1. Parallel ResistorInductor Circuits
Let’s take the same components for our series example circuit and connect them in parallel: (Figure below)
Parallel RL circuit.
Because the power source has the same frequency as the series example circuit, and the resistor and inductor both have the same values of resistance and inductance, respectively, they must also have the same values of impedance. So, we can begin our analysis table with the same "given" values:
The only difference in our analysis technique this time is that we will apply the rules of parallel circuits instead of the rules for series circuits. The approach is fundamentally the same as for DC. We know that voltage is shared uniformly by all components in a parallel circuit, so we can transfer the figure of total voltage (10 volts ∠ 0^{o}) to all components columns:
Now we can apply Ohm’s Law (I=E/Z) vertically to two columns of the table, calculating current through the resistor and current through the inductor:
Just as with DC circuits, branch currents in a parallel AC circuit add to form the total current (Kirchhoff’s Current Law still holds true for AC as it did for DC):
Finally, total impedance can be calculated by using Ohm’s Law (Z=E/I) vertically in the "Total" column. Incidentally, parallel impedance can also be calculated by using a reciprocal formula identical to that used in calculating parallel resistances.
The only problem with using this formula is that it typically involves a lot of calculator keystrokes to carry out. And if you’re determined to run through a formula like this "longhand," be prepared for a very large amount of work! But, just as with DC circuits, we often have multiple options in calculating the quantities in our analysis tables, and this example is no different. No matter which way you calculate total impedance (Ohm’s Law or the reciprocal formula), you will arrive at the same figure:

REVIEW:

Impedances (Z) are managed just like resistances ® in parallel circuit analysis: parallel impedances diminish to form the total impedance, using the reciprocal formula. Just be sure to perform all calculations in complex (not scalar) form! Z_{Total} = 1/(1/Z_{1} + 1/Z_{2} + . . . 1/Z_{n})

Ohm’s Law for AC circuits: E = IZ ; I = E/Z ; Z = E/I

When resistors and inductors are mixed together in parallel circuits (just as in series circuits), the total impedance will have a phase angle somewhere between 0^{o} and +90^{o}. The circuit current will have a phase angle somewhere between 0^{o} and 90^{o}.

Parallel AC circuits exhibit the same fundamental properties as parallel DC circuits: voltage is uniform throughout the circuit, branch currents add to form the total current, and impedances diminish (through the reciprocal formula) to form the total impedance.
2. Parallel ResistorCapacitor Circuits
Using the same value components in our series example circuit, we will connect them in parallel and see what happens: (Figure below)
Parallel RC circuit.
Because the power source has the same frequency as the series example circuit, and the resistor and capacitor both have the same values of resistance and capacitance, respectively, they must also have the same values of impedance. So, we can begin our analysis table with the same "given" values:
This being a parallel circuit now, we know that voltage is shared equally by all components, so we can place the figure for total voltage (10 volts ∠ 0^{o}) in all the columns:
Now we can apply Ohm’s Law (I=E/Z) vertically to two columns in the table, calculating current through the resistor and current through the capacitor:
Just as with DC circuits, branch currents in a parallel AC circuit add up to form the total current (Kirchhoff’s Current Law again):
Finally, total impedance can be calculated by using Ohm’s Law (Z=E/I) vertically in the "Total" column. As we saw in the AC inductance chapter, parallel impedance can also be calculated by using a reciprocal formula identical to that used in calculating parallel resistances. It is noteworthy to mention that this parallel impedance rule holds true regardless of the kind of impedances placed in parallel. In other words, it doesn’t matter if we’re calculating a circuit composed of parallel resistors, parallel inductors, parallel capacitors, or some combination thereof: in the form of impedances (Z), all the terms are common and can be applied uniformly to the same formula. Once again, the parallel impedance formula looks like this:
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The only drawback to using this equation is the significant amount of work required to work it out, especially without the assistance of a calculator capable of manipulating complex quantities. Regardless of how we calculate total impedance for our parallel circuit (either Ohm’s Law or the reciprocal formula), we will arrive at the same figure:

REVIEW:

Impedances (Z) are managed just like resistances ® in parallel circuit analysis: parallel impedances diminish to form the total impedance, using the reciprocal formula. Just be sure to perform all calculations in complex (not scalar) form! Z_{Total} = 1/(1/Z_{1} + 1/Z_{2} + . . . 1/Z_{n})

Ohm’s Law for AC circuits: E = IZ ; I = E/Z ; Z = E/I

When resistors and capacitors are mixed together in parallel circuits (just as in series circuits), the total impedance will have a phase angle somewhere between 0^{o} and 90^{o}. The circuit current will have a phase angle somewhere between 0^{o} and +90^{o}.

Parallel AC circuits exhibit the same fundamental properties as parallel DC circuits: voltage is uniform throughout the circuit, branch currents add to form the total current, and impedances diminish (through the reciprocal formula) to form the total impedance.
3. Parallel R, L, and C Circuits
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.
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ac rlc 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 1e12 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
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freq i(vi) i(vir) i(vil) i(vic) 6.000E+01 6.390E01 4.800E01 4.897E01 6.786E02 freq ip(vi) ip(vir) ip(vil) ip(vic) 6.000E+01 4.131E+01 0.000E+00 9.000E+01 9.000E+01
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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 inductortovoltage 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.
3.1. Parallel RLC Circuits Example
When dealing with a parallel ac circuit, you will find that the concepts presented in this chapter for series ac circuits still apply. There is one major difference between a series circuit and a parallel circuit that must be considered. The difference is that current is the same in all parts of a series circuit, whereas voltage is the same across all branches of a parallel circuit. Because of this difference, the total impedance of a parallel circuit must be computed on the basis of the current in the circuit.
You should remember that in the series RLC circuit the following three formulas were used to find reactance, impedance, and power factor:
When working with a parallel circuit you must use the following formulas instead:
Note

If no value for E is given in a circuit, any value of E can be assumed to find the values of I_{L}, I_{C}, I_{X}, I_{R}, and I_{Z}. The same value of voltage is then used to find impedance. 
For example, find the value of Z in the circuit shown in Figure 1.
The first step in solving for Z is to calculate the individual branch currents.
Using the values for I_{R}, I_{L}, and I_{C}, solve for I_{X} and I_{Z}.
Using this value of I_{Z}, solve for Z.
If the value for E were not given and you were asked to solve for Z, any value of E could be assumed. If, in the example problem above, you assume a value of 50 volts for E, the solution would be:
First solve for the values of current in the same manner as before.
Solve for I_{X} and I_{Z}.
Solve for Z.
When the voltage is given, you can use the values of currents, Ip, Ix, and Iz, to calculate for the true power, reactive power, apparent power, and power factor. For the circuit shown in figure 412, the calculations would be as follows.
To find true power,
To find reactive power, first find the value of reactance (X).
To find apparent power,
The power factor in a parallel circuit is found by either of the following methods.
O31. What is the difference between calculating impedance in a series ac circuit and in a parallel ac circuit?