In Unit 6 we saw that analyzing series AC
circuits involves the same steps as analyzing series DC circuits, but
that at each step you must use complex numbers instead of real numbers.
We'll see in this unit that, as you may have guessed, **parallel** AC
circuits are a lot like parallel DC circuits, except that again you
need to use complex numbers throughout the analysis.

Also, the same rules that are useful in analyzing parallel DC circuits
(such as **Kirchhoff's Current Law** and the **Current-Divider Rule**)
are also useful for parallel AC circuits. Again, though, the math is
more complicated because you have to use complex numbers.

We'll also look at quantities
called **admittance** and **susceptance**, which are similar to
the quantity called **conductance** that you may recall from your
studies of parallel DC circuits. When dealing with parallel circuits,
some people find it more convenient to use conductance, susceptance,
and admittance in place of resistance, reactance, and impedance. I'm
not one of those people, but I do think that you should at least understand
what these terms mean and understand how they relate to each other.

- Let's begin by taking this self-test to review material that you studied in Unit 6.

- In EET 1150 you learned how to analyze parallel DC circuits like
the one shown below. Let's do a quick review of what you learned there.

- You should recall that the basic steps in analyzing a circuit like
this one are:
- Recognize that in a parallel circuit, every component
has the same voltage. Therefore, each resistor's voltage is
equal to the source voltage. For the circuit shown, this means
that
*V*_{S}=*V*_{1}=*V*_{2}=*V*_{3} - Use Ohm's law in the form
*I*=*V*÷*R*to find the current through each resistor. For the circuit shown,*I*_{1}=*V*_{1}÷*R*_{1}and*I*_{2}=*V*_{2}÷*R*_{2}and*I*_{3}=*V*_{3}÷*R*_{3} - Use the reciprocal formula to find the circuit's total resistance.
For the circuit shown,
*R*_{T}= 1 ÷ (1÷*R*_{1}+ 1÷*R*_{2}+ 1÷*R*_{3}) - Use one of the following methods to find the circuit's total
current:
**Either**add together all of the individual resistor currents:*I*_{T}=*I*_{1}+*I*_{2}+*I*_{3}**Or**apply Ohm's law in the form*I*=*V*÷*R*to the entire circuit. In words, the total current produced by the voltage source is equal to the source voltage divided by the total resistance. In symbols,*I*_{T}=*V*_{S}÷*R*_{T}

- Recognize that in a parallel circuit, every component
has the same voltage. Therefore, each resistor's voltage is
equal to the source voltage. For the circuit shown, this means
that

- In EET 1150 you also learned that Kirchhoff's Current Law (KCL) says
that
**the sum of all currents entering a point is equal to the sum of all currents leaving that point**. - In terms of a simple parallel DC circuit like the one you just analyzed, this means that if you add the currents through all of the resistors, the sum must be equal to the value of the total current leaving the voltage source.

- In EET 1150 you also learned that the current-divider rule is one way to find how a current coming into a node in a circuit will split up between the different parallel branches attached to that node.
- The rule says that
**for branches in parallel, the current through any branch equals the ratio of the total parallel resistance to the branch's resistance, multiplied by the total current entering the parallel combination**. - In
equation form, this rule is expressed as:
*I*= (_{x}*R*_{T}÷*R*) ×_{x}*I**I*is the current through the branch you're interested in,_{x}*R*_{T}is the total resistance of all the parallel branches,*R*is the resistance of the branch you're interested in, and_{x}*I*is the total current entering the parallel combination.

- Recall that the two most common types of
circuit problems are
**opens**(breaks) and**shorts**(paths of zero resistance connecting points that should not be connected). - Recall also that
**the current through an open is zero**, and that**the voltage across a short is zero**. - In a parallel DC circuit, an
**open resistor**has no effect on the current passing through the other resistors. But it does increase the circuit's total resistance and therefore decreases the circuit's total current. - A
**shorted resistor**in a parallel DC circuit is basically the same thing as connecting a wire directly from the power supply's positive terminal to its negative terminal. This is a very bad thing to do, and will cause the circuit's total current to increase to an excessive value.- If the circuit is properly protected by a fuse or circuit breaker, the fuse will blow or the breaker will trip, cutting off all current to the circuit.
- If the circuit is not properly protected, the excessive current caused by a short can start a fire or damage the circuit's power supply.

