In this unit we'll pull together a lot of things from earlier units and use them to analyze series circuits containing AC voltage sources. You'll need to remember what you've learned about AC fundamentals, capacitors, inductors, complex numbers, and phasors. Once you take all of that into account, though, you'll find that to analyze a series AC circuit, you follow the same steps that you follow to analyze a series DC circuit.

Also, the same rules that hold for series DC circuits (such as **Ohm's
law**, **Kirchhoff's Voltage Law**, and the **Voltage-Divider
Rule**) also hold for series AC circuits. But the math is a little
more complicated, because each step involves complex numbers
instead of real numbers.

- This unit will build on material that you studied in Unit 4 and Unit 5 . So let's begin by taking these two self-tests to review what you learned in those units.

- In EET 1150 you learned how to analyze series 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:
- Add the resistance values to find the circuit's total resistance,
*R*_{T}. For the circuit shown, this means that*R*_{T}=*R*_{1}+*R*_{2}+*R*_{3} - Apply Ohm's law to the entire circuit to find the circuit's
total current:
*I*_{T}=*V*_{S}÷*R*_{T} - Recognize that, since we're dealing with a series circuit,
each resistor's current is equal to the total current. For the
circuit shown:
*I*_{1}=*I*_{2}=*I*_{3}=*I*_{T} - Apply Ohm's law to each resistor to find the voltage drops:
*V*_{1}=*I*_{1}×*R*_{1}and*V*_{2}=*I*_{2}×*R*_{2}and*V*_{3}=*I*_{3}×*R*_{3}

- Add the resistance values to find the circuit's total resistance,

- In EET 1150 you also learned that Kirchhoff's Voltage Law (KVL) says
that
**the sum of the voltage drops around any closed loop equals the sum of the voltage rises around that loop**. - In terms of a simple series DC circuit like the one you just analyzed, this means that the sum of all the resistor voltage drops must equal the source voltage.

- In EET 1150 you also learned that the voltage-divider rule is a shortcut rule that you can use to find the voltage drop across a resistor in a series circuit.
- The rule says that
**the voltage across any resistance in a series circuit is equal to****the ratio of that resistance to the circuit's total resistance****, multiplied by the source voltage**. - In equation form, this rule is expressed as:
*V*_{x}=*V*×(_{S}*R*_{x}÷*R*_{T})

**Troubleshooting**a non-working circuit means finding the problem that is preventing the circuit from working correctly.- The two most common types of problems are open circuits and short circuits.
- An
**open****circuit**, or "open," is a break in a circuit path. - The most important thing to remember about opens is that
**no current can flow through an open**. - Therefore, no current can flow anywhere in a series circuit containing an open.
- Since no current flows through an open, you can think of the open
as having infinite resistance (
*R*= ∞). - Usually, an open will
**not**have a voltage drop of 0 V. In fact, in a series DC circuit that contains an open,**the entire source voltage will appear across the open, and no voltage will appear across any of the other resistors**. - So if you measure the voltage between any two points in a series
circuit containing an open, you'll measure 0 V if the two points
are on the same side of the open, but you'll measure the entire
source voltage if the points are on opposite sides of the open.
- For example, suppose R3 is open in the circuit shown below.
Then there will be 0 V across R1, across R2, and across
R4. Also,
*V*= 0 V. But there will be 9 V across R3. Also,_{ab}*V*= 9 V, and_{ac}*V*= 9 V._{bc}

- For example, suppose R3 is open in the circuit shown below.
Then there will be 0 V across R1, across R2, and across
R4. Also,
- A
**short circuit**, or "short," is a path of zero resistance connecting two points in a circuit that are not supposed to be connected. - Since a short has zero resistance, the voltage across it must
be zero. This follows from Ohm's law,
*V*=*I*×*R*. - A component is said to be short-circuited, or "shorted out," when
there is a short circuit connected in parallel with it. No current
flows through a short-circuited component. Instead, current is diverted
through the short itself.
- For example, suppose that in the circuit shown below there
is a short between points
*a*and*b*, perhaps caused by a loose wire clipping that connects these two points. Then R2 is short-circuited. No current will flow through R2; instead, current will follow the path of zero resistance through the short itself (the wire clipping).

