent Circuit:
29ECE 202 Ch 13
2
1
2
1
111
111
V
V
L j L j L j
L j L j L j
I
I
C BC
C C A
Similarly, for the network nodal analysis provides:
Equivalent Network
By equating terms in admittance matrices, for the equivalentnetwork we obtain:
M M L L; L
M L M L L ; L
M L M L L L C B A
2
21
1
2
21
2
2
21
M
M L L; L
M L
M L L ; L
M L
M L L L C B A
2
21
1
2
21
2
2
21
13 4 Li T f (11)
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13.4 Linear Transformers (11)Equivalent T or Circuits Summary:
30ECE 202 Ch 13
T Circuit
Circuit
T Circuit
Circuit
13 4 Li T f (12)
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13.4 Linear Transformers (12)Equivalent Circuit:
31ECE 202 Ch 13
Find I1 and I 2 using the T equivalent circuit:
0324012 2111 I I j I j I j
01233 2212 I I j I I j
Mesh I 1:
Mesh I 2:
0123 21 I j jI
06123 21 I j I j
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13.5 Ideal Transformers (1)
32
An ideal transformer has perfect coupling (k=1). It consists of two or more coils with a large number of
turns wound on a common core of high permeability.
ECE 202 Ch 13
Because of the high permeability of the core, the fluxlinks all the turns of both coils, thereby resulting in aperfect coupling.
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13.5 Ideal Transformers (2)(Dots the same polarity)
33
Recall the coupled circuit:
ECE 202 Ch 13
2111 MI j I L jV
2212 I L j MI jV
(1)
(2)
Solving for I 1 in (1):
11
1
2
1
221
1
121
222 nV V L
L
L
I L L j
L
V L L I L jV
(3)
Substituting (3) into (2):
1
211I
L j
MI jV
21 L L M For perfect coupling (k=1):
1
2
2
1
1222
L
I M j
L
MV I L jV
12 nV V Therefore: where 12 L Ln = turns ratio
13 5 Id l T f (3)
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13.5 Ideal Transformers (3)(Dots opposite each other)
34
Mesh equations give the following:
ECE 202 Ch 13
2111 MI j I L jV
2212 I L j MI jV
(1)
(2)
Solving for I 1 in (1):
11
1
2
1
221
1
121
222 nV V L
L
L
I L L j
L
V L L I L jV
(3)
Substituting (3) into (2):
1
21
1I L j
MI jV
21 L L M For perfect coupling (k=1):
1
2
2
1
1
222 L
I M j
L
MV I L jV
12 nV V Therefore: If dot is swapped at output
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13.5 Ideal Transformers (4)
A transformer is said to be ideal if it has thefollowing properties:1. Coils have very large reactances (L 1, L2, M )2. Coupling coefficient is equal to unity (K=1)
3. Primary and secondary coils are lossless (R 1 = R2 = 0) An ideal transformer is a unity-coupled lossless
transformer in which the primary and secondarycoils have infinite self-inductances.
Iron core transformers are close approximations toideal transformers and are used in power systemsand electronics.
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13.5 Ideal Transformers (5)
36
When a sinusoidal voltage is applied to the primarywinding, the same magnetic flux goes through bothwindings.
ECE 202 Ch 13
Using the phasor voltages rather than theinstantaneous voltages:
t d
d N v
11
n N
N
v
v
1
2
1
2
t d
d
N v
22
;
Turns ratio ortransformation ratio
n N
N
V
V
1
2
1
2
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13.5 Ideal Transformers (6)
37
Power conservation:
The energy supplied to the primary must equal theenergy absorbed by the secondary, since there are nolosses in an ideal transformer.
In phasor form:
n=1 isolation transformer (V 2 = V1) n>1 step-up transformer (V 2 > V1) n
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13.5 Ideal Transformers (7)
38
Transformer ratings are usually specified as V 1 / V2
Power companies often generate at some convenientvoltage and use the step-up transformer to increasethe voltage so that the power can be transmitted atvery high voltage and low current over transmissionlines, resulting in significant cost savings. Nearresidential consumers, step-down transformers areused to bring the voltage down to 120 V.
It is important to get the proper polarity of thevoltages and the direction of the currents for thetransformer.
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39ECE 202 Ch 13
13.5 Ideal Transformer (8) Two rules to be followed for voltage polarity and current
direction:1. If V1 and V 2 are both positive or both negative at the
dotted terminals, use the +n. Otherwise, use the n.2. If I1 and I 2 both enter into or both leave the dotted
terminals, use n. Otherwise, use +n.
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13.5 Ideal Transformers (9)
40ECE 202 Ch 13
Expressing V 1 in terms of V 2 and I 1 in terms of I 2 or viceversa:
Complex Power is:
Complex power supplied to the primary is delivered to thesecondary without loss.
The ideal transformer is lossless and absorbs no power .
n
V V 21 12 nV V
n
I I 12
21 nI I
2*
22
*
22*
111 S I V nI n
V I V S
Complex Conjugate of I 2
13 5 Id l T f (10)
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13.5 Ideal Transformers (10)(Reflected Impedance)
41
The input impedance as seen by the source is:
The input impedance is also called the reflected
impedance since it appears as if the load impedanceis reflected to the primary side. The ability of the transformer to transform a given
impedance to another allows impedance matching toensure maximum power transfer.
