REACTANCE. By: Enzo Paterno Date: 03/ PDF Free Download

REACTANCE REACTANCE By: Enzo Paterno Date: 03/2013 5/2007 Enzo Paterno 1

RESISTANCE - R i R (t R A resistor for all practical purposes is unaffected by the frequency of the applied sinusoidal voltage or current (i.e. assumes f < 100 khz As such, R can be treated as a constant and using Ohm s law: Z: Impedance R: Resistance Let v V I R R R : Z i R Z ( t ( t i + v R (t R R ( t I I P I P P [ Ω] sinωt R I Rsinωt sinωt [ A] Rsinωt R 0 [ V ] [ Ω] For a purely resistive element, v R (t and I R (t are in phase, with their peak values related by Ohm s law. 5/2007 Enzo Paterno 2 P

RESISTANCE - R 5/2007 Enzo Paterno 3

RESISTANCE - R 5/2007 Enzo Paterno 4

RESISTOR I Vs V i(t For a resistor V R (t & I R (t are in phase I P V R I P P IN PHASE V P v(t 0 5/2007 Enzo Paterno 5

INDUCTIVE REACTANCE - X L Let : v L V I L L ( t Z i L ( t I ω L L [ H ] i L (t + v L (t P X sinωt [ A] dil ( t L ω L I P cosωt ω L I P sin( ωt + 90 [ V ] dt ω L I P sin( ωt + 90 Z ω L 90 jω L [ Ω] I sinωt P L [ Ω] X L : Inductive Reactance For an inductor, the voltage is: v L ( t dil ( t L dt For an inductor v L (tleads i L (t by 90, or i L (t lags v L (t by 90 with their peak values related by Ohm s law. 5/2007 Enzo Paterno 6

INDUCTIVE REACTANCE - X L The Phase Relationship Between Inductor Voltage and Current: Voltage leads current by 90 Current lags voltage by 90 X L V I P P For an inductor V L leads I L by 90, or I L lags V L by 90 5/2007 Enzo Paterno 7

INDUCTIVE REACTANCE - X L i L (t L [ H ] + v L (t An inductor has a reactance which is a resistance that varies as a function of frequency. As such you might think of an inductor as a variable resistor whose resistance is controlled by the signal frequency applied to the inductor. XL( f 2πfL ωl [ Ω] f 0 f f 0 dc : f 0 X L 0 ( inductor acts as a short for f X L ( inductor acts as an open 5/2007 Enzo Paterno 8

INDUCTIVE REACTANCE - X L X L [ Ω ] Linear function XL { 1 2 L } 2πfL kf π R Increasing L 0 f [ Hz ] 5/2007 Enzo Paterno 9

INDUCTIVE REACTANCE - X L Example: Calculate the total current for the circuit below: I V X S L 12V 1kΩ 5/2007 Enzo Paterno 10

INDUCTIVE REACTANCE - X L Example: Calculate X L for the circuit below f 50 khz L 1 mh X L 2πfL 2π (50x10 3 (1x10 3 314Ω 5/2007 Enzo Paterno 11

INDUCTIVE REACTANCE - X L Example: Calculate the total current for the circuit below 10 V rms f 5 khz L 33 mh 3 3 X L 2πfL 2π (5x10 (33x10 1. 04 I VS 10V 9. 62 X 1.04 kω L ma kω 5/2007 Enzo Paterno 12

Let : dvc ( t ic ( t C ω CVP cosωt ω CVP sin( ωt + 90 dt VC VP sinωt 1 1 Z 90 j I ω CV sin( ωt + 90 ω C ω C C Z v CAPACITIVE REACTANCE - X C C ( t V 1 ω C i C (t P sinωt P X C + [ Ω] [ V ] C [ F ] + v C (t [ V ] [ Ω] For a capacitor i C (t leads v C (t by 90, or v C (t lags i C (t by 90 with their peak values related by Ohm s law. 5/2007 Enzo Paterno 13 For a capacitor, the current is: i ( t X C : Capacitive Reactance C C dv C dt ( t