- For example, suppose that in the circuit shown below there
is a short between points
- A short in a series DC circuit reduces the circuit's total resistance,
causing more current to flow out of the voltage source.
- For example, in the circuit shown above, if R2 is short-circuited
by a wire clipping that connects points
*a*and*b*, then R2's resistance disappears from the circuit, and the circuit's total resistance is equal to*R*_{1}+*R*_{3}+*R*_{4}.

- For example, in the circuit shown above, if R2 is short-circuited
by a wire clipping that connects points
- That ends our quick review of series DC circuits. If you'd like a more thorough review, go to Unit 7 of EET 1150. Now let's get back to AC circuits.

- Here's an important point that we've mentioned a couple of times
before and that is worth repeating:
**When a sinusoidal voltage is applied to any circuit containing resistors, capacitors, and inductors, all of the circuit's current waveforms and voltage waveforms are sinusoids and have the same frequency as the source voltage.** - So, for example, if you're given the circuit shown below, and if
you're told that the source voltage is a sinusoid having a frequency
of 5 kHz, then you can say immediately that the current through
every component is a 5-kHz sinusoidal current, and the voltage drop
across every component is a 5-kHz sinusoidal voltage.

- Where things get a bit tricky is figuring out the peak values and phase shifts of these current and voltage waveforms. But we'll be able to do it, thanks to complex numbers.

- Up to now we have used italicized, non-boldface letters to represent
voltage and current. In particular:
*V*represents voltage.*I*represents current.

- From this point onward, we will usually treat voltage and current
as phasors, which means we'll treat them as complex numbers with both
a magnitude and an angle. We'll use the italic letters listed above
to denote the magnitude of the phasor, and we'll use boldface letters
to represent the total phasor quantity (which includes both the magnitude
and the angle). In particular:
**V**represents a phasor voltage, which has both a magnitude (*V*) and an angle.**I**represents a phasor current, which has both a magnitude (*I*) and an angle.

- In AC circuits, every component or combination of components has
a quantity called an
**impedance**, which we denote with the boldface letter**Z**. - If we know a component's voltage
**V**and its current**I**, we can use the following equation to find the component's impedance:**Z**=**V**÷**I** - According to this equation, impedance
**Z**is equal to the ratio of two complex numbers,**V**and**I**. Therefore impedance itself is a complex number. (That's why we denote it with a boldface letter.) - Like any complex quantity, impedance can be expressed in either polar form or rectangular form. When we express it in polar form, we specify its magnitude and its angle.
- Because impedance is the ratio of a voltage to a current, impedance is measured in ohms.
- For example, suppose
that dividing a component's voltage by its current gives the result

Then the component's impedance**V**÷**I**= 736 ∠26.5° Ω**Z**has a magnitude of 736 Ω and an angle of 26.5°. At times we might be interested in talking just about this impedance's magnitude, in which case we would write*Z*= 736 ΩNote that in this last equation,

*Z*is not written in boldface, because it does not denote a complex number. Rather it just denotes a complex number's magnitude.

- Look again at the equation
**Z**=**V**÷**I** - This equation, which applies to components in AC circuits, is
similar to Ohm's law, which you know from your study of DC circuits:
*R = V ÷ I* - In fact, we'll refer to the equation
**Z**=**V**÷**I**as Ohm's law for AC circuits. In AC circuits, impedance plays a role very similar to the role played by resistance in DC circuits. But you must keep in mind that impedance is a complex quantity, which has both a magnitude and an angle. - Of course, you can also rearrange this equation to solve for voltage
if you know current and impedance, or to solve for current
if you
know voltage and impedance:
**V**=**I**×**Z****I**=**V**÷**Z** - Remember, in each case, all quantities are complex numbers, not real numbers.

