Parallel R, L, and C
We can take the same components from the
series circuit and rearrange them into a parallel configuration for an
easy example circuit:
The fact that these components are
connected in parallel instead of series now has absolutely no effect on
their individual impedances. So long as the power supply is the same
frequency as before, the inductive and capacitive reactances will not
have changed at all:
With all component values expressed as
impedances (Z), we can set up an analysis table and proceed as in the
last example problem, except this time following the rules of parallel
circuits instead of series:
Knowing that voltage is shared equally by
all components in a parallel circuit, we can transfer the figure for
total voltage to all component columns in the table:
Now, we can apply Ohm's Law (I=E/Z)
vertically in each column to determine current through each component:
There are two strategies for calculating
total current and total impedance. First, we could calculate total
impedance from all the individual impedances in parallel (ZTotal
= 1/(1/ZR + 1/ZL + 1/ZC), and then
calculate total current by dividing source voltage by total impedance
(I=E/Z). However, working through the parallel impedance equation with
complex numbers is no easy task, with all the reciprocations (1/Z). This
is especially true if you're unfortunate enough not to have a calculator
that handles complex numbers and are forced to do it all by hand
(reciprocate the individual impedances in polar form, then convert them
all to rectangular form for addition, then convert back to polar form
for the final inversion, then invert). The second way to calculate total
current and total impedance is to add up all the branch currents to
arrive at total current (total current in a parallel circuit -- AC or DC
-- is equal to the sum of the branch currents), then use Ohm's Law to
determine total impedance from total voltage and total current (Z=E/I).
Either method, performed properly, will
provide the correct answers. Let's try analyzing this circuit with SPICE
and see what happens:
ac r-l-c circuit
v1 1 0 ac 120 sin
vi 1 2 ac 0
vir 2 3 ac 0
vil 2 4 ac 0
rbogus 4 5 1e-12
vic 2 6 ac 0
r1 3 0 250
l1 5 0 650m
c1 6 0 1.5u
.ac lin 1 60 60
.print ac i(vi) i(vir) i(vil) i(vic)
.print ac ip(vi) ip(vir) ip(vil) ip(vic)
.end
freq i(vi) i(vir) i(vil) i(vic)
6.000E+01 6.390E-01 4.800E-01 4.897E-01 6.786E-02
freq ip(vi) ip(vir) ip(vil) ip(vic)
6.000E+01 -4.131E+01 0.000E+00 -9.000E+01 9.000E+01
It took a little bit of trickery to get
SPICE working as we would like on this circuit (installing "dummy"
voltage sources in each branch to obtain current figures and installing
the "dummy" resistor in the inductor branch to prevent a direct
inductor-to-voltage source loop, which SPICE cannot tolerate), but we
did get the proper readings. Even more than that, by installing the
dummy voltage sources (current meters) in the proper directions, we were
able to avoid that idiosyncrasy of SPICE of printing current figures 180o
out of phase. This way, our current phase readings came out to exactly
match our hand calculations.
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