CAPACITIVE REACTANCE - X C The Phase Relationship Between Capacitor Current and Voltage: Current leads voltage by 90 Voltage lags current by 90 V X C I P P For a capacitor I C leads V C by 90, or V C lags I C by 90 5/2007 Enzo Paterno 14

CAPACITIVE REACTANCE - X C i(t C [ F ] + f 0 f f 0 + v(t A capacitor has a reactance which is a resistance that varies as a function of frequency. As such you might think of a capacitor as a variable resistor whose resistance is controlled by the signal frequency applied to the capacitor. dc : for f f Xc 0 X c X c 1 2π f C 1 ωc ( capacitor 0 ( capacitor [ Ω] acts act as an open as a short 5/2007 Enzo Paterno 15

CAPACITIVE REACTANCE - X C Xc [ Ω ] Non Linear function Xc 1 k π 2π f C f { 1 2 C R } Increasing C 0 f [ Hz ] 5/2007 Enzo Paterno 16

CAPACITIVE REACTANCE - X C Example: Calculate the total current below 1 X C 2πfC 2π 1 ( 60 Hz( 22 μf 121Ω I C V X S C 10 V 121Ω 8.26 ma 5/2007 Enzo Paterno 17

CAPACITIVE REACTANCE - X C Series and Parallel Values of X C 5/2007 Enzo Paterno 18

REACTANCE X C - Summary Case X v I Purely Resistive R In phase In phase Purely Inductive ωl Leads I by 90 Lags v by 90 Purely Capacitive 1 Lags I by 90 Leads v by 90 ωc 5/2007 Enzo Paterno 19

EXAMPLE The voltage across a resistor is indicated. Find the sinusoidal expression for the current when the resistor is 10Ω. ( 377 + v( t 100sin t 60 Solution : Voltage and current, for a resistor, are in phase v 100sin(377t + 60 100 i( t sin(377t + R 10Ω 10 i( t 10sin(377t + 60 60 5/2007 Enzo Paterno 20

EXAMPLE The current through a 5Ω resistor is indicated. Find the sinusoidal expression for the voltage across the resistor if: ( 377 + i( t 40sin t 30 Solution : For a resistor the voltage and current are in phase v( t ir 40v (5Ωsin(377t v( t 200sin(377t + 30 + 30 200sin(377t + 30 5/2007 Enzo Paterno 21

EXAMPLE The current through a 0.1H coil is indicated. Find the sinusoidal expression for the voltage across the coil if: ( 377 i( t 7sin t 70 Solution : X V L For an inductor, v leads i by 90 v( t v( t (7A (37.7Ω ωl 377 rad/s (0.1H 263.9 sin(377t 263.9 sin(377t V L IL 263.9v 70 + + 20 X 37.7Ω 90 L 5/2007 Enzo Paterno 22

EXAMPLE The voltage across a 1 µf capacitor is indicated. Find the sinusoidal expression for the current through the capacitor if: v( t 30sin 400t Solution : X I C For an capacitor, i leads v by 90 i( t V X 1 ωc C 12x10 400 rad/s (1x 10 30v 2500Ω 3 12mA sin(400t 1 F 2500Ω 5/2007 Enzo Paterno 23-6 + 90 I C V X C C

EXAMPLE Given the voltage across a device and the current through this device, determine whether the device is a resistor, capacitor or inductor. v( t 100sin( ωt + 40 i( t 20sin( ωt + 40 Solution : Since v(t and i(t are in phase, V 100v R 5Ω I 20 A the device is a resistor 5/2007 Enzo Paterno 24

EXAMPLE Given the voltage across a device and the current through this device, determine whether the device is a resistor, capacitor or inductor. v( t 1000sin(377t + 10 i( t 5sin(377t 80 Solution : Since v(t leads i(t by 90, the device is an inductor V 1000v X L 200Ω I 5 A 200Ω X L ωl L 0. 531H 377 rad/s 5/2007 Enzo Paterno 26