- Recall from previous Units that:
- Resistance
*R*represents a resistor's opposition to current. - Capacitive reactance
*X*represents a capacitor's opposition to current._{C} - Inductive reactance
*X*represents an inductor's opposition to current._{L}

- Resistance
- Recall also that each of these quantities is measured in ohms.
- Impedance
**Z**, which is also measured in ohms, can be thought of as a generalization of the concepts of resistance and reactance. It represents*any*component's (or combination of components) opposition to AC current. - Sometimes we attach a subscript to
**Z**to indicate the type of component whose impedance we're talking about:- We might write
**Z**when discussing a resistor's impedance, which is closely related to its resistance_{R}*R*. - Similarly, we might write
**Z**when discussing a capacitor's impedance, which is closely related to its reactance_{C}*X*._{C} - Again, we might write
**Z**when discussing an inductor's impedance, which is closely related to its reactance_{L}*X*._{L}

- We might write
- Let's look at each of these cases more closely.

- A resistor's impedance
**Z**is a complex quantity whose magnitude_{R}*R*is the resistance in ohms and whose angle is 0°.- We use 0° because voltage and current are in phase in resistors. (In other words, there is a 0° phase angle between a resistor's current and its voltage.)

- So in polar notation, a resistor's impedance is
**Z**=_{R}*R*∠0° - We can easily convert this to rectangular notation, to get

or simply**Z**=_{R}*R*+*j*0**Z**=_{R}*R* - Thus,
**Z**for any resistor has a real part but no imaginary part. In the complex plane, a resistor's impedance lies along the positive real axis, as shown in the following diagram representing a resistor whose resistance is 50 Ω:_{R}

- A capacitor's impedance
**Z**is a complex quantity whose magnitude_{C}*X*is 1 ÷ (2p_{C}*fC*), and whose angle is −90°.- We use −90° because voltage lags current by 90° in a capacitor.

- So in polar notation, a capacitor's impedance is
**Z**=_{C}*X*∠−90°_{C} - We can easily convert this to rectangular notation, to get
- Remember, in each of these equations
*X*= 1 ÷ (2p_{C}*fC*), which is also equal to 1 ÷ (ω*C*). - Thus,
**Z**for any capacitor has a negative imaginary part but no real part. In the complex plane, a capacitor's impedance lies along the negative imaginary axis, as shown in the following diagram representing a capacitor whose reactance is 50 Ω:_{C}

.

**Z _{C}** = 0

**Z _{C}** =

- An inductor's impedance
**Z**is a complex quantity whose magnitude_{L}*X*is 2p_{L}*fL*and whose angle is +90°.- We use +90° because voltage leads current by 90° in an inductor.

- So in polar notation, an inductor's impedance is
**Z**=_{L}*X*∠90°_{L} - We can easily convert this to rectangular notation, to get
- Remember, in each of these equations
*X*= 2p_{L}*fL*, which is also equal to ω*L*. - Thus,
**Z**for any inductor has a positive imaginary part but no real part. In the complex plane, an inductor's impedance lies along the positive imaginary axis, as shown in the following diagram representing an inductor whose reactance is 50 Ω:_{L}

.

**Z _{L}** = 0

**Z _{L}** =

- We've seen that Ohm's law for AC circuits can be written in any of
the following forms:
**I**=**V**÷**Z****V**=**I**×**Z****Z**=**V**÷**I** - We've also seen how to calculate a resistor's impedance, or a capacitor's impedance, or an inductor's impedance.
- Let's look at some problems that require us to combine these pieces.

- Suppose we have a resistance in series with a reactance. We'd like
to find the total impedance of these two components, but we can't
simply add resistances and reactances as real numbers. For
example,
**a 1 kΩ resistance in series with a 2 kΩ capacitive reactance does not add up to 3 kΩ.** - Instead, we must add them as complex numbers. So, to find the total
impedance of a 1 kΩ resistance in series with a 2 kΩ capacitive
reactance, we must add
**1∠0° kΩ**plus**2∠−90° kΩ**, which gives us a total of**2.24∠−63.4° kΩ**. - It may seem strange that you can combine a 1 kΩ resistance with a 2 kΩ reactance and come up with a total of only 2.24 kΩ, but that's how it works.

