This impedance is of particular interest because it shows how the transformer affects the impedance of the load as seen from the source.. Without the transformer, the load would be con-n
Trang 1To t h e internal source voltage Vs, the i m p e d a n c e appears as V^/Ii, o r
Vv Z n Z 22 + a> 2 M 2 a> 2 M 2
J " = Am = 7 = Zu + - ^ - - (9.63)
The impedance at the terminals of the source is Zint — Zs, so
2«b = Zu + -Z Z s = R { + ja>L x + — —— (9.64)
Note that the impedance Za b is independent of the magnetic polarity
of the transformer The reason is that the mutual inductance appears in
Eq 9.64 as a squared quantity This impedance is of particular interest
because it shows how the transformer affects the impedance of the load as
seen from the source Without the transformer, the load would be
con-nected directly to the source, and the source would see a load impedance
of ZL; with the transformer, the load is connected to the source through
the transformer, and the source sees a load impedance that is a modified
version of ZL, as seen in the third term of Eq 9.64
Reflected Impedance
The third term in Eq 9.64 is called the reflected impedance (Zr), because
it is the equivalent impedance of the secondary coil and load impedance
transmitted, or reflected, to the primary side of the transformer Note that
the reflected impedance is due solely to the existence of mutual
induc-tance; that is, if the two coils are decoupled, M becomes zero, Zr becomes
zero, and Za b reduces to the self-impedance of the primary coil
To consider reflected impedance in more detail, we first express the
load impedance in rectangular form:
where the load reactance X L carries its own algebraic sign In other words,
X L is a positive number if the load is inductive and a negative number if
the load is capacitive We now use Eq 9.65 to write the reflected
imped-ance in rectangular form:
2
R 2 + R L + j{coL 2 + X L ) _ OJ 2 M 2 [(R 2 + R L ) - ;(o>L2 + X L )]
(R 2 + R L ) 2 + (o>L 2 + X L f
2 A,/2
1^221
T h e derivation of E q 9.66 takes a d v a n t a g e of t h e fact that, when ZL is
written in rectangular form, the self-impedance of the m e s h containing t h e
secondary winding is
Now observe from Eq 9.66 that the self-impedance of the secondary
circuit is reflected into the primary circuit by a scaling factor of
(wM/|Z22|)2, and that the sign of the reactive component (wL2 -I- X L ) is
reversed Thus the linear transformer reflects the conjugate of the
self-impedance of the secondary circuit (Zj2) into the primary winding by a
scalar multiplier Example 9.13 illustrates mesh current analysis for a
cir-cuit containing a linear transformer
Trang 2Example 9.13 Analyzing a Linear Transformer in the Frequency Domain
The parameters of a certain linear transformer are
/?! = 200 ft, R2 = 100 ft, L{ = 9 H , L 2 = 4 H , and
k = 0.5 The transformer couples an impedance
consisting of an 800 O resistor in series with a 1 (x¥
capacitor to a sinusoidal voltage source The 300 V
(rms) source has an internal impedance of
500 + /100 ft and a frequency of 400 rad/s
a) Construct a frequency-domain equivalent circuit
of the system
b) Calculate the self-impedance of the primary
circuit
c) Calculate the self-impedance of the secondary
circuit
d) Calculate the impedance reflected into the
pri-mary winding
e) Calculate the scaling factor for the reflected
impedance
f) Calculate the impedance seen looking into the
primary terminals of the transformer
g) Calculate the Thevenin equivalent with respect
to the terminals
c,d-Solution
a) Figure 9.39 shows the frequency-domain
equiva-lent circuit Note that the internal voltage of the
source serves as the reference phasor, and that
