REACTANCE By: Enzo Paterno Date: 03/2013...For Purely Inductive or Capacitive networks: i(t) and...
Transcript of 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
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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
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RESISTANCE - R
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RESISTANCE - R
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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 =
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][
][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
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INDUCTIVE REACTANCE - XL
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°
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][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 ]
+
INDUCTIVE REACTANCE - XL
iL (t)
vL (t) 0=f0=f ∞=f
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XL [ Ω ]
f [ Hz ]
122 RLkffLXL ∈→= ππLinear function
0
Increasing L
INDUCTIVE REACTANCE - XL
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INDUCTIVE REACTANCE - XL
Example: Calculate the total current for the circuit below:
kΩ1V12
==L
S
XVI
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INDUCTIVE REACTANCE - XL
Example: Calculate XL for the circuit below
f = 50 kHz L
1 mH
Ω=== − 314)101)(1050(22 33 xxfLX L ππ
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INDUCTIVE REACTANCE - XL
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
===
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][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
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The Phase Relationship Between Capacitor Current and Voltage: Current leads voltage by 90° Voltage lags current by 90°
CAPACITIVE REACTANCE - XC
For a capacitor IC leads VC by 90°, or VC lags IC by 90°
P
PC I
VX =
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][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.
CAPACITIVE REACTANCE - XC
C [ F ]
+
i(t)
v(t)
++
0=f0=f ∞=f
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Xc [ Ω ]
f [ Hz ]
122
1 RCfk
CfXc ∈→= π
π
Non Linear function
0
Increasing C
CAPACITIVE REACTANCE - XC
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CAPACITIVE REACTANCE - XC
Example: Calculate the total current below
mA 26.8Ω 121V 10
===C
SC X
VI
( )( ) Ω 121μF 22Hz 602π
12
1≅==
fCX C π
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CAPACITIVE REACTANCE - XC
Series and Parallel Values of XC
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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
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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
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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
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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 =
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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 =
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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
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Given the voltage across a device and the current through this device, determine whether the device is a resistor, capacitor or inductor.
)80377sin(5)()10377sin(1000)(
°−=°+=
ttittv
EXAMPLE
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Given the voltage across a device and the current through this device, determine whether the device is a resistor, capacitor or inductor.
)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
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Given the voltage across a device and the current through this device, determine whether the device is a resistor, capacitor or inductor.
)120157sin(1)()30157sin(500)(
°+=°+=
ttittv
EXAMPLE
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Given the voltage across a device and the current through this device, determine whether the device is a resistor, capacitor or inductor.
)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
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At what frequency will the reactance of a 200 mH inductor match the resistance level of a 5 kΩ resistor?
EXAMPLE
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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
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At what frequency will an inductor of 5 mH have the same reactance as that of a 0.1 µF capacitor?
EXAMPLE
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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
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AVERAGE POWER DUE TO AC SINUSOIDAL INPUTS
AVERAGE POWER
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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)(
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AVERAGE POWER
( ) ( )
++−
−= iviv t θθωθθ 2cos
2IVcos
2IVp
AVGp
constant time-varying f(t) PAVG = 0
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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.
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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)
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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 =°=
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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 )(
=∴
=∴
==−=∴−+=−+=
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POWER FACTOR
POWER FACTOR
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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
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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
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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
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Determine the power factor of the load given below, and indicate whether it is leading or lagging:
i leads the voltage
POWER FACTOR
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Determine the power factor of the load given below, and indicate whether it is leading or lagging:
i lags the voltage
POWER FACTOR
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Determine the power factor of the load given below, and indicate whether it is leading or lagging:
Load is resistive Fp neither leads or lag
POWER FACTOR
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PHASORS
POWER FACTOR
wikipedia.org
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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
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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
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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
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PHASORS
Phasor domain Time domain
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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:
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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
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)17.41377sin()76.54(2
;
17.4176.5405.3622.4137.1861.10)68.1761.30(
°+=
°∠=+=+++=
+=→
tVin
Thus
ljVinjjVin
VbVaVinKVL
PHASORS
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PHASORS
)60377sin(30)()30377sin(50)(°+=°+=
ttvttv
b
a
)17.41377sin()76.54(2 °+= tVin
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COMPUTER ANALYSIS
PSPICE
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COMPUTER ANALYSIS
Plots: PC VC IC