- Now that you know how to treat resistances and reactances as complex
quantities, and how to use the phasor form of Ohm's law, and how to
use complex numbers to find total impedance, you're ready to analyze
any series AC circuit, such as the series
*RLC*circuit shown below. - Here are the steps to follow:
- Use complex addition to find the circuit's total impedance,
**Z**_{T}. - Apply Ohm's law to the entire circuit to find the circuit's total current.
- Recognize that, since we're dealing with a series circuit, each component's current is equal to the total current.
- Apply Ohm's law to each component to find the voltage drops.

- Use complex addition to find the circuit's total impedance,
- These are very similar to the steps that you followed in EET 1150
to analyze a simple series DC circuit containing resistors. The big
difference is that throughout this procedure, we
**must now use complex numbers instead of real numbers**. - Let's look at each step in more detail.

- For
*n*impedances in series, total impedance is given by**Z**_{T}=**Z**_{1}+**Z**_{2}+ … +**Z**_{n} - Each
**Z**on the right-hand side of this equation may be the impedance of a resistor, an inductor, or a capacitor. So this step will require you first to find the reactances of any capacitors or inductors in the circuit. - Remember: we're adding complex numbers here, not real numbers.

- Knowing the source voltage
**V**_{S}and the total impedance**Z**_{T}, you can use Ohm's Law to find the current:**I**_{T}=**V**_{S}÷**Z**_{T} - Again, remember that we're dividing complex numbers, not real numbers.

- This step is the easiest. It simply requires you to remember that
in any series circuit (DC or AC), every component's current is equal
to the total current:
**I**_{T}=**I**_{1}=**I**_{2}= … =**I**_{n}

- Now that you know the current through each component, use Ohm's
Law to find the voltage drop across each component:
**V**_{1}=**I**_{1}×**Z**_{1}**V**_{2}=**I**_{2}×**Z**_{2}**V**_{3}=**I**_{3}×**Z**_{3}and ... - Remember,
**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. - Again, remember that we're multiplying complex numbers, not real numbers.

- Want more practice analyzing series 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 series
*RC*circuits: - And the second one covers series
*RL*circuits:

- After you've analyzed a circuit by finding currents and voltage
drops, you can draw a
**phasor diagram**that shows in graphical form how these quantities relate to each other. A phasor diagram simply shows each voltage and current as a vector in the complex plane, drawn with the appropriate angle and magnitude. - Here is a simple example showing the current and voltage phasors
for a single component:

- Here's another example showing the phasors for all voltages and
currents in a particular series
*RC*circuit.

- From the diagram you can quickly see the relationship between the circuit's current and voltages.
- In every phasor diagram for a series circuit, you should find that
**the current and the resistor's voltage drop have the same angle**, since current and voltage in a resistor are always in phase with each other. - Also, you should find that
**inductor voltage and capacitor voltage are always at a 90° angle to the current**, which should make sense. (Remember ELI the ICEman from Unit 4?)

- As in DC circuits, Kirchhoff's Voltage Law (KVL) says that
**the sum of the voltage drops around any closed loop equals the sum of the voltage rises around that loop**. - As we'll see in later units of this course, KVL applies to all circuits, whether series, parallel, or series-parallel.
- In this unit we're restricting our attention to series circuits containing a single voltage source. In these circuits, KVL simplifies to the following form: the source voltage in a series circuit is equal to the sum of the voltage drops across the circuit's resistors, capacitors, and inductors.
- Whenever you apply KVL to an AC circuit, you must use complex numbers, not real numbers. If you just add the magnitudes of the voltages, instead of adding the magnitudes along with their angles, you won't get good results.

- As in DC circuits, this is a shortcut rule for finding voltage drops in a series circuit.
- The voltage-divider rule says that
**the voltage****V**_{x}across any impedance**Z**_{x}in a series circuit with source voltage**V**_{S}is given by:**V**_{x}= (**Z**_{x}÷**Z**_{T}) ×**V**_{S} - Again, use complex numbers, not real numbers.