- That ends our quick review of parallel DC circuits. If you'd like a more thorough review, go to Unit 8 of EET 1150. Now let's get back to AC circuits.

- Recognize that, since we're dealing with a parallel circuit, each component's voltage is equal to the source voltage.
- Apply Ohm's law separately to each component to find each component's current.
- Use the reciprocal formula to
find the circuit's total impedance,
**Z**_{T}. - Find the circuit's total current either by adding all of the components' currents, or by applying Ohm's law to the entire circuit.

- This step is the easiest. It simply requires you to remember that
in any parallel circuit (DC or AC), every component's voltage is equal
to the source voltage:
**V**_{S}=**V**_{1}=**V**_{2}= … =**V**_{n}

- Now that you know the voltage across each component, use Ohm's
Law to find the current through each component:
**I**_{1}=**V**_{1}÷**Z**_{1}**I**_{2}=**V**_{2}÷**Z**_{2}**I**_{3}=**V**_{3}÷**Z**_{3}and ... - Here
**Z**_{1}is the first component's impedance, which will have an angle of 0° if the first component is a resistor, or an angle of −90° if the first component is a capacitor, or an angle of 90° if the first component is an inductor. Similarly for**Z**_{2},**Z**_{3}, and so on for however many components the circuit contains. So this step will require you first to find the reactances of any capacitors or inductors in the circuit. - Remember: we're using complex numbers here, not real numbers.

- For impedances in parallel, total impedance is given by the reciprocal
formula:
**Z**_{T}= 1 ÷ (1÷**Z**_{1}+ 1÷**Z**_{2}+ … + 1÷**Z**_{n}) - Each
**Z**on the right-hand side of this equation may be the impedance of a resistor, an inductor, or a capacitor. - Remember to treat these quantities as complex numbers, not as real numbers.

- Use one of the following methods to find the circuit's total current:
**Either**add together all of the individual component currents:**I**_{T}=**I**_{1}+**I**_{2}+ … +**I**_{n}**Or**apply Ohm's law to the entire circuit:**I**_{T}=**V**_{S}÷**Z**_{T}

- Whichever method you use, remember to treat all quantities as complex numbers, not as real numbers.

- As in DC circuits, Kirchhoff's Current Law (KCL) says that
**the sum of all currents entering a point is equal to the sum of all currents leaving that point**. - We made use of KCL in Step 4 above when we wrote
**I**_{T}=**I**_{1}+**I**_{2}+ … +**I**_{n}. - Whenever you apply KCL to an AC circuit, you must use complex numbers,
not real numbers. If you just add the magnitudes of the currents,
instead of adding the magnitudes along with their angles, you won't
get good results. For example, in a parallel AC circuit it's usually
not true that
*I*_{T}=*I*_{1}+*I*_{2}+ … +*I*_{n}. - In this unit our main focus is parallel circuits, but KCL
applies to all circuits, whether series, parallel, or series-parallel.
- For example, in the series-parallel
circuit shown below, the capacitor is in series with the voltage source,
so all current leaving the source will pass through the capacitor.
But to the right of the capacitor, this current splits
in two, with some going through the resistor and some passing through the inductor.
Therefore, Kirchhoff's Current Law tells us that the capacitor's
current is equal to the sum of the resistor's current plus the
inductor's current.

- For example, in the series-parallel
circuit shown below, the capacitor is in series with the voltage source,
so all current leaving the source will pass through the capacitor.
But to the right of the capacitor, this current splits
in two, with some going through the resistor and some passing through the inductor.
Therefore, Kirchhoff's Current Law tells us that the capacitor's
current is equal to the sum of the resistor's current plus the
inductor's current.

