AC Circuits 2
1
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
3
Draw voltage phasor diagram for the above circuit
4
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
2
  arctan
2
VL  VC
4
 arctan  arctan0.666  33.690
VR
6
VS  6  j 4 Volts
in Cartesian form
or 7.2V33.690 in Polarform
6
Power in A.C. Circuits
Consider the purely
resistive circuit shown. The
voltage and current are in
phase as shown (on the next
slide)
7
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
8
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
9
Resultant power curve for resistor in a.c. circuit
10
Resultant power curve for resistor in a.c. circuit
ˆIV
ˆ
average power 
watts
2
11
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
12
Voltage and current 900 out of phase
13
Voltage and current 300 out of phase
14
Voltage and current 600 out of phase
15
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)
16
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
17
Series Circuits
18
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
19
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
20
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 ?
21
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
22
And that the current flowing is Maximum
.
f 0  159 Hz
IMAX
Current
IMIN
Note that the Impedance is purely resistive and minimum at the
resonant frequency
23
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
24
25
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
26
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
52
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
27
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
28
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
29
Draw admittance phasor
YT  ( 0.5  j 0.1 ) S
 0.1118   63 .43 0 S
j
-j
30
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
31
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 
32
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
33
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
34