# 2.4.4: Series-Parallel Analysis


Analysis of series-parallel networks involves recognizing those sub-circuits that are in series or that are in parallel among themselves, performing simplifications as needed, and winding up with a simple series-only or parallel-only equivalent. Then the various laws such as Ohm's law, KVL, KCL, VDR and CDR are applied to the various simplified networks to determine the parameters of interest. There is no single solution technique, each circuit being unique.

Figure 5.4.1 : A simple series-parallel circuit.

Let's begin by considering the circuit of Figure 5.4.1 . To review, this is neither just a series circuit nor just a parallel circuit. If it was a series circuit then the current through all components would have to be same, that is, there would no nodes where the current could divide. This is clearly not the case as the current flowing through $$R_1$$ can divide at node $$b$$, with one portion flowing down through $$R_2$$ and the remainder through $$R_3$$. On the other hand, if it was strictly parallel, then all of the components would have to exhibit the same voltage and therefore there would be only two connection points in the circuit. This is also not the case as there are three such points: $$a$$, $$b$$ and ground.

What is true is that resistors $$R_2$$ and $$R_3$$ are in parallel. We know this because both components are attached to the same two nodes, $$b$$ and ground, and must exhibit the same voltage, $$V_b$$. As such, we can find the equivalent resistance of this pair and treat the result as a single resistance, let's call it $$R_x$$. In this newly simplified circuit, Rx is in series with $$R_1$$ and the source, $$E$$. We have reduced the original circuit down to a simple series circuit and thus the series analysis rules may be applied.

There are many solution paths at this point. For example, we could find the total resistance, $$R_t$$, by adding $$R_1$$ to $$R_x$$. Dividing this by $$E$$ yields the total current flowing out of the source, $$I_{total}$$. This current must flow through $$R_1$$ so Ohm's law can be used to find the voltage drop across $$R_1$$. This same current must be flowing through $$R_x$$, so Ohm's law can be used to find the associated voltage ($$V_b$$). The currents through $$R_2$$ and $$R_3$$ may then be found using Ohm's law for each resistor (e.g., the current through $$R_2$$ must be $$V_b/R_2$$). Alternately, these currents may be found by using the current divider rule between $$R_2$$ and $$R_3$$ (e.g., the current through $$R_2$$ must be $$I_{total} \cdot R_1/(R_1 + R_2)$$; remember, the current divider rule uses the ratio of the opposite resistor over the sum).

Another solution path would be to apply the voltage divider rule to $$R_1$$ and $$R_x$$ in order to derive the two voltage drops (or the rule can be applied to find just one of the drops and the other voltage may be found by subtracting that from the source, an application of KVL). Once the voltages are determined, Ohm's law can be used to find the currents. If powers are needed, they are easily determined once the voltages and currents are found.

##### Example 5.4.1

A series-parallel circuit is shown in Figure 5.4.2 . Determine $$V_b$$.

Figure 5.4.2 : Circuit for Example 5.4.1 .

As we are only interested in finding one voltage, the voltage divider rule is a good candidate. The 1 k$$\Omega$$ is in parallel with the 3 k$$\Omega$$, yielding an equivalent resistance of 750 $$\Omega$$. From here we apply VDR.

$V_b = E \frac{R_{eq}}{R_1+R_{eq}} \nonumber$

$V_b = 6V \frac{750 \Omega}{250 \Omega +750 \Omega} \nonumber$

$V_b = 4.5V \nonumber$

To clarify the current directions and voltage polarities, the circuit is redrawn and appropriately labeled in Figure 5.4.3 . We can crosscheck the answer in a variety of ways. For example, Figure 5.4.3 indicates, via KVL, that the voltage drop across the 250 $$\Omega$$ resistor plus $$V_b$$ must equal the source voltage of 6 volts. Consequently, $$V_{250}$$ must be 6 volts minus 4.5 volts, or 1.5 volts. Ohm's law then indicates that the current through the 250 $$\Omega$$ must be 1.5 V/250 $$\Omega$$, or 6 milliamps. Based on the earlier resistive equivalence, the total resistance as seen by the source must be 250 $$\Omega$$ in series with 750 $$\Omega$$, or 1 k$$\Omega$$. This indicates the source current must be 6 V/1 k$$\Omega$$, or 6 milliamps. As the source and the 250 $$\Omega$$ are in series, they must see the same current. This crosscheck shows that they do: happy happy joy joy.