- As in DC circuits, the current-divider rule tells you how a current coming into a node will split up between the different parallel branches attached to that node.
- The current-divider rule says that
**the current****I**_{x}through a branch with impedance Z_{x}in parallel with other branches is given by:

where**I**_{x}= (**Z**_{T}÷**Z**_{x})**I**

**I**is the incoming current and**Z**_{T}is the total impedance of all the parallel branches. - The current-divider rule holds for any combination of components
in parallel, regardless of whether the entire circuit is a parallel
circuit or a series-parallel circuit.
- For instance, in the series-parallel
circuit shown below, the branch containing R1 is in parallel
with the branch containing L1. Suppose
that you have found the current coming into the node at which
C1 meets these two parallel branches. Then the
current-divider rule provides one way to find out how this incoming
current will split up between the two parallel branches.

- For instance, in the series-parallel
circuit shown below, the branch containing R1 is in parallel
with the branch containing L1. Suppose
that you have found the current coming into the node at which
C1 meets these two parallel branches. Then the
current-divider rule provides one way to find out how this incoming
current will split up between the two parallel branches.
- Again, use complex numbers, not real numbers.

- Above we reviewed the basics of troubleshooting parallel DC circuits. Almost all of these same points apply to parallel AC circuits. (But there's one important difference noted below.) In particular:
- A
**short**has zero resistance and zero voltage.- A short in a parallel circuit causes excessive current to flow, resulting in blown fuses, tripped circuit breakers, or damage to the circuit's power supply.

- No current flows through an
**open**.- If a component in a parallel circuit becomes open, this will not have any effect on the current flowing through the circuit's other components.

- So far, everything we've said about shorts and opens in parallel
AC circuits is the same as what we said earlier about shorts and opens
in parallel DC circuits. But here's a difference:
- An open in a parallel
**DC**circuit will always increase the circuit's total resistance, decreasing the circuit's total current. - But in a parallel
**AC**circuit, an open could either increase or decrease the circuit's total impedance, and therefore could either decrease or increase the total current.

- An open in a parallel
- Why this difference between opens in DC circuits and opens in AC
circuits? It's because of the difference between real numbers and
complex numbers.
- To find a parallel
**DC**circuit's total resistance, you use the reciprocal formula on two or more real numbers. If one of the circuit's resistors becomes open, then you remove one of these numbers from the formula, and this will increase the total.- For example, suppose a parallel circuit contains a 100 Ω resistor,
a 150 Ω resistor, and a 200 Ω resistor.
Then the total resistance is
**46.2 Ω**. But if the 200 Ω resistor is open, then you remove its resistance from the reciprocal formula, and so the total resistance increases to**60 Ω**. This increase in total resistance will decrease the circuit's total current.

- For example, suppose a parallel circuit contains a 100 Ω resistor,
a 150 Ω resistor, and a 200 Ω resistor.
Then the total resistance is
- But to find a parallel
**AC**circuit's total impedance, you apply the reciprocal formula to two or more complex numbers. If one of the circuit's components becomes open, then you remove one of these complex numbers from the formula. This could either decrease or increase the total.- For example, suppose a parallel AC circuit contains
a 100∠0° Ω resistive
impedance, a 150∠90° Ω inductive
impedance, and a 200∠−90° Ω capacitive
impedance. Then the total impedance is
**98.6 ∠9.46° Ω**. If the capacitor becomes open, then you remove its impedance from the reciprocal formula, and so the total impedance**decreases to 83.2 ∠33.7° Ω**. This decrease in total impedance will increase the circuit's total current. - On the other hand, suppose that in the same circuit
the capacitor is okay but the resistor becomes open.
Then the circuit's total impedance is
**600 ∠90° Ω**, which is an**increase**from the original total impedance. This increase in total impedance will decrease the circuit's total current. - So an open in a parallel AC circuit may either increase or decrease the total impedance.

- For example, suppose a parallel AC circuit contains
a 100∠0° Ω resistive
impedance, a 150∠90° Ω inductive
impedance, and a 200∠−90° Ω capacitive
impedance. Then the total impedance is

- To find a parallel

- Recall from EET 1150 that a resistor's
**conductance**, which is abbreviated*G*, is the reciprocal of its resistance:*G*= 1 ÷*R* - Recall also that conductance is measured in units called siemens
(abbreviated S).
- For example, a 1 kΩ resistor has a conductance of 1 mS.