Vj and V2 represent the terminal voltages of the
transformer In constructing the circuit in
Fig 9.39, we made the following calculations:
;wL, = /(400)(9) = /3600 ft,
ja>L 2 = /(400)(4) = /1600 ft,
M = 0.5V(9)(4) = 3 H,
joM = /(400)(3) = /1200 ft,
1 106
b) The self-impedance of the primary circuit is
Z n = 500 + /100 + 200 + /3600 = 700 + /3700 ft
c) The self-impedance of the secondary circuit is Z22 = 100 + /1600 + 800 - /2500 = 900 - /900 ft
d) The impedance reflected into the primary winding is
Zr = 1200
|900 - /900| (900 + /900)
= -(900 + /900) = 800 + /800 ft
e) The scaling factor by which Z 22 is reflected is 8/9 f) The impedance seen looking into the primary ter-minals of the transformer is the impedance of the primary winding plus the reflected impedance; thus
Za b = 200 + /3600 + 800 + /800 = 1000 + /4400 ft
g) The Thevenin voltage will equal the open circuit value of Vcd The open circuit value of \cd will
equal /1200 times the open circuit value of l h
The open circuit value of I] is
I, = 300 / 0
jtoC /400 = -/250() ft
700 + /3700
= 79.67/-79.29° mA
Therefore VTh = /1200(79.67/-79.29°) x 10~3
= 95.60/10.71° V
300 /0° V
500 n /ioon a 2ooa innn I O O H soon
_onrv^ * ^/^ ] I ZOO /^V • "WV—
+ •
V,
/3600O-• +
Figure 9.39 • The frequency-domain equivalent circuit for Example 9.13
Trang 3338 Sinusoidal Steady-State Analysis
The Thevenin impedance will be equal to the
imped-ance of the secondary winding plus the impedimped-ance
reflected from the primary when the voltage source is
replaced by a short-circuit Thus
/ 1200 V
ZT h = 100 + /1600 + (| 7 0 0 + J.3 7 ( ) ( ) |j (700 - /3700)
= 171.09 +/1224.26 n
The Thevenin equivalent is shown in Fig 9.40
/1224.26 li
95.60/10.71°/H
V
171.09 O
• W v —
Figure 9.40 A The Thevenin equivalent circuit for Example 9.13
^ A S S E S S M E N T PROBLEM
Objective 4—Be able to analyze circuits containing linear transformers using phasor methods
9.14 A linear transformer couples a load consisting
of a 360 O 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 + /0 Cl and a maximum
volt-age of 245.20 V, and it is operating at 800 rad/s
The transformer parameters are Ri = 100 fl,
L t = 0.5 H, R2 = 40 O, L2 = 0.125 H, and
k = 0.4 Calculate (a) the reflected impedance;
(b) the primary current; and (c) the secondary
current
NOTE: Also try Chapter Problems 9.76 and 9.77
Source Transformer d Load
Answer: (a) 10.24 - /7.68 H;
(b) 0.5 cos(800f - 53.13°) A;
(c) 0.08 cos 800f A
9.11 The Ideal Transformer
An ideal transformer consists of two magnetically coupled coils having N\
and N2 turns, respectively, and exhibiting these three properties:
1 The coefficient of coupling is unity (k = 1)
2 The self-inductance of each coil is infinite (Lj = L 2 = oo)
3 The coil losses, due to parasitic resistance, are negligible
Understanding the behavior of ideal transformers begins with Eq 9.64 which describes the impedance at the terminals of a source connected to a linear transformer We repeat this equation below and examine it further
Exploring Limiting Values
A useful relationship between the input impedance and load impedance,
as given by Za h in Eq 9.68, emerges as L\ and L2 each become infinitely large and, at the same time, the coefficient of coupling approaches unity:
a> 2 M 2
Ri + joL, + o> 2 M 2
(/¾ + jcoL 2 + Z L )' (9.68)
Trang 4Even though such transformers are nonlinear, we can obtain some useful
information by constructing an ideal model that ignores the nonlinearities
To show how Za b changes when k = 1 and L\ and L2 approach
infin-ity, we first introduce the notation
Z2 2 = R2 + R L + KoL 2 + XL ) = R 22 + jX 22
and t h e n r e a r r a n g e E q 9.68:
to 2 M 2 R 72 ( <o2 M 2 X 22
Za|, = Ri -\—5 Y +1 1 oi ^\ 2 2~
«*22 "^ ^ 2 2 ^ -**22 "•" -^-22
A t this point, w e m u s t b e careful with t h e coefficient of ;* in E q 9.69
because, as L\ a n d L2 a p p r o a c h infinity, this coefficient is t h e difference
b e t w e e n t w o large quantities Thus, before letting L] a n d L2 increase, w e
write t h e coefficient as
w h e r e we recognize t h a t , w h e n k = 1, M2 = L^L2 Putting t h e t e r m m u l
-tiplying wLy over a c o m m o n d e n o m i n a t o r gives
(R\y + OiUX V + X}\
Factoring o»L2 o u t of t h e n u m e r a t o r a n d d e n o m i n a t o r of E q 9.71 yields
L , X L + (R 222 + Xl)lo>L 2
L 2 (R 22 /<oL 2 ) 2 +[\ + (XJwL 2 )] 2
As k approaches 1,0, the ratio L]/L2 approaches the constant value of
(N1/N1) 2 , which follows from Eqs 6.54 and 6.55 The reason is that, as the
coupling becomes extremely tight, the two permeances 57^ and SP2 become
equal Equation 9.72 then reduces to
as L j - * 00, L2 —> 00, a n d k —• 1.0
T h e s a m e r e a s o n i n g leads t o simplification of t h e reflected resistance
in E q 9.69:
urM 2 R- >2 L / i V , V
Applying the results given by Eqs 9.73 and 9.74 to Eq 9.69 yields
Zah = * , + ( ^ ) 2 ¾ + ( ^ ) 2 ( « L + / ¾ (9.75)
C o m p a r e this result with t h e result in E q 9.68 H e r e w e see that w h e n t h e
coefficient of coupling a p p r o a c h e s unity a n d the self-inductances of the
coupled coils a p p r o a c h infinity, t h e transformer reflects t h e secondary
winding resistance a n d t h e load i m p e d a n c e t o t h e p r i m a r y side b y a scaling
Trang 5340 Sinusoidal Steady-State Analysis
jcoM
j(oL A j(oL 2
IN,
v,
(a)
jcoM jcoL 2
\N->
(b) Figure 9.41 A The circuits used to verify the
volts-per-turn and ampere-turn relationships for an ideal
transformer
factor equal to the turns ratio (M/A^) squared Hence we may describe the terminal behavior of the ideal transformer in terms of two characteristics
First, the magnitude of the volts per turn is the same for each coil, or
(9.76)
Second, the magnitude of the ampere-turns is the same for each coil, or
i,L J i V l MI = My (9.77)
We are forced to use magnitude signs in Eqs 9.76 and 9.77, because we have not yet established reference polarities for the currents and voltages;
we discuss the removal of the magnitude signs shortly
Figure 9.41 shows two lossless (R^ = R 2 = 0) magnetically coupled
coils We use Fig 9.41 to validate Eqs 9.76 and 9.77 In Fig 9.41(a), coil 2 is open; in Fig 9.41(b), coil 2 is shorted Although we carry out the following analysis in terms of sinusoidal steady-state operation, the results also
apply to instantaneous values of v and L
Determining the Voltage and Current Ratios
Note in Fig 9.41(a) that the voltage at the terminals of the open-circuit coil is entirely the result of the current in coil 1; therefore
The current in coil 1 is
From Eqs 9.78 and 9.79,
V 2 = ja>Ml\
jcoL ]
(9.78)
(9.79)
(9.80)
For unity coupling, the mutual inductance equals VL] L 2 , so Eq 9.80
becomes
For unity coupling, the flux linking coil 1 is the same as the flux linking coil 2, so we n e e d only one p e r m e a n c e to describe t h e self-inductance of each coil Thus E q 9.81 becomes
V, =
/V?SP
A^
or
Voltage relationship for an ideal
transformer •
N 2
(9.83)
Summing the voltages around the shorted coil of Fig 9.41(b) yields
Trang 6from which, for k = 1,
II
VUL2
L 2
(9.85)
Equation 9.85 is equivalent to
I , ^ i - l2 N 2 (9.86) -4 Current relationship for an ideal
transformer
Figure 9.42 shows the graphic symbol for an ideal transformer The
vertical lines in the symbol represent the layers of magnetic material from
