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PARALLEL CIRCUITS
R and XL in Parallel
The analysis of a parallel combination of R and XL proceeds along lines as our
previous analysis of series RC circuits. The figure below shows an ac source
connected to a parallel RL circuit so that VL = Vr
And IR =
Es
R
IL = Es
XL
IT
ES
IR
ES
~
R
IL
XL
(a)
(b)
ES
IT =  IR2 + IL2
Tan = IL
Ir
(c)
(d)
FIGURE 1 (a) Parallel RL circuit. (b) Phasor diagram. (c) Find IT phasor. (d)
Complete phasor diagram showing IT.
(We will assume that the effective resistance Re of the inductor is small enough to be
neglected.) The total current IT is the total current supplied by the source. The value of
IT can be found by taking the vector sum of IR and 1L. That is, IT = IR + IL Part (b) of
the figure above shows the phasor representations of ES, IR, and IL. ES is used as the
reference phasor at 0O since Es is common to all the circuit elements. IR is shown in
phase with Es at 0O. IL will lag E5 by 90O; therefore its phasor is drawn at – 90O. Thus,
IL also lags IR by 90O.
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If IL and IR are known, the total current can be determined using vector addition as
illustrated in part (c) of Fig. 1. The magnitude of IT is easily obtained using
Pythagorean's theorem
IT = IR2 + IL2
As the phasor diagram in 1(c) shows, the IT phasor lags IR by the angle . In other
words IT is at a negative phase angle relative to the 0O reference. The value of  can
be determined using:
tan  = - IL
IR
The negative sign indicates that  is a lagging (negative) phase angle. The complete
phasor diagram for the parallel RL circuit is shown in Fig. 1(d). It shows that the total
current IR lags the applied source voltage by the angle 0o. This is the same relationship
that existed in the series RL circuit. In any inductive circuit, the current delivered by
the source lags the source voltage.
EXAMPLE 19.9 A parallel RL circuit has R = 2 k and XL = 1.5 k
a) Find IR for ES = 120 V; (b) determine 
Solution:
(a) The circuit is drawn in Fig. 2(a).
IT
ES
IR
ES
~
2 k
=
IL
1.5 k
FIGURE 2
Since E5 = 120 V we can calculate IR and IL
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In the parallel RL circuit as was the case for the series RL circuit the total current lags
the applied voltage. However, in the parallel circuit the phase  is negative while for
the series circuit  is positive. The reason for this is our reference is different in each
case. In the series circuit the current chosen as the 0O reference; in the parallel circuit
the applied voltage chosen as the 0O reference. One must be aware of the reference
used in case.
Impedance of Parallel RL Circuit
The total current in a parallel RL circuit can be determined as shown last section.
Once IT is known it is relatively easy to find the total impedance of the parallel
combination. Since Z is the total opposition of the
Z= Es
IT
For example, in the circuit of Fig 2 the applied voltage is 120 V and IT was
determined to be 100 mA. Thus,
Z=
120
100 x 10-3
= 1.2 k
This impedance is the combined opposition of the 2-k resistor and the 1.5-k
inductive reactance. For the parallel RL circuit Z is not equal to the vector sum of R
and XL, because R and XL are not in series. Instead, Z is found by first calculating IT
and then using Z= Es
IT
EXERCISE 1
What is the total Z of a 600-ohm resistor in parallel with 400ohm inductive reactance?
Solution: To find Z we must apply a voltage to the parallel RL circuit and determine
IT. For convenience, a source voltage of 1200 V can be used. Thus,
Keep in mind that this is the impedance value only at the frequency at which XL = 400
ohms. As f changes, XL will change causing Z to take different values.
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R and XC in Parallel
Figure 3(a) shows an ac source connected to a circuit that capacitive reactance XC in
parallel with a resistor R. Because the elements are connected in parallel, each one has
a voltage a cross it equal to the applied voltage ES. In other words, VC = VR = ES.
The current through each branch will depend on the value of E5 and on the amount of
Opposition in each branch. Thus,
IR= ES
and
IC = ES
R
XC
The total current IT is the rms ac current supplied by the Source. We might be tempted
to say that IT can be found by adding the rms currents through each branch. After all,
this is the procedure we followed in our study of parallel resistive circuits.