- Why is this concept of conductance useful? Primarily because in
analyzing parallel circuits, the expression 1 ÷
*R*appears frequently, and so it's useful to have a shorthand way of referring to the reciprocal of resistance.- For example, you're familiar with the reciprocal formula
for finding total resistance of resistors in parallel, which
can be written as:
*R*_{T}= 1 ÷ (1÷*R*_{1}+ 1÷*R*_{2}+ ... + 1÷*R*_{n})*G*_{T}=*G*_{1}+*G*_{2}+ ... +*G*_{n} - People who prefer to work with conductance instead of
resistance also sometimes rewrite Ohm's law in terms of conductance
rather than resistance. For example, you know that one form
of Ohm's law says that a resistor's current is equal to its
voltage divided by its resistance (in symbols,
*I*=*V*÷*R*). Another way of saying the same thing is that a resistor's current is equal to its voltage times its conductance (in symbols,*I*=*V*×*G*). - I find it easier to use resistance instead of conductance--this saves me from having to remember a bunch of new formulas. But if you find it easier to work with conductance, feel free to do so. And you should at least know understand what conductance is and how it's related to resistance.

- For example, you're familiar with the reciprocal formula
for finding total resistance of resistors in parallel, which
can be written as:

- Recall that in AC circuits, capacitors and inductors oppose
the flow of current, and that we use the term "reactance" for this
opposition to current. Recall also that if we know the frequency of
the AC current and the size of the
capacitor
or inductor, we can compute the reactance as follows:
- A capacitor's reactance
*X*is equal to 1 ÷ (2p_{C}*fC*). - An inductor's reactance
*X*is equal to 2p_{L}*fL*.

- A capacitor's reactance
- Just as it is sometimes useful to have a shorthand way of referring
to the reciprocal of a resistor's resistance, it's also sometimes useful
to have a shorthand way of referring to the reciprocal of a capacitor's
or inductor's reactance. The reciprocal of reactance is called
**susceptance**. - The unit of susceptance is the siemens.
- The abbreviation for susceptance is
*B*. - To find a capacitor's susceptance:
*B*= 1 ÷_{C}*X*= 2p_{C}*fC* - To find an inductor's susceptance:
*B*= 1 ÷_{L}*X*= 1 ÷ (2p_{L}*fL*)

- You know from Unit 6 that
**impedance**can be thought of as a generalization of the concepts of resistance and reactance. Impedance is a complex quantity, expressible in either rectangular form or polar form. - Just as it's useful to have a shorthand way of referring to the
reciprocal of resistance and reactance, it's also useful to have a
shorthand way of referring to the reciprocal of impedance. The reciprocal
of impedance is called
**admittance**. - The unit of admittance is the siemens.
- Like impedance, admittance is a complex quantity, expressible in
either rectangular form or polar form.
The abbreviation for an admittance is
**Y**. Therefore we can write**Y**= 1 ÷**Z** - At times we may be interested in the magnitude of the admittance,
which we denote by
*Y*. - When you take the reciprocal of a complex number
with a negative angle, you'll get a complex number with a positive
angle, and vice versa. Therefore, if an impedance has a positive angle,
the corresponding admittance will have a negative angle, and vice
versa.
- For example, if a circuit's total impedance has an angle of +36.4°, then the circuit's total admittance must have an angle of −36.4°.

- If you prefer to work with admittances instead of impedances, you can
use modified forms of Ohm's law: for example, instead of writing
**I**=**V**÷**Z**(which says that current equals voltage divided by impedance) you could instead write**I**=**V**×**Y**(current equals voltage times admittance). But I usually find it easier to stick with impedance rather than using admittance.

- Let's quickly summarize these new terms and abbreviations. Each row
in the table below shows a pair of quantities that are reciprocals
of each other. The quantities in the left column are all measured in
ohms, while those in the right column are all measured in
siemens.

