REACTANCE By: Enzo Paterno Date: 03/2013...For Purely Inductive or Capacitive networks: i(t) and...

5/2007 Enzo Paterno 1

REACTANCE

REACTANCE By: Enzo Paterno Date: 03/2013


RESISTANCE - R

][

][0sin

sin][sin)()(

][sin)( :Let

Ω=⇒

Ω°∠===

==∴=

RZ

RtI

tRIZIV

VtRIRtitvAtIti

P

P

R

R

PRR

PR

ωω

ωω

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:

R iR (t)

+ vR(t)

For a purely resistive element, vR(t) and IR(t) are in phase, with their peak values related by Ohm’s law.

Z: Impedance R: Resistance


RESISTANCE - R


RESISTOR – I Vs V

i(t)

v(t) IN PHASE

0°

IP

VP

For a resistor VR(t) & IR(t) are in phase

P

P

IVR =


][

][90sin

)90sin(

][)90sin(cos)()(

][sin)( :Let

Ω==⇒

Ω=°∠=°+

==

°+===

=

L

P

P

L

L

PPL

L

PL

XLZ

LjLtI

tILZIV

VtILtILdt

tdiLtv

AtIti

ω

ωωωωω

ωωωω

ω

For an inductor vL(t)leads iL(t) by 90°, or iL(t) lags vL(t) by 90° with their peak values related by Ohm’s law.

L [ H ]

+

iL (t)

vL (t) dttdiLtv L

L)()( =

For an inductor, the voltage is:

INDUCTIVE REACTANCE - XL

XL: Inductive Reactance



The Phase Relationship Between Inductor Voltage and Current: Voltage leads current by 90° Current lags voltage by 90°

P

PL I

VX =

For an inductor VL leads IL by 90°, or IL lags VL by 90°


][2)( Ω== LfLfXL ωπ

)()(00:

openanasactsinductorXfforshortaasactsinductorXfdc

L

L

∞=∴∞→=∴=

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.

L [ H ]

+


iL (t)

vL (t) 0=f0=f ∞=f


XL [ Ω ]

f [ Hz ]

122 RLkffLXL ∈→= ππLinear function

0

Increasing L




Example: Calculate the total current for the circuit below:

kΩ1V12

==L

S

XVI



Example: Calculate XL for the circuit below

f = 50 kHz L

1 mH

Ω=== − 314)101)(1050(22 33 xxfLX L ππ



Example: Calculate the total current for the circuit below

10 Vrms f = 5 kHz

L 33 mH

Ω=== − kxxfLX L 04.1)1033)(105(22 33ππ

mAXVI

L

S 62.9kΩ1.04V10

===


][1

][1901)90sin(

sin

][)90sin(cos)()(

][sin)( :Let

Ω==⇒

Ω−=°−∠=°+

==

°+===

=

C

P

P

C

C

PPC

C

PC

XC

Z

Cj

CtVCtVZ

IV

VtVCtVCdt

tdvCti

VtVtv

ω

ωωωωω

ωωωω

ω

For a capacitor iC(t) leads vC(t) by 90°, or vC(t) lags iC(t) by 90° with their peak values related by Ohm’s law.

CAPACITIVE REACTANCE - XC

C [ F ]

+

iC (t)

vC (t)

++

For a capacitor, the current is:

dttdv

Cti CC

)()( =

XC: Capacitive Reactance


The Phase Relationship Between Capacitor Current and Voltage: Current leads voltage by 90° Voltage lags current by 90°


For a capacitor IC leads VC by 90°, or VC lags IC by 90°

P

PC I

VX =


][12

1Ω==

CCfXc

ωπ

)(0)(0:

shortaasactcapacitorXfforopenanasactscapacitorXfdc

c

c

=∴∞→∞=∴=

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.


C [ F ]

+

i(t)

v(t)

++

0=f0=f ∞=f


Xc [ Ω ]

f [ Hz ]

122

1 RCfk

CfXc ∈→= π

π

Non Linear function

0

Increasing C




Example: Calculate the total current below

mA 26.8Ω 121V 10

===C

SC X

VI

( )( ) Ω 121μF 22Hz 602π

12

1≅==

fCX C π



Series and Parallel Values of XC


REACTANCE – XC - 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


EXAMPLE

The voltage across a resistor is indicated. Find the sinusoidal expression for the current when the resistor is 10Ω.

