In Unit 6 and Unit 7 you
learned that to analyze a series AC circuit or a parallel AC circuit,
you follow the same steps as when you analyze a series DC circuit or
a parallel DC circuit, but that at each step you must use complex numbers
instead of real numbers. We’ll see in this unit that, as you’ve probably
guessed, **series-parallel** AC circuits are a lot like
series-parallel DC circuits, except that again you need to use complex
numbers throughout the analysis.

- This unit will build on material that you studied in Unit 7. So let’s begin by taking this self-test to review what you learned in that unit.

- In EET 1150 you learned how to analyze series-parallel DC circuits,
such as the one shown below. Let’s do a quick review of what you learned
there.

- First, realize that it’s not possible to give a step-by-step procedure that will work for all series-parallel DC circuits. There’s too much variety among these circuits, and an approach that works for one circuit may not work for another circuit.
- Having recognized this point, we can describe techniques that will let you analyze series-parallel DC circuits. In any particular case, it will be up to you to decide which of these techniques to use, and in what order.
- The key is practice, practice, and more practice.

- To deal with series-parallel circuits, you absolutely have to be
able to identify which components in a circuit are connected in series
with each other or in parallel with each other. Recall that:
- Two components are
**connected in series**if they are connected to each other at exactly one point and no other component is connected to that point. - Two components are
**connected in parallel**if they are connected to each other at two points.

- Two components are

- Once you’ve figured out how the resistors in a series-parallel DC circuit are connected, a good next step is usually to find the circuit’s total resistance by combining pairs of series-connected or parallel-connected resistors until you’ve combined all the resistors into a single resistance.
- After you have the total resistance, you can use Ohm’s law to find
the circuit’s total current:
*I*_{T}=*V*_{S}÷*R*_{T} - After that, you can apply various rules and laws, in any sequence
that leads to the solution for a particular voltage or current. The
most useful of these rules and laws are:
- Ohm’s law
- voltage/current rules for series or parallel combinations
- Kirchhoff’s Voltage Law
- Kirchhoff’s Current Law

- Let’s review these rules and laws.

- Recall that Ohm’s law relates resistance, current, and voltage. As
you know, the three forms of Ohm’s law are:
*I*=*V*÷*R*

*V*=*I*×*R*

*R*=*V*÷*I* - Ohm’s law can be applied to a
**single resistor**, or to an entire resistive**circuit**, or to a resistive**portion**of a circuit (such as a series path or a parallel combination of resistors).

- Here are two more simple-but-powerful rules. Recall that:
**Components connected in series must have the same current**(but usually they won’t have the same voltage).**Components connected in parallel must have the same voltage**(but usually they won’t have the same current).

- Recall that KVL says that
**the sum of the voltage drops around any closed loop in a circuit equals the sum of the voltage rises around that loop.**

- Recall that KCL says that
**the sum of all currents entering a point is equal to the sum of all currents leaving that point**.

- That ends our quick review of the rules that are most useful for analyzing series-parallel DC circuits. In this review I’ve skipped two other rules that can be useful: the voltage-divider rule and the current-divider rule. I’ve skipped them because in most cases you can get by without them, and because students often make mistakes in trying to apply these two rules to series-parallel circuits.
- If you’d like to review the two rules that I skipped, and to see examples of how they are used incorrectly, go to Unit 9 of EET 1150.

- To troubleshoot a series-parallel circuit, remember these key points
from earlier units:
**An open has zero current**(but usually it does not have zero voltage).- Also, all components in a series path must have the same
current. So
**any component in a series path with an open has zero current.** - Also, Ohm’s law, in the form
*V = I × R*, tells us that a resistor with zero current also has zero voltage. (This is true as long as the resistor itself is not open. If the resistor is open, then its resistance*R*is infinite, and we can’t apply Ohm’s law since we can’t multiply zero times infinity.) So**any good resistor in a series path with an open has zero voltage.**

- Also, all components in a series path must have the same
current. So
**A short has zero voltage**(but usually it does not have zero current).- Also, components connected in parallel must have the same
voltage. So
**any component in parallel with a short has zero voltage.** - Also, Ohm’s law, in the form
*I = V ÷ R*, tells us that a resistor with zero voltage also has zero current. (This is true as long as the resistor itself is not shorted. If the resistor is shorted, then its resistance*R*is zero, and we can’t apply Ohm’s law since we can’t divide zero by zero.) So**any good resistor in parallel with a short has zero current.**

- Also, components connected in parallel must have the same
voltage. So

- Often these facts will be enough to let you see the effect of a short
or open. Sometimes, though, you’ll need to perform a more complete
analysis, as follows:
**To analyze the effect of a shorted resistor in a circuit**, redraw the circuit with a wire replacing the shorted resistor. Then analyze the redrawn circuit.**To analyze the effect of an open resistor in a circuit**, redraw the circuit with a**broken wire**replacing the open resistor. Then analyze the redrawn circuit.

