fowler chapter 9 lecture 9 power in ac circuits
DESCRIPTION
FOWLER CHAPTER 9 LECTURE 9 POWER IN AC CIRCUITS. POWER IN RESISTIVE CIRCUITS, CHAP 9 WITH A RESISTIVE LOAD, CURRENT AND VOLTAGE ARE IN PHASE. F.9.1 THIS COULD BE AN ELECTRIC HEATER, STOVE, LAMP ETC. Ac source drives a purely resistive load. V. I. - PowerPoint PPT PresentationTRANSCRIPT
![Page 1: FOWLER CHAPTER 9 LECTURE 9 POWER IN AC CIRCUITS](https://reader035.vdocument.in/reader035/viewer/2022081506/56813027550346895d95b0b0/html5/thumbnails/1.jpg)
FOWLER CHAPTER 9LECTURE 9 POWER IN AC
CIRCUITS
![Page 2: FOWLER CHAPTER 9 LECTURE 9 POWER IN AC CIRCUITS](https://reader035.vdocument.in/reader035/viewer/2022081506/56813027550346895d95b0b0/html5/thumbnails/2.jpg)
POWER IN RESISTIVE CIRCUITS, CHAP 9WITH A RESISTIVE LOAD, CURRENT AND VOLTAGE ARE IN PHASE. F.9.1THIS COULD BE AN ELECTRIC HEATER, STOVE, LAMP ETC.
Ac source drives a purely resistive load.
In a purely resistive circuit, all circuit power is dissipated by the resistor(s). Voltage and current are in phase with each other.
V
I
![Page 3: FOWLER CHAPTER 9 LECTURE 9 POWER IN AC CIRCUITS](https://reader035.vdocument.in/reader035/viewer/2022081506/56813027550346895d95b0b0/html5/thumbnails/3.jpg)
+POWER= (-CURRENT)X(-VOLTAGE)
POWER IN OUT OF PHASE CIRCUITSREACTANCE IS LIKE RESISTANCE IN AC CIRCUITSWHICH INCULDES CAPACITORS AND INDUCTORS.
R
VRIIVP
22
RESISTORS
INDUCTORS
CAPACITORS
![Page 4: FOWLER CHAPTER 9 LECTURE 9 POWER IN AC CIRCUITS](https://reader035.vdocument.in/reader035/viewer/2022081506/56813027550346895d95b0b0/html5/thumbnails/4.jpg)
WHAT DOES NEGETIVE POWER MEAN?
POWER IS FLOWING BACK FROM THE LOAD TO THE SOURCE. WITH A 90° PHASE SHIFT, NO POWER IS USED BY REACTIVE COMPOENTS,ONLY RESISTANCE USES POWER.
POWER IN RESISTIVE, REACTIVE CIRCUITS CAN BE FOUND FROM,
P =IVCOSØFOR RESISITANCE ONLY CIRCUITS: Ø = 0, SO COS(0°) = 1THEN P= IV(1) =IV
Ø
IT
IR
COSØ =IR/IT
COS IS A MATH FUNCTION THAT CAN VARY FROM 1 TO 0FOR A ANGLE IN DEGREES FROM 0° TO 90°.
THERE IS NO EASY WAY TO CALCULATE POWER WHEN PHASE SHIFTS OCCUR.
WHEN CURRENT AND VOLTAGE ARE OPPOSITE; (Neg. or pos.) voltage times (Neg. or pos.) current = - power
![Page 5: FOWLER CHAPTER 9 LECTURE 9 POWER IN AC CIRCUITS](https://reader035.vdocument.in/reader035/viewer/2022081506/56813027550346895d95b0b0/html5/thumbnails/5.jpg)
FOR RESISTIVE CIRCUITS ONLY: LOAD CURRENT AND VOLTAGE ARE IN PHASE.
![Page 6: FOWLER CHAPTER 9 LECTURE 9 POWER IN AC CIRCUITS](https://reader035.vdocument.in/reader035/viewer/2022081506/56813027550346895d95b0b0/html5/thumbnails/6.jpg)
IN A CIRCUIT WITH CAPACITANCE ONLY: I LEADS V BY 90º
![Page 7: FOWLER CHAPTER 9 LECTURE 9 POWER IN AC CIRCUITS](https://reader035.vdocument.in/reader035/viewer/2022081506/56813027550346895d95b0b0/html5/thumbnails/7.jpg)
IN A CIRCUIT WITH INDUCTANCE V LAGS I BY 90º
![Page 8: FOWLER CHAPTER 9 LECTURE 9 POWER IN AC CIRCUITS](https://reader035.vdocument.in/reader035/viewer/2022081506/56813027550346895d95b0b0/html5/thumbnails/8.jpg)
If a sinusoidal voltage is applied to an resistive circuit with a phase angle of 0o, the resulting voltage and current waveforms will look like this
Given that power is the product of voltage and current (p = i v), let’s look at the waveform for power in this circuit.
![Page 9: FOWLER CHAPTER 9 LECTURE 9 POWER IN AC CIRCUITS](https://reader035.vdocument.in/reader035/viewer/2022081506/56813027550346895d95b0b0/html5/thumbnails/9.jpg)
If the power waveform is plotted for a resistive AC circuit, it will look like this
V
![Page 10: FOWLER CHAPTER 9 LECTURE 9 POWER IN AC CIRCUITS](https://reader035.vdocument.in/reader035/viewer/2022081506/56813027550346895d95b0b0/html5/thumbnails/10.jpg)
Circuit with a capacitor current leads voltage. Circuit with a inductor voltage leads current.
