ASSIGNMENTS DUE
ELECTRIC CIRCUITS
ECSE-2010
Spring 2003
Class 26
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Today (Monday):
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Tuesday/Wednesday:
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Experiment #8 Report Due
Activities 27-1, 27-2 (In Class)
Thursday:
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Activities 28-1, 28-2 (In Class)
IMPEDANCE TRIANGLE
REVIEW
AC Power:
VRMS = Vm
Homework #9 Due
Activities 26-1, 26-2 (In Class)
AC Steady State Impedance
Imaginary
2 ;
I RMS = I m
2
Z = R + jX
X = Z sinθ
Z = Z /θ ; => Impedance Triangle
Real Power = P = VRMS I RMScosθ ; Watts
Z
jX
AC Reactance; [Ohms]
θ
Reactive Power = Q = VRMS I RMSsinθ ; VAr's
R
Complex Power = S = P + jQ; => Power Triangle
AC Resistance; [Ohms]
Power Factor = pf = cosθ ; lagging or leading
POWER TRIANGLE
REVIEW
Complex Power
S = P + jQ
Q = S sinθ
Reactive Power; [VAr's]
jQ
S
Real
R = Z cosθ
Apparent Power = S = VRMS I RMS ; VA
Imaginary
= Z /θ
= S /θ
θ
Real
P
P = S cosθ
Real Power; [Watts]
S = P 2 + Q 2 = VRMS I RMS
Apparent Power; [VA]
P
Power Factor = pf = cosθ =
S
• Power Systems:
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STotal = S1 + S2 + S3 + ….
STotal = (P1+P2+..)+j(Q1+Q2+..); Add P’s & Q’s
Use Power Triangle
Difficult definitions; Easy calculations
• Power Factor Correction:
• Add C or L to make pftotal = 1; Qtotal = 0
• Makes circuit most efficient
• Note that Xtotal = 0 if Qtotal = 0
• Will see this again later
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ACTIVITY 26-1
ACTIVITY 26-1
240 V (RMS)
• Standard Commercial Voltage
• Usually 120 or 240 Volts (RMS)
I hm
I
IC
60 Hz
• Input Source = 240 Volts (RMS), 60 Hz:
Heater
Motor
Ih
30 kW
18 kW
Im
• Heater; Specified as 18 kW
• Heater is Resistive Only
• PH = 18 kW; QH = 0
60 kVA
• Motor; Specified as 30 kW, 60 kVA:
• PM = 30 kW; /SM/ = 60 kVA
Find C to make Total Power Factor = 1
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Choose C such that pftotal = 1.0
ACTIVITY 26-2
ACTIVITY 26-2
• Input = 240 Volts (RMS), 60 Hz:
• Load 1 = Motor:
• Im,RMS = 50 A(RMS); pfm = 0.4 lagging
• Find Pm, Qm, Rm, Xm
Time to do one by yourself
Activity 26-2
• Add a Resistive Heater in Parallel
• Im+h,RMS = 65 A(RMS)
• Find Ph, Ih,RMS using Power Triangle
• Add C in Parallel to make pfsystem = 1:
• Find C and Isystem,RMS
• How many 150 uF C’s should you use?
• Find n and resulting Isystem,RMS
ACTIVITY 26-2
I hm
I
240 V (RMS)
60 Hz
COUPLED INDUCTORS
IC
Heater
I hm = 65 (RMS)
Ih
Motor
I m = 50 A(RMS)
pf m = 0.4 lagging
Im
+
v1 L1
−
+
L2 v
2
−
Find C to make Total Power Factor = 1
How many 150µ F C's should you use
Magnetic Fields of Inductors Link Each Other
to make pf as close to 1.0 as possible?
If Magnetic Field is time varying => Create Voltages
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COUPLED INDUCTORS
• Consider 2 Inductors, L1, L2:
• Inductors can be in same Circuit, or in
different Circuits but close to each other:
• Magnetic Field of L1 can link with L2 and
vice versa:
• If magnetic field changes with time =>Creates
time-varying voltages in linked coils L1, L2
• Good news: Can use to make Transformers
• Bad News: Happens even when we do not
want it to => Mutual Inductance
• Will look at both starting tomorrow
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