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ET 242 Circuit Analysis II
Parallel AC Circuits Analysis
Electrical and Telecommunication
Engineering Technology
Professor Jang
Acknowledgement
I want to express my gratitude to Prentice Hall giving me the permission
to use instructor’s material for developing this module. I would like to
thank the Department of Electrical and Telecommunications Engineering
Technology of NYCCT for giving me support to commence and complete
this module. I hope this module is helpful to enhance our students’
academic performance.
OUTLINES
 Introduction
to Parallel ac Circuits Analysis
 Impedance and Phase Diagram
 Parallel Configuration
 Current Divider Rule
 Frequency Response for Parallel ac Circuits
 Phase Measurements
Key Words: Parallel ac Circuit, Impedance, Phase, Frequency Response
ET 242 Circuit Analysis II – Parallel ac circuits analysis
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Parallel ac Networks
For the representative parallel ac
network in Fig. 15.67, the total
impedance or admittance is determined
as previously described, and the source
current is determined by Ohm’s law as
follows:
E
I
 EYT
ZT
Since the voltage is the same across
parallel elements, the current through
each branch can be found through
another application of Ohm’s law:
Figure 15.67 Parallel ac network.
E
I1 
 EY1
Z1
and
E
I2 
 EY2
Z2
KCL can then be applied in the same manner as used for dc networks with
consideration of the quantities that have both magnitude and direction.
I – I 1 – I2 = 0
or
I = I 1 + I2
The power to the network can be determined by P = EIcosθT
ET 242θCircuit
Analysis
II – Sinusoidal
Alternating E
Waveforms
where
phase
angle between
and I.
T is the
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Parallel ac Networks : R-L
Phasor Notation:
As shown in Fig. 15.69.
YT  YR  YL
1
1
0 
  90
3.33
2.5
 0.3S0  0.4S  90  0.3S  j 0.4S
 G0  BL   90 
Figure 15.68 Parallel R-L network.
 0.5S  53.13
ZT 
1
1

 253.13
YT 0.5S  53.13
Admittance diagram : As shown in Fig .15.70
I
E
Figure 15.69 Applying
 EYT  (20V53.13)(0.5S  53.13)  10 A0 phasor notation to the
ZT
E
 ( E )(G0)
R0
 (20V53.13)(0.3S0)  6 A53.13
network in Fig. 15.68.
IR 
IL 
E
 ( E )( BL   90)
X L 90
ET162
Analysis
 (20Circuit
V53
.13)(–0Ohm’s
.4 SLaw
 90)
Figure 15.70 Admittance diagram
for the parallel R-L network in Fig. 15.68.
 8 A  36.87
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KCL : At node a,
I  I R  I L  0 or
I  IR  IL
10A0  6A53.13  8A  36.87
10A0  (3.60A  j4.8A)  (6.40A  j4.80A)
 10A  j0
and 10A0  10A0 (checks)
Phasor diagram: The
phasor diagram in Fig. 15.71
indicates that the applied
voltage E is in phase with
the current IR and leads the
current IL by 90o.
Power : The total power in watts delivered to the circuit is
PT  EIcosθT
 (20V)(10A)cos53.13  (200W)(0.6 )
 120 W
Power factor : The power factor of the circuit is
Fp  cosT  cos53.13  0.6 lagging
Figure 15.71 Phasor diagram for the parallel R-L network in Fig. 15.68.
ET 242 Circuit Analysis II – Parallel ac circuits analysis
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Parallel ac Networks : R-C
Phasor Notation:
As shown in Fig. 15.73.
YT  YR  YC  G0  BC 90
1
1
0 
90
1.67Ω
1.25Ω
 0.6S0  0.8S90

Figure 15.72 Parallel R-C network.
 0.6S  j0.8S  1.0S53.13
ZT 
1
1

 1Ω  53.13
YT 1.0S53.13
Figure 15.73 Applying phasor
notation to the network in Fig. 15.72.
Admittance diagram : As shown in Fig .15.74.
E  IZ T 
1
10A0

 10V  53.13
YT 1S53.13
I R  (Eθ)(G0 )
 (10V  53.13 )(0.6S0 )  6A  53.13
I C  (Eθ)(BC 90 )
 (10V  53.13 )(0.8S90 )  8A36.87
ET 242 Circuit Analysis II – Parallel ac circuits analysis
Figure 15.74
Admittance diagram
for the parallel R-C
network in Fig. 15.72.
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KCL : At node a,
I  I R  I C  0 or I  I R  I C
Power factor : The power factor of the circuit is
Fp  cosT  cos53.13  0.6 leading
Power : PT  EIcosθT  (10V)(10A)cos53.13  (100W)(0.6S)  60 W
Parallel ac Networks : R-L-C
Phasor Notation:
As shown in Fig. 15.78.
Figure 15.77 Parallel R-L-C network.
YT  YR  YL  YC  G0  BL   90  BC 90
Figure 15.78 Applying phasor notation
to the network in Fig. 15.77.
1
1
1
0 
  90 
90
3.33Ω
1.43Ω
3.33Ω
 0.3S0  0.7S  90  0.3S90

