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Chapter 5: REACTANCE AND IMPEDANCE - R, L, AND C

# 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: (Figure below)

*Example R, L, and C parallel 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: (Figure below)

*Example R, L, and C parallel circuit with impedances replacing component values.*

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 (Z_{Total} = 1/(1/Z_{R} + 1/Z_{L} + 1/Z_{C}),
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: (Figure below)

*Example parallel R, L, and C SPICE circuit. Battery symbols are
"dummy" voltage sources for SPICE to use as current measurement points.
All are set to 0 volts.*

```
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)
6.000E+01 6.390E-01 4.800E-01 4.897E-01
i(vic)
6.786E-02
freq ip(vi) ip(vir) ip(vil)
6.000E+01 -4.131E+01 0.000E+00 -9.000E+01
ip(vic)
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 180^{o} out of phase. This way, our current phase readings came out to exactly match our hand calculations.