single phase ac circuit 02 · eee3405 eep ii-single phase ac circuit_02 example 17-6 a generator...
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27Week © Vocational Training Council, Hong Kong.
EEE3405 EEP II-Single Phase ac circuit_02
From the geometry of the power triangle,
22LQPS +=
In complex form,
S = P + jQL (for inductive circuit)
S = P – jQC (for capacitive circuit)
or S = VI* where I* is the conjugate of current I.
28Week © Vocational Training Council, Hong Kong.
EEE3405 EEP II-Single Phase ac circuit_02
Example 17-5
29Week © Vocational Training Council, Hong Kong.
EEE3405 EEP II-Single Phase ac circuit_02
30Week © Vocational Training Council, Hong Kong.
EEE3405 EEP II-Single Phase ac circuit_02
Example 17-6
A generator supplies power to an electric heater, an inductiveelement, and a capacitor as shown in Fig 17-17(a).
(a) Find P and Q for each load.
(b) Find total active and reactive power supplied by the generator.
(c) Draw the power triangle for the combined loads and determinetotal apparent power.
(d) Find the current supplied by the generator.
31Week © Vocational Training Council, Hong Kong.
EEE3405 EEP II-Single Phase ac circuit_02
VA8.3530811800j2400jQPS)c(
.)(indkVAR8.16002400Q
kW5.2P)b(
.)cap(VAR60024
120
X
VQ0P:Capacitor
.)ind(kVA4.26
120
X
VQ0P:Inductor
0QkW5.2P:Heater)a(
:Solution
0TTT
T
T
2
C
2
CC
2
L
2
LL
HH
∠=+=+=
=−==
====
====
==
32Week © Vocational Training Council, Hong Kong.
EEE3405 EEP II-Single Phase ac circuit_02
QT=1.8kVAR
35.8O
PT=2.5kW
ST=3081VA
The power triangle is drawn as follow:
AV
SId T 7.25
120
3081)( ===
33Week © Vocational Training Council, Hong Kong.
EEE3405 EEP II-Single Phase ac circuit_02
Power Factor
From the power triangle
P = VI cosθ = S cosθ (W)
Q = VI sinθ = S sinθ (VAR)
The quantities cosθ is defined as Power Factor FP
Fp = cosθ, cosθ = P/S.
Fp is expressed as a number(0 ≤ Fp ≤ 1) or a percent.
Power factor angle θ is the angle between applied voltage and
current or the angle between S and P
θ = cos-1(P/S)
34Week © Vocational Training Council, Hong Kong.
EEE3405 EEP II-Single Phase ac circuit_02
For pure resistance, θ = 0o
For pure inductance, θ = 90o
For pure capacitance, θ = -90o
For RL circuit, θ lies between 0o to 90o
For RC circuit, θ lies between 0o to -90o
For RLC circuit, -90o ≤ θ ≤ 90o
Unity,Lagging and Leading Power Factor
for a pure resistive circuit, θ = 0, FP = cos θ = 1
for a load containing RL, the current lags
voltage.(Lagging)
for a load containing RC, the current leads voltage. (Leading)
Thus, an inductive circuit has a lagging power factor while a
capacitive circuit has a leading power factor.
35Week © Vocational Training Council, Hong Kong.
EEE3405 EEP II-Single Phase ac circuit_02
Power Factor Correction
In figure (a), S=P, I=200A
However, in figure (b),
The size of electrical
apparatus required to supply
a load depends on its VA
requirements.
Ax
V
SI
kVA
QPS
3.333600
10200
200
160120
3
2222
===
=+=+=
36Week © Vocational Training Council, Hong Kong.
EEE3405 EEP II-Single Phase ac circuit_02
Power factor correction : To cancel some or all of the reactive
component of power by adding reactance of the opposite type
to the circuit.
If the reactive power is completely cancelled, this
is referred to as unity power factor correction.
Procedure : For a circuit with a fixed active power P,
find out the reactive power of the circuit Q
add an opposite component so that the total
reactive power QT is reduced
find out the new FP.
2T
2FQP
PP
+=
37Week © Vocational Training Council, Hong Kong.
EEE3405 EEP II-Single Phase ac circuit_02
Example 17-7
For the circuit of Fig 17-18(b), a capacitance with QC=160kVAR is added in parallel with the load as in Fig 17-19(a). Determine generator current I.
38Week © Vocational Training Council, Hong Kong.
EEE3405 EEP II-Single Phase ac circuit_02
39Week © Vocational Training Council, Hong Kong.
EEE3405 EEP II-Single Phase ac circuit_02
40Week © Vocational Training Council, Hong Kong.
EEE3405 EEP II-Single Phase ac circuit_02
41Week © Vocational Training Council, Hong Kong.
EEE3405 EEP II-Single Phase ac circuit_02
42Week © Vocational Training Council, Hong Kong.
EEE3405 EEP II-Single Phase ac circuit_02
43Week © Vocational Training Council, Hong Kong.