 0.3S  j0.7S  j0.3S
 0.3S  j0.4S  0.5S  53.13
ET 242 Circuit Analysis II – Parallel ac circuits analysis
1
1

 2Ω53.13
Boylestad
YT 0.5S  53.13
ZT 
2
Admittance diagram : As shown in Fig .15.79.
I
E
 EYT  (100V53.13 )(0.5S  53.13)  50A0
ZT
I R  (Eθ)(G0 )
 (100V53.13 )(0.3S0 )  30A53.13
I L  (Eθ)(BL   90 )
 (100V53.13 )(0.7S  90 )  70A  36.87
I C  (Eθ)(BC 90 )
 (100V53.13 )(0.3S  90 )  30A143.13
KCL : At node a,
I  I R  I L  I C  0 or I  I R  I L  I C
Figure 15.79 Admittance diagram
for the parallel R-L-C network in Fig. 15.77.
Power :
PT  EIcosθT  (100V)(50A)cos53.13  (5000W)(0. 6S)  3000 W
Power factor : The power factor of the circuit is
Fp  cosT  cos53.13  0.6 lagging
ET 242 Circuit Analysis II – Parallel ac circuits analysis
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Current Divider Rule
The basic format for the current divider rule in ac circuit exactly the same as
that dc circuits; for two parallel branches with impedance in Fig. 15.82.
Z 2 IT
Z1 I T
I1 
or I 2 
Z1  Z 2
Z1  Z 2
Figure 15.82 Applying the current divider rule.
Ex. 15-16 Using the current divider rule, find the current through each parallel
branch in Fig. 15.83.
Z L IT
(4Ω90 )(20A0 )
IR 

ZR  ZL
3Ω0  4Ω90
80A90
 16A36.87
553.13
Z R IT
(3Ω0 )(20A0 )
IL 

ZR  ZL
553.13

FIGURE 15.83
ET 242 Circuit Analysis II – Parallel ac circuits analysis
60A0

 12A  53.13
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553.13
2
Ex. 15-17 Using the current divider rule, find the current through each parallel
branch in Fig. 15.84.
I R- L 
Z C IT
(2Ω  90 )(5A30 ) 10A  60


Z C  Z R- L
 j2Ω  1Ω  j8Ω
1  j6
10A - 60
 1.64A - 140.54
6.08380.54
Z R- L IT
(1Ω  j8Ω8Ω)(30 ) 10A  60
IC 


Z R- L  Z C
6.08380.54
1  j6

(8.06A82.87 )(5A30 )
6.08380.54
40.30A112.87

 6.63A32.33
6.08380.54

FIGURE 15.84
Frequency Response of Parallel Elements
For parallel elements, it is important to remember that the smallest parallel
resistor or the smallest parallel reactance will have the most impact on the real
or imaginary component, respectively, of the total impedance.
ET 242 Circuit Analysis II – Parallel ac circuits analysis
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In Fig. 15.85, the frequency response has been included for each element of a
parallel R-L-C combination. At very low frequencies, the importance of the coil
will be less than that of the resistor or capacitor, resulting in an inductive network
in which the reactance of the inductor will have the most impact on the total
impedance. As the frequency increases, the impedance of the inductor will increase
while the impedance of the capacitor will decrease.
FIGURE 15.85 Frequency response for parallel R-L-C elements.
ET 242 Circuit Analysis II – Parallel ac circuits analysis
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Let us now note the impact of frequency
on the total impedance and inductive
current for the parallel R-L network in
Fig. 15.86 for a frequency range through
40 kHz.
FIGURE 15.86 Determining the frequency response of a parallel R-L network.
ZT
In Fig. 15.87, XL is very small at low frequencies compared to R,
establishing XL as the predominant factor in this frequency range. As the frequency
increases, XL increases until it equals the impedance of resistor (220 Ω). The
frequency at which this situation occurs can be determined in the following manner:
X L  2f 2 L  R
and
R
f2 
2L
220

2 (4 10 3 H )
 8.75 kHz
FIGURE
frequency
response
of the
individual elements
of a parallel R-L network.
ET 24215.87
Circuit The
Analysis
II – Sinusoidal
Alternating
Waveforms
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A general equation for the total impedance in vector form can be developed in the
following manner:
ZT 

and
ZRZL
(R0 )(X L 90 )

ZR  ZL
R  jX L
ZT 
RX L 90
and
R 2  X L2 tan1 X L /R
ZT 
RX L
R X
2
2
L
RX L
R 2  X L2
θT  90  tan1 R
 tan
(90  tan1 X L /R)
1
XL
R
XL
IL
Applying the current divider rule to the network in Fig. 15.86 results in the
following:
ZRI
(R0 )(I0 )
IL 

ZR  ZL
R  jX L

and
The magnitude of I L is determined
IT 
RI 0
RI 0
R X
2
R 2  X L2
and the phase angle θ L ,
R 2  X L2 tan1 X L /R
I L  I L  L 
RX L
1
2
L
  tan X L /R
ET 242 Circuit Analysis II – Parallel ac circuits analysis
by which I L leads I is given by
θ L  tan1 X L
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R
13
Equivalent Circuits
In a series circuit, the total impedance of two or more elements in series is often
equivalent to an impedance that can be achieved with fewer elements of different
values, the elements and their values being determined by frequency applied. This is
also true for parallel circuits. For the circuit in Fig. 15.94 (a),
ZC Z L
(5Ω  90 )(10Ω90 ) 50 0
ZT 


