Transcript AC_Circuits2

AC Circuits 2
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For Capacitor
Current
For Inductance
Leads Voltage
Current
Voltage Leads
2
Real in phase quantities
along the real axis
Quadrature Components
Along the j – j axis
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Draw voltage phasor diagram for the above circuit
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VS
VL-VC
θ
VR
VS
Find VS
VS  6 2  4 2  7.2 Volts
VL  VC
4
  arctan
 arctan  arctan0.666  33.690
VR
6
5
VS
VL-VC
θ
VR
VS
VS  6  4  7.2 Volts
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  arctan
2
VL  VC
4
 arctan  arctan0.666  33.690
VR
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VS  6  j 4 Volts
in Cartesian form
or 7.2V33.690 in Polarform
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Power in A.C. Circuits
Consider the purely
resistive circuit shown. The
voltage and current are in
phase as shown (on the next
slide)
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Power in A.C. Circuits
Voltage and
current in phase
ˆ sin t  Iˆ sin ωt
Power  V
2
ˆ
ˆ
power  VI sin t
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Power in A.C. Circuits
ˆ sin t  Iˆ sin ωt
Power  V
2
ˆ
ˆ
power  VI sin t
using cos 2 A  1-2 sin 2 A
1
sin 2 A  (1  cos 2 A)
2
1
thus sin t  1  cos 2t 
2
2
1 ˆˆ
sub into above....... power  VI 1  cos 2ωt 
2
1 ˆˆ 1 ˆˆ
power  VI  VI cos 2ωt
2
2
Fixed component
Time varying
component
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Resultant power curve for resistor in a.c. circuit
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Resultant power curve for resistor in a.c. circuit
ˆIV
ˆ
average power 
watts
2
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Resultant power curve for resistor in a.c. circuit
ˆIV
ˆ
average power 
watts
2
Iˆ  I rms 2, Vˆ  Vrms 2
thus averagepower 
I rms 2 Vrms 2
 I rms Vrms watts
2
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Voltage and current 900 out of phase
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Voltage and current 300 out of phase
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Voltage and current 600 out of phase
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Apparent power (S) = V x I (volt-amperes, VA)
True power (active power) (P) = Apparent power x Cos θ (W)
Reactive power Q = VI sin θ reactive (voltamperes var)
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It can be shown that the power in an AC circuit is equal to
power  VI cos θ
Where  is the phase angle between the current and voltage
cos is known as the power factor p.f.
R
p.f. also equals
Z
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Series Circuits
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In a series circuit consisting of L,C and R if XC = XL the
voltage across each will be equal but in anti-phase. As can
be seen from the phasor diagram they will cancel so that the
series is effectively a pure resistance.
Adding VL and VC
VL
Leaves just VR
VR
This condition is known
as Series resonance
VC
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What happens if the frequency rises or falls ?
The frequency at which this condition occurs
is known as the Resonant Frequency
XC  X L
1
 2fL
2fC
1  4 f LC
2
f 
2
1
2 LC
Hertz
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Rememberthat the impedanceZ  R 2   X L  X C 
2

It should be clear that at resonance when X C  X L that
opposition to current flow will be minimum and only limited
by the ohmic resistance in the circuit
Also, as tan  
XC  X L
R
the phase difference betweenvoltage and current will be 0 0
What is the p.f. of a series resonant circuit ?
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What is the resonant
frequency of the circuit ?
f 
1
2 LC
Hertz
10 Volts
1
f 
 159.15 Hertz
2 1H 1F
Plot the impedance of the above series circuit from 140 Hertz
to 180 Hertz
Plot a graph of current flowing for the same frequency range
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And that the current flowing is Maximum
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f 0  159 Hz
IMAX
Current
IMIN
Note that the Impedance is purely resistive and minimum at the
resonant frequency
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Q factor
Q factor is a measure of merit for a resonant circuit
It is defined as the ratio of reactive power of either the
inductor or the capacitor to the average power of the
resistor at resonance
reactive power
QS 
average power
I 2XL
XL
 2 
R
I R
Note that the quantity is
dimensionless
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Admittance (Y) is defined as 1/Z
a.c. equivalent of the d.c. conductance G = 1/R
Unit is siemens (S)
Product
or for two use
Sum
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Admittance (Y) is defined as 1/Z
e.g. the Admittance of a circuit of impedance (4  j6) 
IS


1
4  6 j
4  6 j 4  6 j



S
4  6 j  4  6 j 4  6 j  16  36
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Note use of complex conjugate
Note also it would have been easy to convert to polar first
4  j 6  7.256.3
0
the reciprocal of which os 0.139S  56.30
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1
Susceptance B is the reciprocal of reactance i.e.
X
for inductance 
for capacitance 
1
X L90
0
1
XC  900
or
1
2fL90
0
siemens, S
or 2fC900 siemens, S
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Find admittance of each branch
Find the total i/p admittance
Calculate the i/p impedance
1
1
00 
0 o
R
20
 0.05S00  ( 0.05  j 0 )S
YR  G00 
YL 
1
1
0



90
X L900 10
0.1S  900  ( 0  j 0.1 )S
YT  YR  YL  0.05  j 0  0  j 0.1 S
 ( 0.5  j0.1 )S  0.1118  63.430 S
ZT 
1
 8.9463.430   ( 4  j 8 )
YT
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Draw admittance phasor
YT  ( 0.5  j 0.1 ) S
 0.1118   63 .43 0 S
j
-j
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Draw impedance vector
ZT 
1
 8.9463.430 
YT
 ( 4  j 8 )
Z=8.94Ω
63.430
It should be clear that
the parallel circuit may
be replaced by a series
circuit of (4+j8)Ω
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Determine the impedance
of the circuit
ZT  Z1 // Z 2
Use product over sum
ZT

6  j 64  j5

6  j 6  4  j5

54  j 610  j 

10  j 10  j 
54  j 6

10  j

534 j114
10  1
2
Or in polar form
2
conjugate
 5.287  j1.129
5.412 0 
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impedance of the circuit
 5.287  j1.128
5.412 0
Determine the current from
the source
VS
IS 
ZT

50V00
5.412
0
9.26 A  120
 9.06  j1.63A
Determine the current through Z1 and Z2
50V00
I Z1 

 4.166  j 4.166 A
6  j 6

IZ 2
50V00


4  j 5

 4.878  6.097A
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I S  9.06  j1.63 A


I Z1  4.166  j 4.166 A
I Z 2  4.878  6.097A
If the calculations up to now had not been rounded at each stage
then if the two branch currents were added they should give the
same value as the source current
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