Thus, we have a shortcut for calculating line currents from phase currents or vice versa for a balanced three-phase A-connected load.. The total average power delivered to the balanced Y
Trang 1e) Calculate the phase voltages at the terminals of
the generator, Van, Vbn, and Vcn
f) Calculate the line voltages Vab, Vbc, and Vca at
the terminals of the generator
g) Repeat (a)-(f) for a negative phase sequence
Solution
a) Figure 11.10 shows the single-phase equivalent
circuit
b) The a-phase line current is
120 / 0 °
dA (0.2 + 0.8 + 39) + /(0.5 + 1.5 + 28)
120 / 0 °
40 + /30
= 2 4 / - 3 6 8 7 ° A
a' 0.2(2 /0-5 O a o.8(2 /1-50 A
f 'VW ^VYY>_
' i A
© 120/0° V V a n
39 n
1/28(1
Figure 11.10 A The single-phase equivalent circuit for
Example 11.1
For a positive phase sequence,
Ib B = 2.4 / - 1 5 6 8 7 ° A,
Ic C = 2.4 /83.13° A
c) The phase voltage at the A terminal of the load is
VAN = (39 + / 2 8 ) ( 2 4 / - 3 6 8 7 ° )
= 115.22/-1.19° V
For a positive phase sequence,
VBN = 115.22/-121.19° V,
VCN = 115.22/118.81° V
d) For a positive phase sequence, the line voltages
lead the phase voltages by 30°; thus
VAB = ( V 3 / 3 0 ° ) V A N
= 199.58 /28.81° V,
VBC = 199.58/-91.19° V,
VCA = 199.58 /148.81° V
e) The phase voltage at the a terminal of the source is
Van = 120 - (0.2 + /0.5)(2.4/-36.87°)
= 120 - 1.29/31.33°
= 118.90 - / 0 6 7
= 118.90/-0.32° V
For a positive phase sequence,
Vbn = 118.90/-120.32° V,
Vcn = 118.90/119.68° V
f) The line voltages at the source terminals are
Vab = ( V 5 / 3 0 " ) Van
- 205.94 /29.68° V, Vbc = 205.94 / - 9 0 3 2 ° V,
Vca = 205.94/149.68° V
g) Changing the phase sequence has no effect on the single-phase equivalent circuit The three line currents are
IaA = 2.4 / - 3 6 8 7 ° A, I,,B = 2.4 /83.13° A,
IcC = 2 4 / - 1 5 6 8 7 ° A
The phase voltages at the load are
VAN = 115.22/-1.19° V,
VBN = 115.22/118.81° V,
VCN = 115.22/-121.19° V
For a negative phase sequence, the line voltages lag the phase voltages by 30°:
VAB = ( V 3 / - 3 0 ° ) VA N
- 199.58 / - 3 1 1 9 ° V,
VBC = 199.58 /88.81° V, VCA = 199.58/-151.19° V
The phase voltages at the terminals of the gener-ator are
Van - 118.90/-0.32° V,
Vbll = 118.90/119.68° V,
Vcn = 118.90/-120.32° V
The line voltages at the terminals of the genera-tor are
Vab = ( V 3 / - 3 0 ° ) Va n
= 205.94/-30.32° V,
Vbc = 205.94 /89.68° V, V,,, = 205.94/-150.32° V
Trang 2^ A S S E S S M E N T P R O B L E M S
Objective 1—Know how to analyze a balanced, three-phase wye-wye circuit
11.1 The voltage from A to N in a balanced
three-phase circuit is 240 /—30° V If the three-phase
sequence is positive, what is the value of VBC?
Answer: 4 1 5 6 9 / - 1 2 0
11.2
V
The c-phase voltage of a balanced three-phase
Y-connected system is 450 / - 2 5 ° V If the
phase sequence is negative, what is the value
ofVA B?