which ferromagnetic cores are often made Thus, the symbol reminds us
that coils wound on a ferromagnetic core behave very much like an ideal
transformer
There are several reasons for this The ferromagnetic material creates
a space with high permeance Thus most of the magnetic flux is trapped
inside the core material, establishing tight magnetic coupling between
coils that share the same core High permeance also means high
self-inductance, because L = N2 V Finally, ferromagnetically coupled coils
efficiently transfer power from one coil to the other Efficiencies in excess
of 95% are common, so neglecting losses is not a crippling approximation
for many applications
Determining the Polarity of the Voltage
and Current Ratios
We now turn to the removal of the magnitude signs from Eqs 9.76 and
9.77 Note that magnitude signs did not show up in the derivations of
Eqs 9.83 and 9.86 We did not need them there because we had established
reference polarities for voltages and reference directions for currents In
addition, we knew the magnetic polarity dots of the two coupled coils
The rules for assigning the proper algebraic sign to Eqs 9.76 and 9.77
are as follows:
If the coil voltages V! and V2 are both positive or negative at the
dot-marked terminal, use a plus sign in Eq 9.76 Otherwise, use a
nega-tive sign
If the coil currents I] and I2 are both directed into or out of the
dot-marked terminal, use a minus sign in Eq 9.77 Otherwise, use a
plus sign
The four circuits shown in Fig 9.43 illustrate these rules
A", • N,
Ideal
Figure 9.42 • The graphic symbol for an ideal
transformer
A Dot convention for ideal transformers
+ • JV, N 2 \ • +
\
V, I
>
Ideal
vi = v_2
/V, /V 2 "
Ni\i = ~N 2 l 2
(a)
I 2 V 2
+ • yv, N 2 \ +
f
Ideal I •
W,I, = N 2 l 2
(b)
+ •
V, I,)
J
|N, N 2 \ • +
Ideal
V , V 2
/V, N 2
A^I, = M,I 2
(c)
l/V, N 2 \
V t I,
>'
I, V 2
Ideal
V ] = _V_ 2
A7,I, = -N 2 l 2
Trang 7342 Sinusoidal Steady-State Analysis
AT,-v,
= 500 N 2 =
•1
i
J Ideal L
(a)
2500 +
V 2
-+ n i: 5
nr^-Ideal (b)
V,
+ ni/5 :irr^
V, Ideal (c)
Figure 9.44 • Three ways to show that the turns ratio
of an ideal transformer is 5
The ratio of the turns on the two windings is an important parameter
of the ideal transformer The turns ratio is defined as either N]/N 2 or
M/N,; both ratios appear in various writings In this text, we use a to denote the ratio N2 /N], or
Figure 9.44 shows three ways to represent the turns ratio of an ideal transformer Figure 9.44(a) shows the number of turns in each coil
explic-itly Figure 9.44(b) shows that the ratio Ni/N] is 5 to 1, and Fig 9.44(c)
shows that the ratio /V2//V, is 1 to | Example 9.14 illustrates the analysis of a circuit containing an ideal transformer
Example 9.14 Analyzing an Ideal Transformer Circuit in the Frequency Domain
The load impedance connected to the secondary
winding of the ideal transformer in Fig 9.45 consists of
a 237.5 m i l resistor in series with a 125 /xH inductor
If the sinusoidal voltage source (-yg) is
generat-ing the voltage 2500 cos 400/ V, find the
steady-state expressions for: (a) /,; (b) V\; (c) /2; and (d) v 2
O
10:1
Ideal
; 125 yaH
Figure 9.45 • The circuit for Example 9.14
Solution
a) We begin by constructing the phasor domain
equivalent circuit The voltage source becomes
2500/0° V; the 5 mH inductor converts to an
impedance of / 2 D; and the 125 fxH inductor
converts to an impedance of/0.05 ft The phasor
domain equivalent circuit is shown in Fig 9.46
It follows directly from Fig 9.46 that
2500/0° = (0.25 + /2)1] + V,,
and
V, = 10V2 = 10[(0.2375 + /0.05)I2]
Because
I2 = 101,
we have
Vi = 10(0.2375 +/0.05)101!
= (23.75 + /5)1,
0.25 II /2 ft
AAA, / - Y V W - 0.2375 ft
/ T N 2500/0!