Unfortunately, this cannot be done here because of the phase difference between IR
and IC.
The current IR will be in phase with the applied voltage Es. The current IC will lead
by 90o the applied voltage (see phasor diagram in figure). Thus IC leads IR by 90o.
Since IR and IC are out of phase, we must resort to a vector addition to find the total
resultant current IT. This is exactly the same situation we encountered when
combining VR and VC in the series RC circuit .
Finding IT
To find the total current phasor IT we must combine the IR and IC
phasors using vector addition. To help do this we can draw the phasor diagram
showing the phasors representing ES, IC and IR This is done in Fig. 3(a). In drawing
this phasor diagram, the phasor ES is chosen as the reference phasor at 0o. This is
because ES is common to all the branches in a parallel circuit. Recall that in the series
RC circuit, the circuit current I was chosen as the reference phasor because it was
common to all the series elements.
IT
ES
ES
~
IR
IC
R
XC
ES
FIGURE 3 Vector addition of IC and IR produces total current phasor IT
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The phasor representing IR is drawn in phase with ES at 0o. The magnitude of the
resistor current phasor is, of course, ES/R. The phasor representing IC is drawn at a
phase angle of +90O relative to ES (capacitor current leads capacitor voltage by 90o).
The magnitude of the capacitor current phasor is of course, Es/Xc.
The phasor representing the total source current IT is obtained by adding the IC and IR
phasors vectorially. This is shown in part (b) of Fig.3 Since IC and IR are at right
angles to each other, we can use Pythagorean's theorem to determine the magnitude of
IT. Thus,
IT = IR2 + IC2
The phasor right triangle in Fig. 3(b) can also be used to determine the angle .
tan  =
IL
IC
As the diagram shows,  is the angle by which the total current leads the applied
voltage ES. The complete phasor diagram is shown in part (c) of the figure.
EXAMPLE 18.21 A parallel RC circuit has R = 3 k and Xc = 4 k. (a) Find the
total current for ES = 12 V; (b) determine the phase angle between the total current
and the applied voltage; (c) find the phase angle between the capacitor current and the
total current.
Solution: (a) The circuit diagram is shown in Fig. 4(a). Since E5 = 12 V we can
calculate the rms values of IC and IC.
FIGURE 4
ES
ES
~
IR
IC
3 k
4k
To find IT we must add these currents vectorially. The phasors are drawn in part (b) of
the figure. The magnitude of IT is
IT = IR2 + IC2
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so that  = 37o .Thus, IT leads ES by 37o . In the parallel RC circuit the total
current always leads the applied voltage.
(c)
It is obvious from the phasor diagram that IC leads IT by
90o – 37o = 53o
In the parallel RC circuit as was the case for the series RC circuit9 the total current
leads the applied voltage. However, in the parallel circuit the phase angle  is positive
while in the series circuit the phase angle  is negative. The reason for this is the fact
that the reference is different in each case. In the series circuit the current was chosen
as the 0o reference; in the parallel circuit the applied voltage was chosen as the 0o
reference. One must always be aware of the reference used in each case.
Impedance of Parallel RC Circuit
The total current in a parallel RC circuit can be determined as shown in the last
section. Once IT is known it is relatively simple to find the total impedance Z of the
RC parallel combination. Since Z is the total opposition to current:
Z =-E5
IT
For example, for the circuit of Fig.4 the applied voltage is 12 V and IT was determined
to be 5 mA. Thus,
Z = -12V
5 x 10-3
= 2.4 k
This impedance is the combined opposition of the 4-k resistor in parallel with 3-k
capacitive reactance. For the parallel RC circuit Z is not equal to the vector sum of R
and Xc because R and XC are not in series. Z is found by finding IT and then using
Z =-ES
IT.
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EXERCISE What is the total Z of a 600- resistor in parallel with a 300- Xc?
Solution: To find Z we must apply a voltage to the parallel RC circuit and determine
IR. Since no voltage was specified, we can assume any value we wish. A good value
to use in this example is 600 V so that the branch currents come out as whole
numbers.
Keep in mind that this value of Z is only the impedance at the particular frequency at
which Xc =300 . As f changes, Xc will change causing Z to take on different values.
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