Answer: 779.42 / 6 5 ° V
11.3 The phase voltage at the terminals of a
bal-anced three-phase Y-connected load is 2400 V
The load has an impedance of 16 + /12 il/cp
and is fed from a line having an impedance of
0.10 + /0.80 Cl/cj) The Y-connected source at
the sending end of the line has a phase
NOTE: Also try Chapter Problems 11.7, 11.9, and 11.11
sequence of acb and an internal impedance of 0.02 -I- /0.16 n/<£ Use the a-phase voltage at the load as the reference and calculate (a) the line currents Ia A, IbB, and IcC; (b) the line volt-ages at the source, Vab, Vbc, and Vca; and (c) the internal phase-to-neutral voltages at the source,
Va<n, V b - n , and V c - n Answer: (a) I aA = 120 / - 3 6 8 7 ° A,
Ib B = 120 /83.13° A, and
Ic C = 120/-156.87° A;
(b) Va b = 4275.02 / - 2 8 3 8 ° V,
Vbc = 4275.02 /91.62° V, and
Vca = 4275.02 / - 1 4 8 3 8 ° V;
(c) Va.n = 2482.05 /1.93° V,
Vb-n = 2482.05 /121.93° V, and
Vc'n = 2482.05/-118.07° V
11.4 Analysis of the Wye-Delta Circuit
If the load in a three-phase circuit is connected in a delta, it can be
trans-formed into a wye by using the delta-to-wye transformation discussed in
Section 9.6 When the load is balanced, the impedance of each leg of the
wye is one third the impedance of each leg of the delta, or
Relationship between three-phase delta-connected and wye-connected
-4 impedance
z - ~±
which follows directly from Eqs 9.51-9.53 After the A load has been
replaced by its Y equivalent, the a-phase can be modeled by the
single-phase equivalent circuit shown in Fig 11.11
We use this circuit to calculate the line currents, and we then use the
line currents to find the currents in each leg of the original A load The
relationship between the line currents and the currents in each leg of the
delta can be derived using the circuit shown in Fig 11.12
When a load (or source) is connected in a delta, the current in each leg
of the delta is the phase current, and the voltage across each leg is the
phase voltage Figure 11.12 shows that, in the A configuration, the phase
voltage is identical to the line voltage
To demonstrate the relationship between the phase currents and line
currents, we assume a positive phase sequence and let A/> represent the
magnitude of the phase current Then
A B cr
Ipc = /A/ - H O "
lr A = At/120"
Va-n
Zga a
Zu
c
A
zA
N
Figure 11.11 A A single-phase equivalent circuit
(11.22)
(11.23) Figure 11.12 • A circuit used to establish the
relationship between line currents and phase currents in (11-24) a balanced A load
Trang 3Figure 11.13 • Phasor diagrams showing the
relationship between line currents and phase currents in
a A-connected load, (a) The positive sequence, (b) The
negative sequence
In writing these equations, we arbitrarily selected IA B as the reference phasor
We can write the line currents in terms of the phase currents by direct application of Kirchhoff s current law:
IaA = IAB ~ ICA
= I A /Q° -U/120*
= V3I 4 / - 3 0 ° ,
IbB = I B C - IAB
= U / - 1 2 0 ° - U
= ^ / ^ , / - 1 5 0 %
(11.25)
(11.26) IcC _ IcA ~~ I B C
= / , , / 1 2 0 ° - / ^ / - 1 2 0 °
Comparing Eqs 11.25-11.27 with Eqs 11.22-11.24 reveals that the magni-tude of the line currents is V3 times the magnimagni-tude of the phase currents and that the set of line currents lags the set of phase currents by 30°
We leave to you to verify that, for a negative phase sequence, the line currents are V 3 times larger than the phase currents and lead the phase currents by 30° Thus, we have a shortcut for calculating line currents from phase currents (or vice versa) for a balanced three-phase A-connected load Figure 11.13 summarizes this shortcut graphically Example 11.2 illustrates the calculations involved in analyzing a balanced three-phase circuit having a Y-connected source and a A-connected load
Example 11.2 Analyzing a Wye-Delta Circuit
The Y-connected source in Example 11.1 feeds a
A-connected load through a distribution line
hav-ing an impedance of 0.3 + /0.9 Q,/4> The load
impedance is 118.5 + /85.8 Cl/4> Use the a-phase
internal voltage of the generator as the reference
a) Construct a single-phase equivalent circuit of the
three-phase system
b) Calculate the line currents Ia A, Ib B, and Ic C