- That ends our quick review of series-parallel DC circuits. If you’d like a more complete review, go to Unit 9 and Unit 10 of EET 1150. Now let’s get back to AC circuits.

- As with series-parallel DC circuits, it’s impossible to give a single step-by-step procedure that will work for all series-parallel AC circuits. There’s too much variety among these circuits, and an approach that works for one circuit may not work for another circuit.
- But the same general techniques that work for series-parallel DC circuits also work for series-parallel AC circuits, except that you’ve got to use complex numbers instead of real numbers.

- When it comes to recognizing that components are connected in series
or in parallel, there’s no difference at all between DC and AC circuits.
Remember:
- Two components are
**connected in series**if they’re connected to each other at exactly one point and no other component is connected to that point. - Two components are
**connected in parallel**if they’re connected to each other at two points.

- Two components are
- For example, in the circuit shown below, R2 and L1 are connected
in series, and this series path is connected in parallel with R1, and
this entire combination of R1, L1, and R2 is connected in series with
C1.
- Using the symbol + to indicate a series connection, and the symbol
|| to indicate a parallel connection, we can write the following
shorthand description of the connections in this circuit:
C1 + (R1 || (R2 + L1))

- Using the symbol + to indicate a series connection, and the symbol
|| to indicate a parallel connection, we can write the following
shorthand description of the connections in this circuit:

- Once you’ve figured out how the components in a series-parallel AC circuit are connected, a good next step is usually to find the circuit’s total impedance by combining pairs of series-connected or parallel-connected components until you’ve combined all the components into a single impedance.
- Of course, as you’re doing this you have to remember to treat each component’s impedance as a complex number, not as a real number.
- After you’ve found the total impedance, you can use Ohm’s law to
find the circuit’s total current:
**I**_{T}=**V**_{S}÷**Z**_{T} - After that, you can apply various rules and laws, in any sequence
that leads to the solution for a particular voltage or current. The
most useful of these rules and laws are:
- Ohm’s law
- voltage/current rules for series or parallel combinations
- Kirchhoff’s Voltage Law
- Kirchhoff’s Current Law

- Let’s review these rules and laws.

- Here are two more rules that are simple, but powerful. Recall that:
**Components connected in series must have the same current**(but usually they won’t have the same voltage).**Components connected in parallel must have the same voltage**(but usually they won’t have the same current).

- These two rules apply not only to individual components but also
to portions of circuits.
- In our example circuit, for instance, R1 is connected in parallel
with the series path containing R2 and L1. Therefore, we know that
the voltage across R1 must be equal to the voltage across the series
path. In symbols:
**V**_{R1}=**V**_{R2+L1} - Also, in the same circuit, since R2 and L1 form a series path,
we know that these two components must carry the same current.
In symbols:
**I**_{R2}=**I**_{L1}

- In our example circuit, for instance, R1 is connected in parallel
with the series path containing R2 and L1. Therefore, we know that
the voltage across R1 must be equal to the voltage across the series
path. In symbols:

- In AC circuits, Ohm’s law relates current, voltage, and impedance.
The three forms of Ohm’s law are:
**I**=**V**÷**Z**

**V = I × Z**

**Z = V ÷ I** - In each of these forms, impedance
**Z**could be a resistor’s impedance (in which case we could also write it as**Z**), or a capacitor’s impedance (_{R}**Z**), or an inductor’s impedance (_{C}**Z**). Or it could be the combined impedance of two or more resistors, capacitors, and inductors. Regardless, we’re using the boldface notation here to remind ourselves that all of these quantities are complex numbers, not real numbers._{L} - Ohm’s law can be applied to a
**single component**, or to an entire**circuit**, or to a**portion**of a circuit (such as a series path or a parallel combination of components).- For instance, in our example circuit (repeated below), Ohm’s
law can be applied to any one of the components by itself.