( )°+= 60377sin100)( ttv

)60377sin(10)(

)60377sin(10

10010

)60377sin(100)(

phasein are resistor, afor current, and Voltage:Solution

°+=

°+=Ω

°+==

tti

ttRvti


The current through a 5Ω resistor is indicated. Find the sinusoidal expression for the voltage across the resistor if:

( )°+= 30377sin40)( tti

)30377sin(200)()30377sin(200)30377sin()5(40)(

phasein arecurrent and voltageheresistor t aFor :Solution

°+=°+=°+Ω==

ttvttviRtv

EXAMPLE


The current through a 0.1H coil is indicated. Find the sinusoidal expression for the voltage across the coil if:

( )°−= 70377sin7)( tti

)20377sin( 9.263)()9070377sin( 9.263)(

90by i leads vinductor,an For 263.9v)(37.7 )7(

7.37)1.0( rad/s 377X:Solution

L

°+=°+°−=°

=Ω=Ω===

ttvttv

AVHLω

EXAMPLE

LLL XIV =


The voltage across a 1 µF capacitor is indicated. Find the sinusoidal expression for the current through the capacitor if:

ttv 400sin30)( =

)90400sin( 1012)(90by vleads i capacitor,an For

122500

30

2500)10 x (1 rad/s 400

11X

:Solution

3

6-C

°+=

°

=Ω

==

Ω===

− txti

mAvXVI

FC

C

ω

EXAMPLE

C

CC

XVI =


Given the voltage across a device and the current through this device, determine whether the device is a resistor, capacitor or inductor.

)40sin(20)()40sin(100)(

°+=°+=

ttittv

ωω

Ω=== 5A 20

100R

resistor a is device thephase,in are and Since:Solution

vIV

i(t)v(t)

EXAMPLE



)80377sin(5)()10377sin(1000)(

°−=°+=

ttittv

EXAMPLE



)80377sin(5)()10377sin(1000)(

°−=°+=

ttittv

HLLX

vIV

i(t)v(t)

L 531.0rad/s 377

200

200A 5

1000X

inductoran is device the,90by leads Since:Solution

L

=Ω

=→=

Ω===

°

ω

EXAMPLE



)120157sin(1)()30157sin(500)(

°+=°+=

ttittv

EXAMPLE



)120157sin(1)()30157sin(500)(

°+=°+=

ttittv

FCC

X

vIV

v(t)i(t)

C µω

74.12)(500rad/s 157

11

500A 1

500X

capacitor a is device the,90by leads Since:Solution

C

=Ω

=→=

Ω===

°

EXAMPLE


At what frequency will the reactance of a 200 mH inductor match the resistance level of a 5 kΩ resistor?

EXAMPLE


At what frequency will the reactance of a 200 mH inductor match the resistance level of a 5 kΩ resistor?

kHz 98.31.257

5000L2

Xf

50002:Solution

L =Ω

==

Ω==

π

πfLX L

EXAMPLE


At what frequency will an inductor of 5 mH have the same reactance as that of a 0.1 µF capacitor?

EXAMPLE


At what frequency will an inductor of 5 mH have the same reactance as that of a 0.1 µF capacitor?

kHz 12.72

1LC4

12

12

X:Solution

22

L

==

=

=

=

LCf

f

fCfL

X C

π

π

ππ

EXAMPLE


AVERAGE POWER DUE TO AC SINUSOIDAL INPUTS

AVERAGE POWER


AVERAGE POWER

L O A D

i(t)

+

v(t) -

P

( )( )

( ) ( )

( ) ( )[ ]

( ) ( )

++−

−=

+−−=

++==

+=+=

iviv

vi

v

i

t

ttIVivp

tVtvtIti

θθωθθ

βαβαβα

θωθω

θωθω

2cos2

IVcos2

IVp

:get we

coscos21sinsin

:identity trig. theuse Wesinsin

sin)(sin)(


AVERAGE POWER

( ) ( )

++−

−= iviv t θθωθθ 2cos

2IVcos

2IVp

AVGp

constant time-varying f(t) PAVG = 0


AVERAGE POWER

For Purely Resistive networks: i(t) and v(t) are in phase Get a maximum PAVG : PAVG = IV / 2

αθθ cos2

IVcos2

IVpAVG =−= iv

Phase difference between I(t) and v(t)

For Purely Inductive or Capacitive networks: i(t) and v(t) are out of phase by 90° Get a minimum PAVG : PAVG = 0

The average power provides a net transfer of energy – It represents the power delivered to and dissipated by the load.