0 1 1 0 : 1 / 0 - v v ^ ~
V 2 /0.05 ft;
Ideal
Figure 9.46 A Phasor domain circuit for Example 9.14
Therefore
2500 / 0 ° = (24 + / 7 ) 1 , ,
or
I, = 100/-16.26° A
Thus the steady-state expression for i1 is
/, = 100cos(400r - 16.26°) A
b) V, = 2500/0° - (100 / - 1 6 2 6 " )(0.25 + /2)
= 2500 - 80 - /185
= 2420 - /185 = 2427.06/-4.37° V, Hence
v ] = 2427.06 cos (400r - 4.37°) V
c) I2 = 101, = 1000/-16.26° A
Therefore
i 2 = 1000 cos (400f - 16.26°) A
d) V2 = 0.1V, = 242.71 / - 4 3 7 ° V, giving
y2 = 242.71 cos (400/ - 4.37°) V
Trang 8The Use of an Ideal Transformer for Impedance Matching
Ideal transformers can also be used to raise or lower the impedance level
of a load.Tlie circuit shown in Fig 9.47 illustrates this.The impedance seen
by the practical voltage source (Vv in series with Zv) is Vi/Ij The voltage
and current at the terminals of the load impedance (V2 and I2) are related
to V] and Ij by the transformer turns ratio; thus
and
v, = - ,
I, = a\2
Therefore the impedance seen by the practical source is
Z I N J — _ — - T
1 Y>
'IN
Ii a2 h
but the ratio V2/I2 is the load impedance ZL, so Eq 9.90 becomes
(9.88)
(9.89)
(9.90)
1 :a
deal
• +
— 0 —
zL
Figure 9.47 A Using an ideal transformer to couple a
load to a source
Thus, the ideal transformer's secondary coil reflects the load impedance
back to the primary coil, with the scaling factor \/a~
Note that the ideal transformer changes the magnitude of ZL but does
not affect its phase angle Whether Z)N is greater or less than ZL depends
on the turns ratio a
The ideal transformer —or its practical counterpart, the
ferromag-netic core transformer—can be used to match the magnitude of ZL to
the magnitude of Zv We will discuss why this may be desirable in
Chapter 10
^ A S S E S S M E N T P R O B L E M
Objective 5—Be able to analyze circuits with ideal transformers
9.15 The source voltage in the phasor domain circuit
in the accompanying figure is 25 / 0 ° kV Find
the amplitude and phase angle of V2 a nd
^2-Answer: V2 = 1868.15 /142.39° V;
I2 = 125 /216.87° A
NOTE: Also try Chapter Problem 9.83
'•©
4 f t
^ 7 1 2 5 : 1
V,
Ideal
-/14.4 ft
As we shall see, ideal transformers are used to increase or decrease
voltages from a source to a load Thus, ideal transformers are used widely
in the electric utility industry, where it is desirable to decrease, or step
down, the voltage level at the power line to safer residential voltage levels
Trang 9344 Sinusoidal Steady-State Analysis
9.12 Phasor Diagrams
2 / 1 5 0 ° ^ - ^
8/-170°
s^-150t
-170*-5
30°
i
/ -45°
/-45°
10/30°
Figure 9.48 A A graphic representation of phasors
Figure 9.49 A The complex number
- 7 - /3 = 7.62 /-156.80°
When we are using the phasor method to analyze the steady-state sinu-soidal operation of a circuit, a diagram of the phasor currents and voltages may give further insight into the behavior of the circuit A phasor diagram shows the magnitude and phase angle of each phasor quantity in the complex-number plane Phase angles are measured counterclockwise from the positive real axis, and magnitudes are measured from the origin of the axes For example, Fig 9.48 shows the phasor quantities 1 0 / 3 0 ° , 12 /150%
5 / - 4 5 ° , and 8 / - 1 7 0 ° Constructing phasor diagrams of circuit quantities generally involves both currents and voltages As a result, two different magnitude scales are necessary, one for currents and one for voltages The ability to visualize a pha-sor quantity on the complex-number plane can be useful when you are check-ing pocket calculator calculations The typical pocket calculator doesn't offer
a printout of the data entered But when the calculated angle is displayed, you can compare it to your mental image as a check on whether you keyed in the appropriate values For example, suppose that you are to compute the polar form of - 7 - / 3 Without making any calculations, you should anticipate a magnitude greater than 7 and an angle in the third quadrant that is more neg-ative than —135° or less positive than 225°, as illustrated in Fig 9.49
Examples 9.15 and 9.16 illustrate the construction and use of phasor diagrams We use such diagrams in subsequent chapters whenever they give additional insight into the steady-state sinusoidal operation of the cir-cuit under investigation Problem 9.84 shows how a phasor diagram can help explain the operation of a phase-shifting circuit
Example 9.15 Using Phasor Diagrams to Analyze a Circuit