![Page 11: FOWLER CHAPTER 9 LECTURE 9 POWER IN AC CIRCUITS](https://reader035.vdocument.in/reader035/viewer/2022081506/56813027550346895d95b0b0/html5/thumbnails/11.jpg)
Current
Current (red)
Current
Current
Voltage (green)
Voltage
VoltageVoltage
Power Power
Power Power
30º
90º60º
(a) No phase shift(b) 30º phase shift
(c) 60º phase shift (d) 90º phase shift
Fig. 9-3 Power in phase-shifted circuits. At 90º of phase shift, the power is zero.
+ +
++
- -
--
![Page 12: FOWLER CHAPTER 9 LECTURE 9 POWER IN AC CIRCUITS](https://reader035.vdocument.in/reader035/viewer/2022081506/56813027550346895d95b0b0/html5/thumbnails/12.jpg)
RESISTIVE ONLY
INDUCTIVE ONLY
CAPACITIVE ONLY
ALL 3 COMBINED
![Page 13: FOWLER CHAPTER 9 LECTURE 9 POWER IN AC CIRCUITS](https://reader035.vdocument.in/reader035/viewer/2022081506/56813027550346895d95b0b0/html5/thumbnails/13.jpg)
TRUE POWERIS THE ACTUAL POWER USED BY THE CIRCUIT.IT IS MEASURED WITH A WATTMETER.
APPARENT POWERPOWER IN A CIRCUIT WHEN VOLTAGE AND POWER ARE MEASURED SEPARATELYIT IS CALCULATED IN UNITS OF VOLTAMPERE. (VA)
PTRUE = IVCOSØPAPPARENT =IV
POWER FACTORIS THE RATIO OF TRUE POWER/APPARENT POWER.WHEN CURRENT AND VOLTAGE ARE IN PHASE THE POWER FACTOR = 1.IF 90° OUT OF PHASE THE POWER FACTOR = 0.POWER FACTOR OF A CIRCUIT CAN VARY BETWEEN O AND 1.
![Page 14: FOWLER CHAPTER 9 LECTURE 9 POWER IN AC CIRCUITS](https://reader035.vdocument.in/reader035/viewer/2022081506/56813027550346895d95b0b0/html5/thumbnails/14.jpg)
Power Factor (PF)Power factor is defined as the ratio of true power (measured in watts) to apparent power (measured in Volt Amps). It measures how effectively AC power is being used by a device. The difference between true power and apparent power is expressed as the power factor and results from the way true power and apparent power are measured. Ideally we would like to have true power and apparent power equal to one another, which would result in a PF of 1.00 or 100% effective power utilization. AC Volts x AC Amps = VA (Volt Amp) Purely Resistive AC Load: VA = Watts (same as DC circuits)
Inductive/Reactive AC Load: VA x PF = WattsAC Volts x AC Amps x PF = Watts
![Page 15: FOWLER CHAPTER 9 LECTURE 9 POWER IN AC CIRCUITS](https://reader035.vdocument.in/reader035/viewer/2022081506/56813027550346895d95b0b0/html5/thumbnails/15.jpg)
A Power Factor of 0.75, means that an installation is using 75% of the power being supplied to it.
![Page 16: FOWLER CHAPTER 9 LECTURE 9 POWER IN AC CIRCUITS](https://reader035.vdocument.in/reader035/viewer/2022081506/56813027550346895d95b0b0/html5/thumbnails/16.jpg)
Aerovox manufactures both single and three-phase power factor correction capacitors up to 4,800 VAC
![Page 17: FOWLER CHAPTER 9 LECTURE 9 POWER IN AC CIRCUITS](https://reader035.vdocument.in/reader035/viewer/2022081506/56813027550346895d95b0b0/html5/thumbnails/17.jpg)
GROUND
NEUTRAL
LINE 1
LINE 2
LINE 3
PHASE 1 277 V
PHASE 2 277 V
PHASE 3 277 V
277 V
277 V
277 V 480 V
480 V
480 V
3 PHASE 277 V/480V, 4 WIRE WYE SYSTEM
TO 3 PHASE LOADS
UNDER LOAD:LINE AND PHASE CURRENTS ARE NOT EQUAL.SINCE 2 PHASE VOLTAGES ARE SEPARATED BY 120º , THEY CANNOT BE ADDED TOGETHERILINE = 1.732IPHASE
VLINE = 1.732VPHASE
VLINE2 = 1.732(277V) =480 VSINGLE PHASE 277 V ARE CONNECTED BETWEEN THE NEUTRAL AND ANY LINE.SINGLE PHASE480 V CIRCUITS ARE CONNECTED BETWEEN ANY 2 OF THE 3 LINES.3 PHASE 480 V ARE CONNECTED ACROSS 3LINES.
![Page 18: FOWLER CHAPTER 9 LECTURE 9 POWER IN AC CIRCUITS](https://reader035.vdocument.in/reader035/viewer/2022081506/56813027550346895d95b0b0/html5/thumbnails/18.jpg)
RESISTIVE LOAD
IF WE ADD A REACTIVE ELEMENT (CAPACITOR OR INDUCTOR) TO THIS CIRCUIT, THE PF WOULD BE REDUCED.
![Page 19: FOWLER CHAPTER 9 LECTURE 9 POWER IN AC CIRCUITS](https://reader035.vdocument.in/reader035/viewer/2022081506/56813027550346895d95b0b0/html5/thumbnails/19.jpg)
METER USED TO MEASURE 3 PHASE POWER FACTOR