ECE 202 Ch 13
2
2
2
2
1
1
n
Z
I n
V
I
V Z Lin
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13.5 Ideal Transformers (11)
42
In analyzing a circuit containing an ideal transformer,it is common practice to eliminate the transformer byreflecting impedances and sources from one side ofthe transformer to the other.
Suppose we want to reflect the secondary side of the
circuit to the primary side:
ECE 202 Ch 13
2
2
n
Z Z R
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13.5 Ideal Transformers (12)
43
We find the Thevenin equivalent of the circuit to the right of a-b: Obtaining V th from open circuit voltage:
Obtaining Z Th (remove the voltage source in the secondary andinsert a unit source at a-b terminals.)
ECE 202 Ch 13
n
V
n
V V V
V V
I I
s
Th
s
22
1
22
21 0 Since a-b is open
Equivalent CircuitReflecting the secondary to the primary
2
2
2
2
2
1
1
n
Z
I n
V
I
V Z Th
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13.5 Ideal Transformers (13)
44
The general rule for eliminating thetransformer and reflecting the secondary
circuit to the primary side: Divide the secondary impedance by n 2 Divide the secondary voltage by n Multiply the secondary current by n
The general rule for eliminating thetransformer and reflecting the primarycircuit to the secondary side:
Multiply the primary impedance by n 2 Multiply the primary voltage by n Divide the primary current by n
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13.5 Ideal Transformer (14)
Example 13.7 (Textbook)
An ideal transformer is rated at 2400/120V, 9.6 kVA, and has 50turns on the secondary side.
Calculate:(a) the turns ratio,(b) the number of turns on the primary side, and(c) the current ratings for the primary and secondary windings.
45
Ans:(a) This is a step-down transformer, n=0.05(b) N 1 = 1000 turns(c) I1 = 4A and I 2 = 80A
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13.5 Ideal Transformer (15)
Example 13.8 (Textbook)
For the ideal transformer, find: (a) the source current I 1, (b) theoutput voltage V 0, and (c ) the complex power supplied by thesource.
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13.6 Ideal Auto-Transformers (1)
47ECE 202 Ch 13
An autotransformer is atransformer in which both theprimary and the secondaryare in a single winding
A connection point called atap is separates the primaryand secondary.
The tap is often adjustable toprovide a desired turns ratio.
An adjustable tap provides a
variable voltage to the load A disadvantage of the
autotransformer is it providesno electrical isolation
Step Down Auto-Transformer
Step Up Auto-Transformer
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13.6 Ideal Auto-Transformers (2)
48ECE 202 Ch 13
The voltage / current relationships for the lossless ideal
autotransformer are as follows:
Step Down Auto-Transformer Step Up Auto-Transformer
1
21
22 V
N N
N V 1
1
212 V
N
N N V
1
21
12 I
N N
N I 1
2
212 I
N
N N I
Lin Z N
N N Z
2
2
21
Lin Z
N N
N Z
2
21
1
Similar toVoltage Divider
equation
Inverse Relation
Derive from V/I
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13.8 Multisim
ECE 202 Ch 13 49
REMEMBER:k=M/sqrt(L 1L2)-add ground!-freq=omega/2pi
The dot is always on the left-hand
terminal of the inductor when theinductor component L is placed(horizontally) without rotation onthe schematic. Thus the dot willbe at the top after 90 clockwiserotation and at the bottom after90 counterclockwise rotation.
For Magnetically Coupled Coils or Linear Transformer use:Multisim TRANSFORMER / INDUCTOR_COUPLING
For Ideal Transformer use:Multisim TRANSFORMER / 1P1S
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13.9 Applications (1)
50
Transformer as an Isolation Device to isolate ac supply
from a rectifier
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13.9 Applications (2)
51
Transformer as an Isolation Device to isolate dc betweentwo amplifier stages.
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13.9 Applications (3)
52
Transformer as a Matching Device
Using an ideal transformer to mat chthe speaker to the amplifier
Equivalent circuit
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Practice Problem 13.16 (Textbook)
Calculate the turns ratio of an ideal transformer
required to match a 400 load to a source withinternal impedance of 2.5k. Find the loadvoltage when the source voltage is 30V.
53
Ans: n = 0.4; V L = 6V
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13.9 Applications (4)
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13.9 Applications (5)
54
A typical power distribution system
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Chapter 13 - Equation / Analysis Summary Series Aiding / Opposing Dot Convention Model Coupling coefficient k Linear Transformer
Input Impedance:
Reflected Impedance:
Equivalent T or Circuits:
Ideal Transformer K = 1, L1, L2, Lossless (S 1 = S2) Voltage / Current Relationship:
Dots same = +n, Dots diff = -n Complex Power:
Autotransformer Adjustable tap No electrical Isolation Voltage Divider like relationship
221 M L L L 221 M L L L
21 L Lk M
Secondary
primaryin Z
M Z Z
22
Secondary
reflected Z
M Z
22
12 nV V n
I I 12 2
n
Z Z Lin
1
21
22 V
N N
N V
1
2
212 I
N
N N I
2
*
22
*
111 S I V I V S Lrmsave
R I P 2
2