Figure 5.4.3 : Circuit for Example 5.4.1 with polarities and directions included.

We can go further and verify the currents via KCL. For the first parallel resistor we find:

$I_{1k} = \frac{V_b}{1 k \Omega} \nonumber$

$I_{1k} = \frac{4.5V}{1 k \Omega} \nonumber$

$I_{1k} = 4.5 mA \nonumber$

And for the second resistor we see:

$I_{3k} = \frac{V_b}{1 k \Omega} \nonumber$

$I_{3k} = \frac{4.5V}{3 k \Omega} \nonumber$

$I_{3k} = 1.5mA \nonumber$

Obviously, these sum to the entering current of 6 mA.

##### Example 5.4.2

Determine $$V_b$$ and the source current in the circuit of Figure 5.4.4 .

Figure 5.4.4 : Circuit for Example 5.4.2 .

In this circuit the 3 k$$\Omega$$ resistor is in parallel with the series combination of the 2 k$$\Omega$$ and 1 k$$\Omega$$. This leads to an equivalent resistance of 3 k$$\Omega$$ in parallel with 3 k$$\Omega$$, or 1.5 k$$\Omega$$. From here we can find the source current.

$I_s = \frac{E}{R_{Total}} \nonumber$

$I_s = \frac{30 V}{1.5k \Omega} \nonumber$

$I_s = 20mA \nonumber$

This current should split evenly down the two vertical paths as they each present 3 k$$\Omega$$ of resistance (10 mA each achieves 30 volts for $$V_a$$, which is the source).

The voltage divider rule is a good choice for $$V_b$$ as we know the applied voltage.

$V_b = E \frac{R_{eq}}{R_1+R_{eq}} \nonumber$

$V_b = 30 V \frac{1k \Omega}{1k \Omega +2k \Omega} \nonumber$

$V_b = 10 V \nonumber$

For clarity, the circuit is redrawn and relabeled with current directions and voltage polarities in Figure 5.4.5 . As a further exercise, try to verify the answers above using alternate means, as illustrated in Exercise 5.4.

Figure 5.4.5 : Circuit for Example 5.4.2 with polarities and directions included.

As the circuit grows, more and more solution paths exist. Consider the circuit of Figure 5.4.6 . In this case, $$R_3$$ and $$R_4$$ are in parallel. This parallel combination is in series with $$R_2$$. Finally, this set of three resistors is in parallel with $$R_1$$ and $$E$$, reducing to a parallel circuit. Consequently, we know that the voltage across $$R_1$$ must be equal to $$E$$. Also, the currents through $$R_1$$ and $$R_2$$ must add up to the current exiting the source (KCL). Further, the currents through $$R_3$$ and $$R_4$$ must add up to the current flowing through $$R_2$$ (KCL). Also, the voltages across $$R_3$$ and $$R_4$$ must be the same (they are in parallel) and that this voltage plus the voltage across $$R_2$$ must equal $$E$$ (KVL). One solution path would be to find the total resistance as seen by the source, $$R_1 || (R_2 + (R_3 || R_4))$$, and use this to find the total current flowing out of the source. The current divider rule can then be used between $$R_1$$ and the equivalent resistance $$(R_2 + (R_3 || R_4))$$. Ohm's law can be used subsequently to find various voltage drops. Alternately, the voltage divider rule can be used between $$R_2$$ and the parallel equivalent $$(R_3 || R_4)$$ as this combination is driven by $$E$$. Knowing the voltages, the currents may be determined. With so many possible solution paths for large circuits, it is often worthwhile to take a moment to map out a strategy rather than just “diving in” and hoping it will all work out.