 10Ω  90
ZC  Z L
5Ω  90  10Ω90
590
The total impedance at the frequency applied is equivalent to a capacitor with a
reactance of 10 Ω, as shown in Fig. 15.94 (b).
FIGURE 15.94 Defining the equivalence between two networks at a specific frequency.
Boylestad
ET 242 Circuit Analysis II – Parallel ac circuits analysis
14
Another interesting development appears if the impedance of a parallel circuit, such
as the one in Fig. 15.95(a), is found in rectangular form. In this case,
ZLZR
(4Ω90 )(3Ω0 )
ZT 

ZL  ZR
4Ω90  3Ω0
1290

 2.40Ω36.87
553.13
 1.92Ω  j1.44Ω
There is an alternative method to find same result
by using formulas
(3)( 4) 2
48Ω
Rs  2


 1.92Ω
2
2
2
X p  R p (4)  (3)
25
R p X p2
and
(3) 2 (4)
36Ω
Xs  2


 1.44Ω
2
2
2
X p  R p (4)  (3)
25
R p2 X p
FIGURE 15.95 Finding the series
equivalent circuit for a parallel R-L
network.
Rs2  X s2 (1.92) 2  (1.44) 2 5.76Ω
Rp 


 3.0Ω and
Rs
1.92
1.92
ET 242 Circuit Analysis II – Parallel ac circuits analysis
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Rs2  X s2 5.76
Xp 

 4.0Ω
Xs
1.44
15
Ex. 15-18 Determine the series equivalent circuit for the network in Fig. 15.97.
R p  8k
X p (resultant )  X L  X C  9k  4k  5k
and
(8k)(5k) 2
200kΩ
Rs  2


 2.25kΩ
2
2
2
X p  R p (5k)  (8k)
89
with
(8k) 2 (5k)
320kΩ
Xs  2


 3.6kΩ (inductive)
2
2
2
X p  R p (5k)  (8k)
89
R p X p2
R p2 X p
FIGURE
Example
15.18.
ET 242 Circuit
Analysis II 15.97
– Sinusoidal
Alternating
Waveforms
FIGURE 15.98 The equivalent series circuit
forBoylestad
the parallel network in Fig. 15.97.
16
Phase Measurement
Measuring the phase angle between quantities is one of the most important
functions that an oscilloscope can perform. Whenever you are using the dual-trace
capability of an oscilloscope, the most important thing to remember is that both
channel of a dual-trace oscilloscope must be connected to the same ground.
Measuring ZT and θT
For ac parallel networks, the total impedance can be found in the same manner as
described for dc circuits: Simply remove the source and place an ohmmeter across
the network terminals. However,
For parallel ac networks with reactive elements, the total impedance cannot be
measured with an ohmmeter.
The phase angle between the applied voltage and the resulting source current is one
of the most important because (a) it is also the phase angle associated with the total
impedance; (b) it provides an instant indication of whether the network is resistive
or reactive; (c) it reveals whether a network is inductive or capacitive; and (d) it can
be used to find the power delivered to the network.
ET 242 Circuit Analysis II – Parallel ac circuits analysis
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In Fig. 15.104, a resistor has been added to the configuration between the source
and the network to permit measuring the current and finding the phase angle
between the applied voltage and the source current. In Fig. 15.104, channel 1 is
displaying the applied voltage, and channel 2 the voltage across the sensing
resistor. Sensitivities for each channel are chosen to establishes the waveforms
appearing on the screen in Fig. 15.105.
FIGURE 15.104 Using an oscilloscope to
measure ZT and θT.
FIGURE 15.105 e and vR for the
configuration in Fig. 15.104.
ET 242 Circuit Analysis II – Parallel ac circuits analysis
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Using the sensitivities, the peak voltage
across the sensing resistor is
Em = (4div.)(2V/div.) = 8 V
while the peak value of the voltage
across the sensing resistor is
VR,(peak) = (2div.)(10mV/div.) = 20 mV
Using Ohm' s law, the peak value of the current is
VRs ( peak ) 20 mV
IT 

 2 mA
Rs
10
The magnitude of the input impedance is then
V
E
8V
ZT  x  
 4 k
I s I s 2 mA
For the chosen horizontal sensitivity, each
waveform in Fig. 15.105 has a period T
defined by ten horizontal divisions, and the
phase angle between the two waveforms is
1.7 divisions. Using the fact that each
period of a sinusoidal waveform
encompasses 360o, the following ratios can
be set up to determine the phase angle θ:
10div. 1.7div.

360

 1.7 
and   
360  61.2
 10 
In general ,

(div. for  )
 360
(div. for T )
Therefore, the total impedance is
Z T  4kΩ61.2  1.93kΩ  j3.51kΩ  R  jX L
ET 242 Circuit Analysis II – Parallel ac circuits analysis
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