EXAMPLE

Pavg = (Vm Im / 2) cos | θv − θi | Let α = | θv − θi | We get Which expresses Pavg in terms of the effective RMS values

αα cos22

cos2

mmmmavg

IVIVP ==

αcosrmsrmsavg IVP =

RMS value of power equals peak since we are Dealing with average power (dc)


Find the average power dissipated in a network having the following input voltage and current:

EXAMPLE

)70377sin(20)()40377sin(100)(

°+=°+=

ttittv

αθθ cos2

IVcos2

IVpAVG =−= iv

°=°−°= 307040α

W 86630cos2

(100)(20)pAVG =°=


A series RLC circuit has the following impedance: Z = R + JXL – JXC Determine the frequency at which the average power is maximum. Solution: It occurs when the network is purely resistive with Z = R

EXAMPLE

LCf

fCfL

XXXXXXJRJXJXRZ

CLCL

CLCL

π

ππ

212

12

0 R for Z )(

=∴

=∴

==−=∴−+=−+=


POWER FACTOR

POWER FACTOR


In the equation Pavg = (Vm Im / 2) cos α, (i.e. α = | θv − θi |), 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 Pavg = Pmin = 0 When cos α = 1, then Pavg = Pmax = (Vm Im / 2) Since cos α controls the power, the expression is named power factor and is defined by:

POWER FACTOR

rmsrms

avg

mm

avgp IV

PIV

PFfactorPower ====

2cosα


For a purely resistive load, the phase angle between v and i is α = 0° giving a power factor of 1 (Fp cos α = cos 0° = 1). The delivered power is then the maximum value of (Vm Im / 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 ( Fp 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.

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.

POWER FACTOR


Determine the power factor of the load given below, and indicate whether it is leading or lagging:

i leads the voltage

POWER FACTOR



i lags the voltage

POWER FACTOR



Load is resistive Fp neither leads or lag

POWER FACTOR


PHASORS

POWER FACTOR

wikipedia.org


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.

PHASORS

Long & Tedious process


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

( )

( ) °∠=→+=

°∠=→+=

θθω

θθω

2sinv(t)

2sini(t)

:

mm

mm

VvtV

IitI

Example

Time domain Phasor domain

21707.0: =Note

Peak value RMS value

PHASORS


Phasors can also be converted back to their time domain format:

( )

( )θωθ

θωθ

+→°∠==

+=→°∠=

tVVv

tIIi

Example

m

m

sin2v(t)

sin2i(t)

:

Phasor domain Time domain

707.012: =Note

RMS value Peak value

PHASORS


PHASORS

Phasor domain Time domain


Convert the following from the time to the phasor domain:

PHASORS

Convert the following from the phasor to the time domain when f = 60 Hz:


Find the input voltage of the circuit below if:

)60377sin(30)()30377sin(50)(°+=°+=

ttvttv

b

a

+

Vin -

+ Va - +

Vb -

37.1861.106021.216030)707(.68.1761.303035.353050)707(.

:Solution

jVbjVa

VbVaVinKVL

+=°∠=°∠=+=°∠=°∠=

+=→

PHASORS


)17.41377sin()76.54(2

;

17.4176.5405.3622.4137.1861.10)68.1761.30(

°+=

°∠=+=+++=

+=→

tVin

Thus

ljVinjjVin

VbVaVinKVL

PHASORS


PHASORS

)60377sin(30)()30377sin(50)(°+=°+=

ttvttv

b

a

)17.41377sin()76.54(2 °+= tVin


COMPUTER ANALYSIS

PSPICE


COMPUTER ANALYSIS

Plots: PC VC IC

REACTANCE By: Enzo Paterno Date: 03/2013...For Purely Inductive or Capacitive networks: i(t) and...

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