In Unit 9 you studied concepts and techniques for analyzing series-parallel circuits in general. In this unit we’ll look at several specific examples of series-parallel circuits that technicians frequently encounter. We’ll also look at techniques for troubleshooting series-parallel circuits.

- This unit will build on material that you studied in Unit 9. So let’s begin by taking this self-test to review what you learned in that unit.

- Some electrical devices require two or more different supply voltages
to work properly.
- For instance, a battery-powered smoke detector might
contain a buzzer that runs off of 9 V, as well as a light bulb
that requires 3 V. Does this mean that our smoke detector must
contain a 9 V battery for the buzzer and also two 1.5 V
batteries connected in series for the light bulb?
**No!**We can use just the 9 V battery if we also include a voltage divider in the design of our smoke detector.

- For instance, a battery-powered smoke detector might
contain a buzzer that runs off of 9 V, as well as a light bulb
that requires 3 V. Does this mean that our smoke detector must
contain a 9 V battery for the buzzer and also two 1.5 V
batteries connected in series for the light bulb?
- Recall from Unit
7 that a string of series resistors is often called a
**voltage divider**because the total voltage across the entire string is divided among the various resistors in direct proportion to the resistance of each one. - Take a look at the circuit shown below. This is just a series circuit
with two resistors and a voltage source. It looks a bit different
from series circuits that you’ve studied before, because this one
has three labeled wires connected between the components. But since
the other ends of those wires are not connected to anything, no current
will flow through the wires, and the wires will not have any effect
on the voltages or currents in the circuit. So we’re still dealing
with a simple series circuit.

- Using your knowledge of series circuits, you can easily find that
the voltage across R2 is 3 V. Therefore, we see that

and*V*= 9 V_{AC}*V*= 3 V_{BC} - So we can think of the entire circuit shown above as a voltage supply
that provides two different output voltages. It provides 9 V between
points
*A*and*C*, and it provides 3 V between points*B*and*C*.- Going back to our smoke-detector example, we could now connect the
buzzer to points
*A*and*C*, so that it gets the 9 V that it needs to work properly, and we could connect the light bulb to points*B*and*C*, so that it gets the 3 V that it requires.

- Going back to our smoke-detector example, we could now connect the
buzzer to points

- The circuit that we looked at above is an
**unloaded voltage divider**.**Unloaded**means that nothing is connected to the output points*B*and*C*. If we connect one or more components (such as a light bulb) to those points, we will have a**loaded voltage divider**. (The components that we connect, such as the bulb, are called**loads**.) - After we connect one or more loads, we no longer have a simple series
circuit. Rather, we have a series-parallel circuit. The diagram below
shows our voltage divider with a load resistor, R
_{L}, connected to points*B*and*C*. We now have a series-parallel circuit, since the parallel combination of R2 and R_{L}is in series with R1.

- Depending on the value of
*R*_{L}, the voltage across the load may be very close to the original unloaded voltage between points*B*and*C*, or it may be very different.- In particular, if
*R*_{L}is much larger than R2, then the voltage across the load will be very close to the original voltage. - But if
*R*_{L}is not much larger than R2, then the voltage between points*B*and*C*will decrease significantly when you connect the load.

- In particular, if
- In the following self-test you will look at some examples of this.

- An ideal (perfect) measuring instrument does not change the value
of the quantity that it is used to measure. For instance, if you were
to connect an ideal voltmeter across a resistor in a circuit, the voltage
across that resistor after you connect the voltmeter would be
**exactly**the same as the voltage before you connected the meter. - It’s impossible to build an ideal voltmeter. Any real voltmeter
will have some effect on the voltage that it’s being used to measure.
This effect is called the
**loading effect**of the voltmeter. - Often,
the meter’s loading effect is so small that we can safely ignore
it.
- For instance, suppose the voltage across a resistor in a circuit is 9.00 V when no voltmeter is connected, and suppose that connecting a meter changes the voltage slightly to 8.99 V. This is such a small change that, most of the time, we could safely ignore it, and we could pretend that connecting the voltmeter made no change in the actual voltage.

- To understand why connecting a voltmeter (or a multimeter) causes
the voltage to change, take a look at the following diagram.
- Ignoring the voltmeter, we have a simple series circuit containing three resistors and a voltage source.
- But in order to use the voltmeter to measure R2’s voltage, we connect the meter in parallel with R2. So if you think of the voltmeter as being a component added to the circuit, we no longer have a simple series circuit once the meter is connected. Rather, we have a series-parallel circuit in which R2 and the meter are connected in parallel, and then this parallel combination is connected in series with R1 and R3.

- A voltmeter or multimeter is itself a complicated circuit containing
many components, such as resistors and capacitors. But for many purposes,
when we want to consider the effect of attaching a meter to
a circuit, we can simply replace the meter with its
**internal resistance**, which is a single resistance value that lets us estimate the meter’s loading effect on a circuit. - A meter’s instruction manual will usually specify the value of the
meter’s internal resistance (or, as it may be called, the meter’s
**internal impedance**).- For instance, if you look in the instruction manual for the Fluke 8050A digital multimeters that we use in Sinclair’s labs, you’ll find that when the meter is set to measure voltage, its internal impedance is 10 MΩ.

- We’ll use the symbol
*R*to represent the meter’s internal resistance._{M} - The following diagram shows the same series circuit again, but this
time we’ve replaced the voltmeter with a resistor whose value is equal
to the meter’s internal resistance,
*R*. As you can clearly see, we now have a series-parallel circuit whose total resistance is different from the total resistance of R1, R2, and R3 by themselves. This means that the circuit’s total current and its voltage drops will also be different from what they would be without the meter attached._{M}

- In the following self-test you will look at some examples of this.