- Above we reviewed the basics of troubleshooting series DC circuits. Almost all of these same points apply to series AC circuits. (But there's one important difference noted below.) In particular:
- No current flows through an
**open**. - Therefore, no current flows anywhere in a series circuit containing an open.
- Since no current flows through an open, you can think of the open
as having infinite resistance (
*R*= ∞). - Usually, an open will
**not**have a voltage drop of 0 V. In fact, in a series circuit that contains an open,**the entire source voltage will appear across the open, and no voltage will appear across any of the other resistors, capacitors, or inductors**. - So if you measure the voltage between any two points in a series circuit containing an open, you'll measure 0 V if the two points are on the same side of the open, but you'll measure the entire source voltage if the points are on opposite sides of the open.
- A
**short**has zero resistance and zero voltage.- A component is said to be short-circuited, or "shorted out," when there is a short connected in parallel with it. No current flows through a short-circuited component. Instead, current is diverted through the short itself.

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

- A short in a series
- Why this difference between shorts in DC circuits and shorts in
AC circuits? It's because of the difference between real numbers
and complex numbers.
- To find a series
**DC**circuit's total resistance, you add two or more real numbers. If one of the circuit's resistors is shorted out, then you replace one of these numbers with zero, and this will decrease the total.- For example, suppose a series circuit contains a 100 Ω resistor,
a 150 Ω resistor, and a 200 Ω resistor.
Then the total resistance is
**450 Ω**. But if the 200 Ω resistor is shorted out, then you replace its resistance with zero, and so the total resistance decreases to**250 Ω**. This decrease in total resistance will increase the circuit's total current.

- For example, suppose a series circuit contains a 100 Ω resistor,
a 150 Ω resistor, and a 200 Ω resistor.
Then the total resistance is
- But to find a series
**AC**circuit's total impedance, you add two or more complex numbers. If one of the circuit's components is shorted out, then you replace one of these complex numbers with zero. When you're adding complex numbers, replacing one of them by zero could either decrease or increase the total.- For example, suppose a series 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
**112∠−26.6° Ω**. If the capacitor is shorted out, then you replace its impedance with zero, and so the total impedance**increases to 180∠56.3° Ω**. This increase in total impedance will decrease the circuit's total current. - On the other hand, suppose that in the same circuit the
capacitor is okay but the resistor is shorted out.
Then the circuit's total impedance is
**50∠−90° Ω**, which is a**decrease**from the original total impedance. This decrease in total impedance will increase the circuit's total current. - So a short in a series AC circuit may either increase or decrease the total impedance.

- For example, suppose a series 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 series

- In your electronics courses up to now, you've considered a circuit as a collection of individual components such as resistors, capacitors, and inductors. Most of the circuits you've studied have been very small, with just a few components.
- But real-world circuits are usually much more complex, with hundreds or even thousands of components. To study and analyze such circuits, you have to shift your way of looking at circuits. Instead of concentrating on individual components, it's more useful to think of the circuit as a system made up of parts that perform certain functions. The "parts" that I'm referring to here are not individual components. Rather, they are sub-circuits that contain many components connected together to perform some function.
- For example, in your later courses you'll study amplifier circuits.
These circuits contain transistors as well as resistors, capacitors,
and inductors, so we're not ready to understand the details now. But
there are a number of standard designs for amplifier circuits, and
rather than focusing on the details you might just want to consider
the entire amplifier circuit as a "box." This box has two
input terminals to which you can connect an input voltage, and two
output terminals at which the amplifier's output voltage will appear.
Here's a diagram:

- The amplifier circuit contains many components (resistors, capacitors, transistors). But in this diagram we're not showing those details.
- A more complete circuit might consist of an oscillator connected
to an amplifier connected to a power amplifier, as shown here:

- In this diagram we have three boxes representing three complicated sub-circuits. We're not showing the details of these sub-circuits, but we are showing how the sub-circuits are connected to each other. (For example, the two lines between the oscillator and the amplifier show that the oscillator's output voltage is also the amplifier's input voltage.)
- Next we'll look at a few simple series
*RC*and*RL*circuits that we can think of as "boxes" that perform a certain function.