Applying it to the inductor L1, we can write:**I**_{L1}=**V**_{L1}÷**Z**_{L1} - Or we can apply Ohm’s law (in any of its forms) to the entire
circuit, letting us write:
**I**_{T}=**V**_{S}÷**Z**_{T} - Or we can apply Ohm’s law to a portion of the circuit. For instace,
our example circuit contains a series path made up of R2 and L1.
Applying Ohm’s law just to this series path, we can say that the
current through the series path is equal to the voltage across
the series path divided by the impedance of the series path. In
symbols:

Here we’re using the subscript**I**_{R2+L1}=**V**_{R2+L1}÷**Z**_{R2+L1}_{R2+L1}to denote the series combination of R2 and L1. So**Z**_{R2+L1}is the impedance of this series path, and**V**_{R2+L1}is the voltage across this series path, and**I**_{R2+L1}is the current through this series path.

- For instance, in our example circuit (repeated below), Ohm’s
law can be applied to any one of the components by itself.

- Recall that KVL says that
**the sum of the voltage drops around any closed loop in a circuit equals the sum of the voltage rises around that loop.** - When you use KVL in an AC circuit, it’s crucial that you remember to treat voltages as complex numbers with both a magnitude and an angle. If you just add the magnitudes of the voltages, and you ignore their angles, you’ll usually get wrong answers.
- We originally studied KVL in the context of series circuits, but
KVL holds for all circuits.
- In our example circuit, for instance, one closed loop contains
the voltage source, C1, and R1. Therefore KVL tells us that the
voltage rise across the source must be equal to the sum of the
voltage drops across C1 and R1. In symbols:
**V**_{S}=**V**_{C1}+**V**_{R1} - Another closed loop in this circuit contains the voltage source,
C1, R2, and L1. Therefore KVL tells us that the voltage rise across
the source must be equal to the sum of the voltage drops across
those three components. In symbols:
**V**_{S}=**V**_{C1}+**V**_{R2}+**V**_{L1}

- In our example circuit, for instance, one closed loop contains
the voltage source, C1, and R1. Therefore KVL tells us that the
voltage rise across the source must be equal to the sum of the
voltage drops across C1 and R1. In symbols:

- Recall that KCL says that
**the sum of all currents entering a point is equal to the sum of all currents leaving that point**. - When you use KCL in an AC circuit, you must remember to treat currents as complex numbers with both a magnitude and an angle. If you just add the magnitudes of the currents, and you ignore their angles, the answers you get will probably be wrong.
- We originally studied KCL in the context of parallel circuits, but
KCL holds for all circuits.
- In our example circuit, for instance, consider the point where
C1 meets R1 and R2. KCL tells us that the current flowing into
this point from C1 must be equal to the sum of the currents flowing
out of this point through R1 and R2. In symbols:
**I**_{C1}=**I**_{R1}+**I**_{R2}

- In our example circuit, for instance, consider the point where
C1 meets R1 and R2. KCL tells us that the current flowing into
this point from C1 must be equal to the sum of the currents flowing
out of this point through R1 and R2. In symbols:

- If you apply them correctly, the voltage-divider rule and the current-divider rule can be useful in analyzing series-parallel AC circuits. But as I noted earlier, students often make mistakes in trying to apply these two rules to series-parallel DC circuits. And they’re even more difficult to use correctly in AC circuits, since you have to remember to use complex numbers instead of real numbers.
- For these reasons, and because you can almost always succeed using just the rules we discussed above, I recommend that you avoid using the voltage-divder and current-divider rules in analyzing series-parallel AC circuits.

- In the Self-Test Questions above you’ve used Ohm’s law, KVL, and
KCL to analyze the circuit shown below.