EXAMPLE Given the voltage across a device and the current through this device, determine whether the device is a resistor, capacitor or inductor. v( t 500sin(157t + 30 i( t 1sin(157t + 120 Solution : Since i(t leads v(t by 90, the device is a capacitor V 500v X C 500Ω I 1A 1 1 X C C 12.74µ F ωc 157 rad/s (500Ω 5/2007 Enzo Paterno 28

EXAMPLE At what frequency will the reactance of a 200 mh inductor match the resistance level of a 5 kω resistor? 5/2007 Enzo Paterno 29

EXAMPLE At what frequency will the reactance of a 200 mh inductor match the resistance level of a 5 kω resistor? Solution : X L f X 2πfL L 2πL 5000 Ω 5000 Ω 1.257 3.98 khz 5/2007 Enzo Paterno 30

EXAMPLE At what frequency will an inductor of 5 mh have the same reactance as that of a 0.1 µf capacitor? 5/2007 Enzo Paterno 31

EXAMPLE At what frequency will an inductor of 5 mh have the same reactance as that of a 0.1 µf capacitor? Solution : X L 1 2πfL 2πfC f 2 f X C 1 2 4π LC 1 2π LC 7.12 khz 5/2007 Enzo Paterno 32

AVERAGE POWER AVERAGE POWER DUE TO AC SINUSOIDAL INPUTS 5/2007 Enzo Paterno 33

AVERAGE POWER i(t i( t I sin v( t V sin ( ωt + θ ( ωt + θ i v { P - + v(t L O A D p iv IV sinα sin β we get : sin We use the trig.identity : 1 2 ( ωt + θ sin( ωt + θ i [ cos( α β cos( α + β ] v p IV 2 cos IV 2 ( ( θ θ cos 2ωt + θ + θ v i v i 5/2007 Enzo Paterno 34

AVERAGE POWER p IV cos 2 IV v i t v θ 2 ( θ θ cos( 2ω + θ + i constant time-varying f(t P AVG 0 p AVG 5/2007 Enzo Paterno 35

AVERAGE POWER IV p AVG cos θ v θ i 2 IV 2 cosα Phase difference between I(t and v(t For Purely Resistive networks: i(t and v(t are in phase Get a maximum P AVG : P AVG IV / 2 For Purely Inductive or Capacitive networks: i(t and v(t are out of phase by 90 Get a minimum P AVG : P AVG 0 The average power provides a net transfer of energy It represents the power delivered to and dissipated by the load. 5/2007 Enzo Paterno 36

EXAMPLE P avg (V m I m / 2 cos θ v θ i Let α θ v θ i We get VmI m Vm I m P avg cosα cosα 2 2 2 RMS value of power equals peak since we are Dealing with average power (dc Which expresses P avg in terms of the effective RMS values P V avg rms I rms cosα 5/2007 Enzo Paterno 37

EXAMPLE Find the average power dissipated in a network having the following input voltage and current: v( t 100sin(377t + 40 i( t 20sin(377t + 70 α 40 70 30 IV p AVG cos θ v θi 2 IV 2 (100(20 p AVG cos30 2 cosα 866 5/2007 Enzo Paterno 38 W

EXAMPLE A series RLC circuit has the following impedance: Z R + JX L JX C Determine the frequency at which the average power is maximum. Solution: It occurs when the network is purely resistive with Z R Z R + for Z JX R L 1 2πfL 2πfC 1 f 2π LC JX X C L R X 5/2007 Enzo Paterno 39 C + J ( X 0 L X X L C X C

POWER FACTOR POWER FACTOR 5/2007 Enzo Paterno 40

POWER FACTOR In the equation P avg (V m I m / 2 cos α, (i.e. α θ v θ i, the factor that has significant control over the delivered power level is the cos α. No matter how large the voltage or current. When cos α 0, then P avg P min 0 When cos α 1, then P avg P max (V m I m / 2 Since cos α controls the power, the expression is named power factor and is defined by: 2Pavg Power factor Fp cosα V I m m V P rms avg I rms 5/2007 Enzo Paterno 41