c) Calculate the phase voltages at the load terminals
d) Calculate the phase currents of the load
e) Calculate the line voltages at the source terminals
Solution
a) Figure 11.14 shows the single-phase equivalent
circuit The load impedance of the Y equivalent is
118.5 + /85.8
= 39.5 + /28.6 ft/0
a' 0.212 /0.511 a 0.3 ft /0.9 0 A
C_J 120/0°
39.5 ft
/28.6 ft
N
Figure 11.14 • The single-phase equivalent circuit for
Example 11.2
b) The a-phase line current is
120 /C
IaA =
(0.2 + 0.3 + 39.5) + /(0.5 + 0.9 + 28.6)
1 2 0 / 0 °
rr ~ - = 2.4 / - 3 6 8 7 ° A
40 + /30
Trang 4Hence
IbB = 2 4 / - 1 5 6 8 7 ° A,
IcC = 2.4 /83.13° A
c) Because the load is A connected, the phase
volt-ages are the same as the line voltvolt-ages To
calcu-late the line voltages, we first calcucalcu-late VA N:
VAN = (39.5 + /28.6)(2.4/-36.87°)
= 117.04/-0.96° V
Because the phase sequence is positive, the line
voltage VAB *S
VAB = ( V 5 / 3 0 ' ) V A N
= 202.72 /29.04° V
Therefore
VB C = 202.72 / - 9 0 9 6 ° V,
VC A = 202.72 /149.04° V
d) The phase currents of the load may be calculated
directly from the line currents:
l AB
1
V3 /30' aA
= 1.39/-6.87° A
Once we know IA B, we also know the other load phase currents:
IB C = 1.39/-126.87° A,
IC A = 1.39/113.13° A
Note that we can check the calculation of IA B by using the previously calculated VAB ar>d the impedance of the A-connected load; that is,
VA B _ 202.72/29.04°
I A B ~" ~ Z ^ ' 118.5 + / 8 5 8
= 1.39/-6.87° A
e) To calculate the line voltage at the terminals of the source, we first calculate Van Figure 11.14 shows that Van is the voltage drop across the line impedance plus the load impedance, so
Van = (39.8 + /29.5)(2.4/-36.87°)
= 118.90/-0.32° V
The line voltage Va b is
Vab = (V3~/30°)Va n,
or
Vab = 205.94 /29.68° V
Therefore
Vbc = 205.94/-90.32° V,
Vca = 205.94/149.68° V
^ASSESSMENT PROBLEMS
Objective 2—Know how to analyze a balanced, three-phase wye-delta connected circuit
11.4 The current IC A in a balanced three-phase
A-connected load is 8 /—15° A If the phase
sequence is positive, what is the value of IcC?
Answer: 1 3 8 6 / - 4 5 ° A
11.5 A balanced three-phase A-connected load is
fed from a balanced three-phase circuit The
reference for the b-phase line current is toward
the load The value of the current in the
b-phase is 12 / 6 5 ° A If the phase sequence is
negative, what is the value of I A B ?
Answer: 6.93 / - 8 5 ° A
11.6 The line voltage VAB at the terminals of a bal-anced three-phase A-connected load is
4160 / 0 ° V The line current Ia A is
6 9 2 8 / - 1 0 ° A
a) Calculate the per-phase impedance of the load if the phase sequence is positive
b) Repeat (a) for a negative phase sequence
Answer: (a) 104 / - 2 0 ° H;
(b) 1 0 4 / + 4 0 ° a 11.7 The line voltage at the terminals of a balanced A-connected load is 110 V Each phase of the load consists of a 3.667 II resistor in parallel with
a 2.75 Cl inductive impedance What is the
mag-nitude of the current in the line feeding the load? Answer: 86.60 A
NOTE: Also try Chapter Problems 11.12,11.13, and 11.16
Trang 5IaA
%B
* •
Ice'
B + v BN
^B
zA
—
C
i Ts
» i\
Z c
Figure 11.15 • A balanced Y load used to introduce
average power calculations in three-phase circuits
11.5 Power Calculations in Balanced
Three-Phase Circuits
So far, we have limited our analysis of balanced three-phase circuits to determining currents and voltages We now discuss three-phase power calculations We begin by considering the average power delivered to a balanced Y-connected load
Average Power in a Balanced Wye Load
Figure 11.15 shows a Y-connected load, along with its pertinent currents and voltages We calculate the average power associated with any one phase by using the techniques introduced in Chapter 10 With Eq 10.21 as a starting point, we express the average power associated with the a-phase as
^A = | VA N| | Ia A| c o s ( t 9v A- M , (11.28)
where 8„ A and 0,A denote the phase angles of VA N and IuA, respectively Using the notation introduced in Eq 11.28, we can find the power associ-ated with the b- and c-phases:
Pa = |VB Nl|IbBl«>s(0vB-0/B);
Pc = ! VC N| | Ic C| c o s ( f \c- 0 ;C)
(11.29) (11.30)