- In particular, you:
- Found the circuit’s total impedance to be
**Z**_{T}= 120.4 ∠−45.2º Ω. - Used Ohm’s law (
**I**_{T}=**V**_{S}÷**Z**_{T}) to find that the circuit’s total current is**I**_{T}= 41.54 ∠45.2º mA. - Recognized that, since C1 is in series with the voltage source,
C1’s current must equal the total current, so that
**I**_{C1}= 41.54 ∠45.2º mA. - Used Ohm’s law (
**V**_{C1}=**I**_{C1}×**Z**_{C1}) to find that the capacitor’s voltage drop is**V**_{C1}= 4.007 ∠−44.8º V. - Used KVL to find that R1’s voltage drop is
**V**_{R1}= 3.554 ∠52.6º V. - Used Ohm’s law (
**I**_{R1 }=**V**_{R1 }÷**Z**_{R1}) to find that R1’s current is**I**_{R1}= 35.54 ∠52.6º mA. - Used KCL to to find that R2’s current is
**I**_{R2 }= 7.792 ∠9.10º mA. - Used Ohm’s law (
**V**_{R2}=**I**_{R2}×**Z**_{R2}) to find that R2’s voltage drop is**V**_{R2}= 2.572 ∠9.10º V. - Recognized that, since R2 and L1 are in series, L1’s current
must equal R2’s current, so that
**I**_{L1}= 7.792 ∠9.10º mA. - Used Ohm’s law (
**V**_{L1}=**I**_{L1}×**Z**_{L1}) to find that the inductor’s voltage drop is**V**_{L1}= 2.448 ∠99.1º V.

- Found the circuit’s total impedance to be
- To analyze other series-parallel AC circuits, you’ll use these same rules and laws, but you’ll probably have to apply them in a different order. To get good at playing this game, you’ve got to practice, practice, practice!
- Even in the circuit above, we could have arrived at the same answers
by applying the rules in a different order. For example, once we had
found that
**V**_{R1}= 3.554 ∠52.6º V and that**V**_{R2}= 2.572 ∠9.10º V, we could have found**V**_{L1}by noticing that, in this circuit,**V**_{L1}=**V**_{R1}−**V**_{R2}. This would have given us the same answer for**V**_{L1}(plus or minus a small rounding error) that we got above.

- To troubleshoot a series-parallel AC circuit, remember these key
points:
**An open has zero current**(but usually it does not have zero voltage).- Also, all components in a series path must have the same
current. So
**any component in a series path with an open has zero current.** - Also, Ohm’s law, in the form
**V**=**I**×**Z**, tells us that a component with zero current also has zero voltage. (This is true as long as the component itself is not open. If the component is open, then its impedance**Z**is infinite, and we can’t take**I**×**Z**since we can’t multiply zero times infinity.) So**any good component in a series path with an open has zero voltage.**

- Also, all components in a series path must have the same
current. So
**A short has zero voltage**(but usually it does not have zero current).- Also, components connected in parallel must have the same
voltage. So
**any component in parallel with a short has zero voltage.** - Also, Ohm’s law, in the form
**I**=**V**÷**Z**, tells us that a component with zero voltage also has zero current. (This is true as long as the component itself is not shorted. If the component is shorted, then its impedance**Z**is zero, and we can’t take**V**÷**Z**since we can’t divide zero by zero.) So**any good component in parallel with a short has zero current.**

- Also, components connected in parallel must have the same
voltage. So

- Often these facts will be enough to let you see the effect of a short or open. Sometimes, though, you’ll need to perform a more complete analysis. Let’s see how to do this.

**To analyze the effect of a shorted component in a circuit**, redraw the circuit with a wire replacing the shorted component. Then analyze the redrawn circuit.- Example: Suppose you want to know the values of currents and voltages
in the circuit shown below if R1 is shorted.

To answer this question, first redraw the circuit as shown below, and then find the currents and voltages in this redrawn circuit.- Note: In this redrawn circuit, all current passing through C1 will take the path of zero resistance through the short, and no current will flow through R2 or L1. So the effect of this short is to remove R2 and L1 from the circuit.

**To analyze the effect of an open component in a circuit**, redraw the circuit with a**broken wire**replacing the open component. Then analyze the redrawn circuit.- Example: Suppose you want to know the values of currents and voltages
in the circuit shown below if resistor R1 is open.

To answer this question, first redraw the circuit as shown below, and then find the currents and voltages in this redrawn circuit.- Note: In this redrawn circuit, no current will pass through either of the dead-end wire stubs. So the effect of this open is to convert the circuit into a series circuit containing just the voltage source, C1, R2, and L1.

- This e-Lesson has covered several important topics, including:
- series-parallel AC circuits
- troubleshooting series-parallel AC 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.