**Quantity measured in ohms****Reciprocal quantity measured in siemens**resistance*R*conductance*G*reactance*X*susceptance*B*impedance**Z**admittance**Y** - As a memory aid, notice that the terms in the left-hand column come
from English words that imply opposing or working against something:
to
**resist**or**impede**someone means to try to stop him, and to**react**against something means to work against it. - Conversely, the terms in the right-hand column come from English
words that imply helping or working with something: to
**conduct**or**admit**someone means to help her progress, and to be**susceptible**to something means that you go along with it. - These similarities to English words make sense, because all of these
terms describe how much (or how little) a component or
circuit opposes the flow of current.
- If you increase a circuit's resistance, reactance, or impedance, then you're increasing the circuit's opposition to current, and therefore less current will flow.
- Conversely, if you increase a circuit's conductance, susceptance,
or admittance, then you're
**decreasing**the circuit's opposition to current, and therefore**more**current will flow.

- From Unit 6 you know that in AC circuits a resistor's impedance
**Z**is a complex quantity whose magnitude in ohms is_{R}*R*and whose angle is 0°. - Therefore, the resistor's admittance is
a complex quantity whose
magnitude in siemens is 1/
*R*(which is just the conductance*G*) and whose angle is 0°. - So in polar notation, a resistor's
admittance is
**Y**=_{R}*G*∠0° - We can easily convert this to rectangular notation, to get

or simply**Y**=_{R}*G*+*j*0**Y**=_{R}*G* - Thus,
**Y**for any resistor has a real part but no imaginary part._{R} - For example, consider a 2 kΩ resistor. Its conductance
*G*is 500 µS, so we can write its admittance (in polar form) as**Y**= 500 ∠0° µS_{R}

- From Unit 6 you know that in AC circuits a capacitor's impedance
**Z**is a complex quantity whose magnitude in ohms is 1/(2p_{R}*fC*) and whose angle is −90°. - Therefore, the same capacitor's admittance is
a complex quantity whose
magnitude in siemens is 2p
*fC*and whose angle is +90°. - So in polar notation, a capacitor's
admittance is
**Y**= 2p_{C}*fC*∠90° - We can easily convert this to rectangular notation, to get

or simply**Y**= 0 +_{C}*j*2p*fC***Y**=_{C}*j*2p*fC* - Thus,
**Y**for any capacitor has a positive imaginary part but no real part._{C} - For example, consider a 1 µF capacitor in a circuit whose frequency
is 500 Hz. The capacitor's reactance
*X*is equal to 1/(2p*fC*), which works out to 318.3 Ω. Therefore its susceptance*B*is 3.14 mS, so we can write its admittance (in polar form) as**Y**= 3.14 ∠90° mS_{C}

- From Unit 6 you know that in AC circuits an indutor's impedance
**Z**is a complex quantity whose magnitude in ohms is 2p_{L}*fL*and whose angle is 90°. - Therefore, the indutor's admittance is
a complex quantity whose
magnitude in siemens is 1/(2p
*fL*) and whose angle is −90°. - So in polar notation, an indutor's
admittance is
**Y**= 1/(2p_{L}*fL*)∠−90° - We can easily convert this to rectangular notation, to get

or simply**Y**= 0 −_{L}*j*/(2p*fL*)**Y**= −_{L}*j*/(2p*fL*) - Thus,
**Y**for any indutor has a negative imaginary part but no real part._{L} - For example, consider a 10 mH inductor in a circuit whose frequency
is 200 Hz. The inductor's reactance
*X*is equal to 2p*fL*, which works out to 12.6 Ω. Therefore its susceptance*B*is 79.6 mS, so we can write its admittance (in polar form) as**Y**= 79.6 ∠−90° µS_{L}

- Want more practice analyzing parallel AC circuits? Here are a couple of lessons that will generate as many practice problems as you want, and then let you check your answers against the correct answers.
- The first one covers parallel
*RC*circuits: - And the second one covers parallel
*RL*circuits:

- This e-Lesson has covered several important topics, including:
- parallel AC circuits
- Kirchhoff's Current Law
- current-divider rule
- troubleshooting parallel AC circuits
- conductance, susceptance, and admittance.

- To finish the e-Lesson, take this self-test to check your understanding of these topics.

Congratulations! You've completed the e-Lesson for this unit.