POWER FACTOR For a purely resistive load, the phase angle between v and i is α 0 giving a power factor of 1 (F p cos α cos 0 1. The delivered power is then the maximum value of (V m I m / 2 watts. For a purely reactive load, (inductive or capacitive, the phase angle between v and i is α 90 giving a power factor of 0 ( F p cos α cos 90 0. The delivered power is then the minimum value of zero watts. When the load is a combination of resistive and reactive elements, the power factor will vary between 0 and 1. The more resistive the total impedance, the closer the power factor is to 1; the more reactive the total impedance, the closer the power factor is to 0. 5/2007 Enzo Paterno 42

POWER FACTOR The terms leading and lagging are often written in conjunction with the power factor. They are defined by the current through the load. If the current leads the voltage across a load, the load has a leading power factor. If the current lags the voltage across the load, the load has a lagging power factor. In general: Capacitive networks have leading power factors Inductive networks have lagging power factors. 5/2007 Enzo Paterno 43

POWER FACTOR Determine the power factor of the load given below, and indicate whether it is leading or lagging: i leads the voltage 5/2007 Enzo Paterno 44

POWER FACTOR Determine the power factor of the load given below, and indicate whether it is leading or lagging: i lags the voltage 5/2007 Enzo Paterno 45

POWER FACTOR Determine the power factor of the load given below, and indicate whether it is leading or lagging: Load is resistive F p neither leads or lag 5/2007 Enzo Paterno 46

POWER FACTOR wikipedia.org PHASORS 5/2007 Enzo Paterno 47

PHASORS As part of performing the analysis of an AC circuit, it will be required to perform mathematical operations with sinusoidal voltages and/or currents represented in the time domain. One lengthy but valid method of performing this operation is to place both sinusoidal waveforms on the same set of axes and add algebraically the magnitudes of each at every point along the abscissa. Long & Tedious process 5/2007 Enzo Paterno 48

PHASORS To alleviate this long and tedious process, one technique to perform such mathematical operations is to use PHASORS. A phasor is the polar form representation of the time domain voltage or current: Time domain Phasor domain Magnitude: The peak values I or V are converted to RMS Angle: Phase θ Time domain i(t I m v(t V m sin Peak value sin Example : ( ωt + θ m ( ωt + θ v θ 5/2007 Enzo Paterno 49 i RMS value Note : 0.707 I m θ 2 V 2 1 2 Phasor domain

PHASORS Phasors can also be converted back to their time domain format: Phasor domain i v(t I θ RMS value v V θ Example : i(t 2 I 2 m V sin Peak value m ( ωt + θ sin Note : 2 ( ωt + θ 1 0.707 Time domain 5/2007 Enzo Paterno 50

PHASORS Phasor domain Time domain 5/2007 Enzo Paterno 51

PHASORS Convert the following from the time to the phasor domain: Convert the following from the phasor to the time domain when f 60 Hz: 5/2007 Enzo Paterno 52

PHASORS Find the input voltage of the circuit below if: v v a b ( t ( t 50sin(377t 30sin(377t + 30 + 60 + + Va - + Vin Vb - Solution : KVL Vin Va + Vb Va (.70750 30 35.35 30 Vb (.70730 60 21.21 60-30.61+ 10.61+ j17.68 j18.37 5/2007 Enzo Paterno 53

PHASORS KVL Vin Va + Vb Vin Vin (30.61+ j17.68 + 10.61+ j18.37 41.22 + j36. l05 54.76 41.17 Thus; Vin 2(54.76 sin(377t + 41.17 5/2007 Enzo Paterno 54

PHASORS v v a b ( t ( t 50sin(377t 30sin(377t + 30 + 60 Vin 2 (54.76sin(377t + 41.17 5/2007 Enzo Paterno 55

COMPUTER ANALYSIS PSPICE 5/2007 Enzo Paterno 56

COMPUTER ANALYSIS Plots: P C V C I C 5/2007 Enzo Paterno 57

REACTANCE. By: Enzo Paterno Date: 03/2013