EEE3405 EEP II-Single Phase ac circuit_02
AC Series-Parallel Circuits (Ch. 18)
OBJECTIVES
• Compute impedance, currents and voltages in simple ac circuits
• Draw phasor diagrams for voltages and current
• Apply Ohm’s law to analyze simple series circuits
• Apply Kirchhoff’s laws in simple ac circuits
• Determine the series or parallel equivalent of any network consisting of a combination of resistors, inductors and capacitors.
44Week © Vocational Training Council, Hong Kong.
EEE3405 EEP II-Single Phase ac circuit_02
RL series circuit
IVR=IR
VL=IXLVS
θ R
fL2tan
R
Xtan
IR
IXtan
,IandVbetweenanglePhase
impedance)fL2(RXRZwhere
IZ
XRI
)IX()IR(
VVV
1L1L1
S
222L
2L
L
2L
2
2L
2
2L
2RS
π===θ
=π+=+=
=+=
+=
+=
−−−
XL
RVS
VL
VR
45Week © Vocational Training Council, Hong Kong.
EEE3405 EEP II-Single Phase ac circuit_02
RC series circuit
IVR=IR
VC=IXCVS
θ
fCR2
1tan
R
Xtan
IR
IXtan
,IandVbetweenanglePhase
impedance)fC2
1(RXRZwhere
IZ
XRI
)IX()IR(
VVV
1C1C1
S
222C
2C
C
2C
2
2C
2
2C
2RS
π===θ
=π
+=+=
=+=
+=
+=
−−−
RVS
VC
VR
XC
46Week © Vocational Training Council, Hong Kong.
EEE3405 EEP II-Single Phase ac circuit_02
RLC series circuit
R
XXtan
IR
)XX(Itan
,IandVbetweenanglePhase
impedance)XX(RZwhere
IZ
)XX(RI
)IXIX()IR(
)VV(VV
CL1CL1
S
2CL
2
2CL
2
2CL
2
2CL
2RS
−=−=θ
=−+=
=−+=
−+=
−+=
−−
IVR=IR
VL=IXL
VS
θ
VC =IXC
VL-VC=I(X L-XC)
XL
RVS
VL
VR
XC VC
47Week © Vocational Training Council, Hong Kong.
EEE3405 EEP II-Single Phase ac circuit_02
Ohm’s law for AC circuits
In Complex form,
Circuit Impedance
R-L in series R + jXL
R-C in series R – jXC
R-L-C in series R + j(XL – XC)
Z1, Z2, ….Zn in series Z1 + Z2 + …….. + Zn
48Week © Vocational Training Council, Hong Kong.
EEE3405 EEP II-Single Phase ac circuit_02
Example 18-5
For the circuit shown in Fig 18-20, find
a) the total impedance
b) the circuit current
c) voltage across each element of the circuit
Draw the phasor diagram
49Week © Vocational Training Council, Hong Kong.
EEE3405 EEP II-Single Phase ac circuit_02
Solution:
(a) ZT = 25 + j(200-225) = 25 – j25= 35.56∠-45o Ω
(b) I = V/ZT = 10 ∠0o/ 35.56∠-45o = 0.283 ∠45o A
(c) VR = IR = (0.283 ∠45o) (25) = 7.07 ∠45o V
VL = IXL = (0.283 ∠45o) (200 ∠90o) = 56.6 ∠135o V
VC = IXC = (0.283 ∠45o) (225 ∠-90o) = 63.6 ∠-45o VI
VR
VS
VL
VC
VC - VL
50Week © Vocational Training Council, Hong Kong.
EEE3405 EEP II-Single Phase ac circuit_02
AC Parallel Circuits
The admittance Y of any impedance is the reciprocal of the
impedance Z.
-jXC
jXL
G
Impedance
BC = 1/-jXC = j/XCC
BL = 1/jXL = -j/XLL
G=1/RR
AdmittanceCircuit element
51Week © Vocational Training Council, Hong Kong.
EEE3405 EEP II-Single Phase ac circuit_02
For n impedances connected in parallel
IZ
ZI
Y
YIor
VYZ
VI
Y.....YYY
x
T
T
xx
xx
x
n21T
==
==
+++=
V
52Week © Vocational Training Council, Hong Kong.
EEE3405 EEP II-Single Phase ac circuit_02
Example 18-16
Refer to the circuit of Fig 18-45, find:
(a) the total impedance, ZT(b) the supply current, IT(c) the branch currents, I1, I2 and I3.
53Week © Vocational Training Council, Hong Kong.
EEE3405 EEP II-Single Phase ac circuit_02
mA025.0)05)(010x05.0(EYI)b(
k020RY
1Z
mS005.0YYYY
mS9011000
jY
mS9011000
jY
mS005.0010x20
1Y)a(
:Solution
0003TT
0
TT
0321T
03
02
0031
∠=∠∠==
Ω∠===
∠=++=
∠==
−∠=−=
∠=∠
=
−
mA905)05)(90001.0(EYI
mA905)05)(90001.0(EYI
mA025.0IEYI)c(
00033
00022
0T11
∠=∠∠==
−∠=∠−∠==
∠===
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