In Eqs 11.28-11.30, all phasor currents and voltages are written in terms
of the rms value of the sinusoidal function they represent
In a balanced three-phase system, the magnitude of each line-to-neutral voltage is the same, as is the magnitude of each phase current The argu-ment of the cosine functions is also the same for all three phases We emphasize these observations by introducing the following notation:
and
(11.32)
(11.33) Moreover, for a balanced system, the power delivered to each phase of the load is the same, so
PA = PB = PC = P* = V ^ cos ^ , (11.34)
where P^ represents the average power per phase
The total average power delivered to the balanced Y-connected load
is simply three times the power per phase, or
Expressing the total power in terms of the rms magnitudes of the line
volt-age and current is also desirable If we let V L and I L represent the rms magnitudes of the line voltage and current, respectively, we can modify
Eq 11.35 as follows:
Total real power in a balanced
V 3 VL/Lc o s ^ (11.36)
Trang 6In deriving Eq 11.36, we recognized that, for a balanced Y-connected
load, the magnitude of the phase voltage is the magnitude of the line
volt-age divided by V5, and that the magnitude of the line current is equal to
the magnitude of the phase current When using Eq 11.36 to calculate the
total power delivered to the load, remember that 0$ is the phase angle
between the phase voltage and current
Complex Power in a Balanced Wye Load
We can also calculate the reactive power and complex power associated
with any one phase of a Y-connected load by using the techniques
intro-duced in Chapter 10 For a balanced load, the expressions for the reactive
power are
2 r = 3Q^ = V3V L / L sin^
(11.37)
(11.38) Equation 10.29 is the basis for expressing the complex power associated
with any phase For a balanced load,
S<t> - ^AN^aA _ ^BNlbB ~ VCNIcC v; (11.39)
where V^ and 1^, represent a phase voltage and current taken from the
same phase Thus, in general,
(11.40)
(11.41)
< Total reactive power in a balanced
three-phase load
-4 Total complex power in a balanced
three-phase load
Power Calculations in a Balanced Delta Load
If the load is A-connected, the calculation of power—reactive or
complex-is basically the same as that for a Y-connected load Figure 11.16 shows a
A-connected load, along with its pertinent currents and voltages The
power associated with each phase is
^ A = |V A BI|IAB|COS(0 V A B - 0 /AB ),
PB = |VBCI|IBC|C°S(0V BC " 0 '-BC)»
^ c = | V C A I | I C A | C O S ( 0 V C A - 0 ( C A )
For a balanced load,
| V A B | = IVBCI = IVCAI = V * HABI = IIBCI = IICAI = / *
0<AB = #t>BC ~ ^/BC = ^«CA
and
P A = PB = Pc = Pd> = K ^ c o s ^
Note that Eq 11.48 is the same as Eq 11.34 Thus, in a balanced load,
regardless of whether it is Y- or A-connected, the average power per phase
is equal to the product of the rms magnitude of the phase voltage, the rms
magnitude of the phase current, and the cosine of the angle between the
phase voltage and current
B L ZA V,-CA
(11.42) (11.43)
(11.45)
Figure 11.16 • A A-connected load used to discuss
(11.46) power calculations
(11.47)
(11.48)
Trang 7Pr = 3 ^ = 3 1 ^ cos 0*
Note that Eq 11.49 is the same as Eq 11.36 The expressions for reactive
power and complex power also have the same form as those developed for
the Y load:
Q^ = V 4> I lf> sme 4> ; (11.50)
Qr = 3Q* = 3 V V0 sin $+; (11.51)
S<i> = P<i> + jQ^> = V ^ l J ; (11.52)
S T = 3S lb = V3V L I L /$£ (11.53)
Instantaneous Power in Three-Phase Circuits
Although we are primarily interested in average, reactive, and complex
power calculations, the computation of the total instantaneous power is
also important In a balanced three-phase circuit, this power has an
inter-esting property: It is invariant with time! Thus the torque developed at the
shaft of a three-phase motor is constant, which in turn means less
vibra-tion in machinery powered by three-phase motors
Let the instantaneous line-to-neutral voltage vA N be the reference,
and, as before, 0^ is the phase angle 0„A — 0/A Then, for a positive phase
sequence, the instantaneous power in each phase is
PA = v AN i aA = Vm I m cos art cos (a>t - 0 4 ),