- A resistive
**ladder network**is a group of resistors connected in a particular series-parallel configuration. An example is contained in the circuit shown below. It’s called a ladder network because if you turn this diagram on its side, it looks like a ladder that you could climb up. (R2, R4, and R6 form the rungs that you would step on, while R1, R3, and R5 form one side that you would hold onto as you climb.)

- This ladder network could be extended by adding more resistors (two at a time) in the same pattern to the right-hand end of the circuit.
- You can analyze such a circuit using the same techniques that you’ve
learned for series-parallel circuits in general:
- Determine which resistors or combinations of resistors are in series or in parallel with each other.
- Combine the resistors together to get the circuit’s total resistance.
- Use Ohm’s law to find the circuit’s total current. (
*I*_{T}=*V*_{S}÷*R*_{T}) - Using a combination of Ohm’s law, Kirchhoff’s Voltage Law, Kirchhoff’s Current Law, the voltage-divider rule, and the current-divider rule, find the voltage across or the current through any of the circuit’s resistors.

- In general, circuits containing ladder
networks are no easier or harder to analyze than other series-parallel
circuits with the same number of resistors. One special case, however,
is fairly common and is also easy to analyze because of a shortcut
that you can use. This special case is called an
**R/2R ladder network**. It is used in digital-to-analog converters, which you’ll study in a later course. In an R/2R ladder network, some of the resistors have one value (called R), and the other resistors have twice this value (2R).

- The
**Wheatstone bridge**, shown below, is a circuit that is used to make precise resistance measurements.

- As you can see from the diagram,
it’s a fairly simple series-parallel circuit, containing four resistors
and a voltage source. As long as nothing is connected to points
*A*and*B*, those two short wires have no effect on the circuit. So R1 and R3 are connected in series with each other, and R2 and R4 are connected in series with each other. These two series combinations are then connected in parallel with each other, so we have (R1 + R3) || (R2 + R4). - The voltage that we’re
primarily interested in is
*V*, the voltage between the two points labeled_{AB}*A*and*B*in the diagram. Notice that this voltage is**not**equal to the voltage drop across any one of the circuit’s resistors.

- If the output voltage
is equal to zero
**balanced Wheatstone bridge**. It’s easy to show that in such a case, the ratio of R1 to R3 must be equal to the ratio of R2 to R4:If

*V*_{OUT}= 0 V then R1 ÷ R3 = R2 ÷ R4 - So suppose that you know the values of R2 and R4 in a Wheatstone bridge,
and suppose that R1 has an unknown resistance that you’re trying to
find. Suppose also that R3 is a high-precision variable resistor.
By varying R3 until you get an output voltage of precisely zero volts
, you know that the ratio of R1 to R3 must be equal to the ratio of
R2 to R4: in symbols, R1 ÷ R3 = R2 ÷ R4.
Now you can rearrange this equation to solve for R1 in terms of the
three known resistances:
R1 = R3 × (R2 ÷ R4)

- If you start out with a balanced Wheatstone bridge (
*V*_{OUT }= 0 V), and if one of the resistor values then changes slightly, the output voltage will no longer be zero volts. Then you’ll have what’s called an**unbalanced Wheatstone bridge**. By seeing how far the voltage changes from zero volts, you can work backwards and figure out how much the resistance has changed from its initial value. The mathematical details can get a bit tricky, but here we’re mainly interested in the idea that by adjusting the known resistors until the bridge is balanced, you can find the value of an unknown resistance, using the equation given above.

- The Wheatstone bridge is useful because
many types of transducers convert physical quantities (such as temperature
or pressure) into resistance. Since the Wheatstone bridge lets us
make precise resistance measurements, it can be used in combination
with such transducers to let us make precise measurements of those
physical quantities (temperature, pressure, and so on).
So in the circuit diagram of a Wheatstone bridge shown earlier, the
unknown resistor (R1) would be replaced by a transducer whose
resistance we wish to find.
- Example: One widely used transducer is called a
**strain gauge**. This is a device whose resistance changes as the device is subjected to different pressures or forces. As a simple example of its use, imagine that a strain gauge has been embedded in the concrete beneath a huge tank for holding liquids. Also imagine that two wires run through the concrete to the strain gauge so that we can connect it to three resistors to form a Wheatstone bridge. When the tank is filled with liquid, it is very heavy, and so it exerts a great deal of pressure on the strain gauge beneath it. As the tank is emptied, it becomes lighter, so there is less pressure on the strain gauge. Since the strain gauge’s resistance varies as the weight above it varies, we can indirectly measure how full the tank is by measuring the strain gauge’s resistance. But since the change in the strain gauge’s resistance is not very large, we need to be able to measure its resistance very precisely. That’s exactly what the Wheatstone bridge lets us do.

- Example: One widely used transducer is called a

- 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. Let’s see how to do this.

**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.- Example: Suppose you want to know the values of currents and voltages
in the circuit shown below if resistor R2 is shorted.

To answer this question, first redraw the circuit as shown below, and then find the currents and voltages in this 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.- Example: Suppose you want to know the values of currents and voltages
in the circuit shown below if resistor R2 is open.

To answer this question, first redraw the circuit as shown below, and then find the currents and voltages in this redrawn circuit.

- This e-Lesson has covered several important topics, including:
- loaded voltage dividers
- voltmeter loading
- ladder networks
- the Wheatstone bridge
- troubleshooting series-parallel 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.