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. 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

+POWER= (-CURRENT)X(-VOLTAGE) POWER IN OUT OF PHASE CIRCUITS REACTANCE IS LIKE RESISTANCE IN AC CIRCUITS WHICH INCULDES CAPACITORS AND INDUCTORS. RESISTORS INDUCTORS CAPACITORS

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 =IVCOSØ FOR RESISITANCE ONLY CIRCUITS: Ø = 0, SO COS(0°) = 1 THEN P= IV(1) =IV Ø ITIT IRIR COSØ = I R/ I T COS IS A MATH FUNCTION THAT CAN VARY FROM 1 TO 0 FOR 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

FOR RESISTIVE CIRCUITS ONLY: LOAD CURRENT AND VOLTAGE ARE IN PHASE.

IN A CIRCUIT WITH CAPACITANCE ONLY: I LEADS V BY 90º

IN A CIRCUIT WITH INDUCTANCE V LAGS I BY 90º

If a sinusoidal voltage is applied to an resistive circuit with a phase angle of 0 o, 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.

If the power waveform is plotted for a resistive AC circuit, it will look like this V

Circuit with a capacitor current leads voltage.Circuit with a inductor voltage leads current.

Current (red) Current Voltage (green) Voltage Power 30º 90º 60º (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

RESISTIVE ONLY INDUCTIVE ONLY CAPACITIVE ONLY ALL 3 COMBINED

TRUE POWER IS THE ACTUAL POWER USED BY THE CIRCUIT. IT IS MEASURED WITH A WATTMETER. APPARENT POWER POWER IN A CIRCUIT WHEN VOLTAGE AND POWER ARE MEASURED SEPARATELY IT IS CALCULATED IN UNITS OF VOLTAMPERE. (VA) P TRUE = IVCOSØ P APPARENT =IV POWER FACTOR IS 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.

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 = Watts AC Volts x AC Amps x PF = Watts

A Power Factor of 0.75, means that an installation is using 75% of the power being supplied to it.

Aerovox manufactures both single and three-phase power factor correction capacitors up to 4,800 VAC

GROUND NEUTRAL LINE 1 LINE 2 LINE 3 PHASE V PHASE V PHASE 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 TOGETHER I LINE = 1.732I PHASE V LINE = 1.732V PHASE V LINE 2 = 1.732(277V) =480 V SINGLE 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.

RESISTIVE LOAD IF WE ADD A REACTIVE ELEMENT (CAPACITOR OR INDUCTOR) TO THIS CIRCUIT, THE PF WOULD BE REDUCED.

METER USED TO MEASURE 3 PHASE POWER FACTOR