PB = VBNA>B = V,„I m cos (art - 120°) cos (a>t - 0 , / , - 120°),
Pc = UCN'CC = V in I m cos(a)t + 120°) cos (art - 0 ^ , + 120°),
where Vm and Im represent the maximum amplitude of the phase voltage
and line current, respectively The total instantaneous power is the sum of
the instantaneous phase powers, which reduces to 1.5V m I m cos 9^, that is,
PT = PA + PB + Pc = 1.5VlfI/mcos00
Note this result is consistent with Eq 11.35 since Vm = V2~V'$ and
I m = V 2 / ^ (see Problem 11.26)
Examples 11.3-11.5 illustrate power calculations in balanced
three-phase circuits
V3 cos0rf
Trang 8Example 11.3 Calculating Power in a Three-Phase Wye-Wye Circuit
a) Calculate the average power per phase delivered
to the Y-connected load of Example 11.1
b) Calculate the total average power delivered to
the load
c) Calculate the total average power lost in the line
d) Calculate the total average power lost in the
generator
e) Calculate the total number of magnetizing vars
absorbed by the load
f) Calculate the total complex power delivered by
the source
Solution
a) From Example 11.1, V^ = 115.22 V, I 4 = 2.4 A,
and 0^ = -1.19 - (-36.87) = 35.68° Therefore
?$ = (115.22)(2.4) cos 35.68'
= 224.64 W
The power per phase may also be calculated
from l\R^ or
P+ = (2.4)2(39) = 224.64 W
b) The total average power delivered to the load is
p T = 3 / ^ = 673.92 W We calculated the line
voltage in Example 11.1, so we may also use
Eq 11.36:
P T = V3(199.58)(2.4) cos 35.68°
= 673.92 W
c) The total power lost in the line is
P Vme = 3(2.4)2(0.8) = 13.824 W
d) The total internal power lost in the generator is
gen 3(2.4)2(0.2) = 3.456 W
c) The total number of magnetizing vars absorbed
by the load is
Q T = V5( 199.58)(2.4) sin 35.68°
= 483.84 VAR
f) The total complex power associated with the source is
S T = 3St = -3(120)(2.4) /36.87°
= -691.20 - / 5 1 8 4 0 VA
The minus sign indicates that the internal power and magnetizing reactive power are being deliv-ered to the circuit We check this result by calcu-lating the total and reactive power absorbed by the circuit:
P = 673.92 + 13.824 + 3.456
= 691.20 W (check),
Q = 483.84 + 3(2.4)2(1.5) + 3(2.4)2(0.5)
= 483.84 + 25.92 + 8.64
= 518.40 VAR(check)
Example 11.4 Calculating Power in a Three-Phase Wye-Delta Circuit
a) Calculate the total complex power delivered to
the A-connected load of Example 11.2
b) What percentage of the average power at the
sending end of the line is delivered to the load?
Solution
a) Using the a-phase values from the solution of
Example 11.2, we obtain
V^ = VA B = 202.72 /29.04° V,
Ia = IA B = 1 3 9 / - 6 8 7aA
Using Eqs 11.52 and 11.53, we have
S T = 3(202.72 /29.04° )(1.39 /6.87°)
= 682.56 + /494.21 VA
b) The total power at the sending end of the distri-bution line equals the total power delivered to the load plus the total power lost in the line; therefore
/»input = 682.56 + 3(2.4)2(0.3)
= 687.74 W
The percentage of the average power reaching the load is 682.56/687.74, or 99.25% Nearly 100% of the average power at the input is delivered to the load because the impedance of the line is quite small compared to the load impedance
Trang 9Example 11.5 Calculating Three-Phase Power with an Unspecified Load
A balanced three-phase load requires 480 kW at a
lagging power factor of 0.8 The load is fed from a
line having an impedance of 0.005 + /0.025 0/<£
The line voltage at the terminals of the load is 600 V
a) Construct a single-phase equivalent circuit of
the system
b) Calculate the magnitude of the line current
c) Calculate the magnitude of the line voltage at
the sending end of the line
d) Calculate the power factor at the sending end of
the line
Solution
a) Figure 11.17 shows the single-phase equivalent
circuit We arbitrarily selected the line-to-neutral
voltage at the load as the reference
0.005 n /0.025 n
a • 'VW— —<'Y^'Y-^ 9 A
+
fa*
• —
— « »N
160 kW at 0.8 lag
Figure 11.17 • The single-phase equivalent circuit for
Example 11.5
b) The line current I*A is given by
600
V3 IaA = (160 + /120)103,
or
IaA = 577.35 /36.87° A
Therefore, Ia A = 577.35 / - 3 6 8 7 ° A The
mag-nitude of the line current is the magmag-nitude of IaA:
I L = 577.35 A
We obtain an alternative solution for IL from