- In some applications, a designer needs to shift a voltage's phase angle by a certain amount. In such cases the designer uses a circuit that introduces a phase shift between the circuit's output voltage and its input voltage.
- There are two basic possibilities here. Either:
- The circuit is designed so that its output voltage lags its input
voltage, in which case we're dealing with a
**lag circuit**. Or: - The circuit is designed so that its output voltage leads its input
voltage, in which case we're dealing with a
**lead circuit**.

- A simple series
*RC*circuit can serve as either a lag circuit or a lead circuit, depending on whether you take the output voltage across the resistor or across the capacitor. In particular:- In any series
*RC*circuit, the capacitor's voltage lags the source voltage, so you'll have a**lag circuit**if you take the output voltage across the capacitor, as shown here:

- Also, in any series
*RC*circuit, the resistor's voltage leads the source voltage, so you'll have a**lead circuit**if you take the output voltage across the resistor, as shown here:

- In any series
- If you remember
**ELI the ICEman**, you'll be able to quickly identify circuits like the ones above as either lead circuits or lag circuits.**ICE**reminds you that a capacitor's voltage tends to lag everything else in the circuit, so you've got a lag circuit if you're taking the output voltage across the capacitor.

- Whenever you encounter circuits like these, you can always use the
general techniques you learned above to analyze the circuit and figure
out how far the output voltage is shifted from the input voltage.
Or you can remember the following formulas.
- For an
*RC*lag circuit, the phase angle φ between the input and output isφ = −tan

^{−1}(*R*÷*X*)_{C} - For an
*RC*lead circuit, the phase angle φ between the input and output isφ = tan

^{−1}(*X*_{C}*÷ R*)

- For an
- Using these formulas, and by choosing appropriate values of
*R*and*C*, you can design a lead circuit or lag circuit to shift the voltage by any desired angle. - Here's a nice memory trick to help you remember those two formulas:
Notice that the order of the
*R*and the*X*in each formula is the same as the order of the resistor and the capacitor in the corresponding schematic diagram._{C}

- A simple series
*RL*circuit can also serve as either a lag circuit or a lead circuit, depending on whether you take the output voltage across the resistor or across the inductor. In particular:- In any series
*RL*circuit, the resistor's voltage lags the source voltage, so you'll have a**lag circuit**if you take the output voltage across the resistor, as shown here:

- Also, in any series
*RL*circuit, the inductor's voltage leads the source voltage, so you'll have a**lead circuit**if you take the output voltage across the inductor, as shown here:

- In any series
- If you remember
**ELI the ICEman**, you'll be able to quickly identify circuits like the ones above as either lead circuits or lag circuits.**ELI**reminds you that an inductor's voltage tends to lead everything else in the circuit, so you've got a lead circuit if you're taking the output voltage across the inductor.

- Whenever you encounter circuits like these, you can always use the
general techniques you learned above to analyze the circuit and figure
out how far the output voltage is shifted from the input voltage.
Or you can remember the following formulas.
- For an
*RL*lag circuit, the phase angle φ between the input and output isφ = −tan

^{−1}(*X*_{L}*÷ R*) - For an
*RL*lead circuit, the phase angle φ between the input and output isφ = tan

^{−1}(*R ÷ X*)_{L}

- For an
- Here's a nice memory trick to help you remember those two formulas:
Notice that the order of the
*R*and the*X*in each formula is the same as the order of the resistor and the inductor in the corresponding schematic diagram._{L}

- This e-Lesson has covered several important topics, including:
- series AC circuits
- phasor diagrams
- Kirchhoff's Voltage Law
- voltage-divider rule
- troubleshooting series AC circuits
- lag circuits and lead circuits.

- 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.