For the circuit in Fig 9.50, use a phasor diagram to
find the value of R that will cause the current
through that resistor, iR , to lag the source current, („
by 45° when <o = 5 krad/s
h = vm/v_ \V m / 9 0 ° ,
-//(5000)(800 X 10~6) and the current phasor for the resistor is given by
I V,„ / 0 y
= ^ / 0 °
Figure 9.50 A The circuit for Example 9.15
Solution
By Kirchhoff s current law, the sum of the currents
l R , l L , and Ic must equal the source current I5 If we
assume that the phase angle of the voltage V,„ is
zero, we can draw the current phasors for each of
the components The current phasor for the
induc-tor is given by
These phasors are shown in Fig 9.51 The phasor diagram also shows the source current phasor, sketched as a dotted line, which must be the sum of the current phasors of the three circuit components and must be at an angle that is 45 ° more positive than the current phasor for the resistor As you can see, summing the phasors makes an isosceles triangle, so the length of the current phasor for the resistor must
equal 3V,„ Therefore, the value of the resistor is | Cl
Ic = / 4 V >
1/ = vm / o£
whereas the current phasor for the capacitor is
given by
>
/
A
h = VJR
Figure 9.51 A The phasor diagram for the currents in Fig 9.50
Trang 10Example 9.16 Using Phasor Diagrams to Analyze Capadtive Loading Effects
The circuit in Fig 9.52 has a load consisting of the
parallel combination of the resistor and inductor
Use phasor diagrams to explore the effect of
adding a capacitor across the terminals of the load
on the amplitude of Vs if we adjust \ s so that the
amplitude of V L remains constant Utility
compa-nies use this technique to control the voltage drop
on their lines
R?
Figure 9.52 • The circuit for Example 9.16
For convenience, we place this phasor on the pos-itive real axis
b) We know that Ia is in phase with VL and that its magnitude is |VL|/#2- (On the phasor diagram, the magnitude scale for the current phasors is independent of the magnitude scale for the volt-age phasors.)
c) We know that Ib lags behind VL by 90° and that its magnitude is |VL|/wL2
d) The line current I is equal to the sum of Ia and Ib
e) The voltage drop across Ry is in phase with the line current, and the voltage drop across jo)L]
leads the line current by 90°
f) The source voltage is the sum of the load voltage and the drop along the line; that is, Vs = VL
+ (R x + jcoL { )\
Solution
We begin by assuming zero capacitance across the
load After constructing the phasor diagram for the
zero-capacitance case, we can add the capacitor and
study its effect on the amplitude of Y,, holding the
amplitude of VL constant Figure 9.53 shows the
fre-quency-domain equivalent of the circuit shown in
Fig 9.52 We added the phasor branch currents I, Ia,
and Ib to Fig 9.53 to aid discussion
rY-v-v> —» a
- — V ,
VL R 2 i\h J<oL 2 i\l u
-• *
Figure 9.53 • The frequency-domain equivalent of the circuit
in Fig 9.52
Figure 9.54 shows the stepwise evolution of
the phasor diagram Keep in mind that we are not
interested in specific phasor values and positions
in this example, but rather in the general effect of
adding a capacitor across the terminals of the
load Thus, we want to develop the relative
posi-tions of the phasors before and after the capacitor
has been added
Relating the phasor diagram to the circuit
shown in Fig 9.53 reveals the following points:
a) Because we are holding the amplitude of the load
voltage constant, we choose VL as our reference
Figure 9.54 • The step-by-step evolution of the phasor
diagram for the circuit in Fig 9.53
Note that the completed phasor diagram shown in step 6 of Fig 9.54 clearly shows the amplitude and phase angle relationships among all the currents and voltages
in Fig 9.53
Now add the capacitor branch shown in Fig 9.55 We are holding VL constant, so we construct the phasor dia-gram for the circuit in Fig 9.55 following the same steps
as those in Fig 9.54, except that, in step 4, we add the capacitor current Ic to the diagram In so doing, Ic leads
VL by 90°, with its magnitude being |VLwC| Figure 9.56 shows the effect of Ic on the line current: Both the magni-tude and phase angle of the line current I change with changes in the magnitude of Ic As I changes, so do the magnitude and phase angle of the voltage drop along the line As the drop along the line changes, the magnitude and phase angle of V? change The phasor diagram shown