the expression
p r
h
= V3V L I L cosB p
= V3(600)/L(0.8)
= 480,000 W;
480,000 V3(600)(0.8)
1000 V3
= 577.35 A
c) To calculate the magnitude of the line voltage at the sending end, we first calculate Van From Fig 11.17,
Vn AN
600
+ Z,l (HA
-z= + (0.005 + /0.025)(577.35/-36.87°)
= 357.51 /1.57° V
Thus
V L = V 3 | V:J
= 619.23 V
d) The power factor at the sending end of the line
is the cosine of the phase angle between Van and Ia A:
pf = cos [1.57° - (-36.87°)]
= cos 38.44°
= 0.783 lagging
An alternative method for calculating the power factor is to first calculate the complex power at the sending end of the line:
S (b = (160 + /12())103 + (577.35)2(0.005 + /0.025)
= 161.67 +/128.33 kVA
= 206.41 /38.44° kVA
The power factor is
pf = cos 38.44°
= 0.783 lagging
Finally, if we calculate the total complex power
at the sending end, after first calculating the magnitude of the line current, we may use this
value to calculate V L That is,
V3VJ L = 3(206.41) x 103,
3(206.41) X 103
Vi =
V3(577.35) '
= 619.23 V
Trang 10^ A S S E S S M E N T PROBLEMS
Objective 3—Be able to calculate power (average, reactive, and complex) in any three-phase circuit
11.8 The three-phase average power rating of the
central processing unit (CPU) on a mainframe
digital computer is 22,659 W The three-phase
line supplying the computer has a line voltage
rating of 208 V (rms) The line current is 73.8 A
(rms).The computer absorbs magnetizing VARs
a) Calculate the total magnetizing reactive
power absorbed by the CPU
b) Calculate the power factor
Answer: (a) 13,909.50 VAR;
(b) 0.852 lagging
NOTE: Also try Chapter Problems 11.22 and 11.24
11.9 The complex power associated with each phase
of a balanced load is 144 + /192 kVA The line voltage at the terminals of the load is 2450 V
a) What is the magnitude of the line current feeding the load?
b) The load is delta connected, and the imped-ance of each phase consists of a resistimped-ance in
parallel with a reactance Calculate R and X
c) The load is wye connected, and the imped-ance of each phase consists of a resistimped-ance in
series with a reactance Calculate R and X
Answer: (a) 169.67 A;
(b)R = 41.68 ft, X = 31.26 ft;
(c) R = 5 ft, X = 6.67 ft
11.6 Measuring Average Power
in Three-Phase Circuits
The basic instrument used to measure power in three-phase circuits is the
electrodynamometer wattmeter It contains two coils One coil, called the
current coil, is stationary and is designed to carry a current proportional to
the load current The second coil, called the potential coil, is movable and
carries a current proportional to the load voltage The important features
of the wattmeter are shown in Fig 11.18
The average deflection of the pointer attached to the movable coil is
proportional to the product of the effective value of the current in the
cur-rent coil, the effective value of the voltage impressed on the potential coil,
and the cosine of the phase angle between the voltage and current The
direction in which the pointer deflects depends on the instantaneous
polar-ity of the current-coil current and the potential-coil voltage.Therefore each
coil has one terminal with a polarity mark —usually a plus sign—but
some-times the double polarity mark ± is used The wattmeter deflects upscale
when (1) the polarity-marked terminal of the current coil is toward the
source, and (2) the polarity-marked terminal of the potential coil is
con-nected to the same line in which the current coil has been inserted
The Two-Wattmeter Method
Consider a general network inside a box to which power is supplied by
n conducting lines Such a system is shown in Fig 11.19
If we wish to measure the total power at the terminals of the box, we
need to know n — 1 currents and voltages This follows because if we
choose one terminal as a reference, there are only n - 1 independent
voltages Likewise, only n — 1 independent currents can exist in the n
con-ductors entering the box Thus the total power is the sum of n — 1 product
terms; that is, p = v x i\ + v 2 t2 + • • • + t>fl_rL_j
Watt scale
Current-coil terminals Potential-coil
terminals Pointer
Figure 11.18 • The key features of the
electrodynamometer wattmeter