Figure 5.4.6 : A slightly more complex series-parallel circuit.

##### Example 5.4.3

The circuit of Figure 5.4.7 is the same as the resistive network presented in Example 5.3.1 but with the addition of a voltage source. Determine $$V_b$$, $$V_c$$, $$V_d$$, the source current $$I_s$$, and the current flowing through the 40 $$\Omega$$ resistor, $$I_{40}$$.

Figure 5.4.7 : Circuit for Example 5.4.3 .

Thanks to Example 5.3.1, the equivalent resistances of various portions have already been determined, saving some time. To reiterate, the pair of 100 $$\Omega$$ resistors are in parallel for 50 $$\Omega$$. That's in series with the 30 $$\Omega$$ for 80 $$\Omega$$. In the middle path, the 200 $$\Omega$$ and 40 $$\Omega$$ are in series, yielding 240 $$\Omega$$. 240 $$||$$ 80 = 60 $$\Omega$$. This is in series with the 10 $$\Omega$$, yielding a total resistance loading the source of 70 $$\Omega$$.

$I_s = \frac{E}{R_{Total}} \nonumber$

$I_s = \frac{ 16V}{70 \Omega} \nonumber$

$I_s \approx 228.6mA \nonumber$

The three voltages can be quickly found via VDR as we know all of the associated sub-circuit resistances.

$V_b = E \frac{R_x}{R_x+R_y} \nonumber$

$V_b = 16V \frac{60 \Omega}{60 \Omega +10 \Omega} \nonumber$

$V_b \approx 13.71V \nonumber$

$V_c = V_b \frac{R_x}{R_x+R_y} \nonumber$

$V_c = 13.71 V \frac{50 \Omega}{50 \Omega +30 \Omega} \nonumber$

$V_c \approx 8.57 V \nonumber$

$V_d = V_b \frac{R_x}{R_x+R_y} \nonumber$

$V_d = 13.71V \frac{200 \Omega}{200 \Omega +40 \Omega} \nonumber$

$V_d \approx 11.43V \nonumber$

Lastly, the current through the 40 $$\Omega$$ resistor can be found by dividing its voltage by its resistance. Its voltage is $$V_b − V_d$$. A quicker method is to note that because it is in series with the 240 $$\Omega$$, their currents are the same.

$I_{40} = \frac{V_b}{R_{Series}} \nonumber$

$I_{40} = \frac{13.71 V}{240 \Omega} \nonumber$

$I_{40} \approx 57.1mA \nonumber$

Once again the circuit is redrawn in Figure 5.4.8 to illustrate the current directions and voltage polarities. The usefulness of this will become even more apparent in the next example.

Figure 5.4.8 : Circuit for Example 5.4.3 with directions and polarities.

##### Example 5.4.4

Determine $$V_{ac}$$ and the voltages across each resistor in the circuit of Figure 5.4.9 .

Figure 5.4.9 : Circuit for Example 5.4.3 .

This circuit is perhaps the most complex offered so far. It may not be immediately apparent what the voltage polarities are for the various resistors or the proper current directions. Without these data, it will be impossible to determine $$V_{ac}$$. To complicate matters, the voltage of interest is not ground referenced, but instead refers to two nodes located in arbitrary points of the circuit. It is worth recognizing that, by definition, $$V_{ac} = V_a − V_c$$, but how are those voltages determined?

To assist with this minor quandary, the circuit has been redrawn in Figure 5.4.10 . Note that the current source and associated 1 k$$\Omega$$ resistor have been flipped to the side, essentially trading places with the 2 k$$\Omega$$ and 7 k$$\Omega$$ series combination. This is entirely acceptable as both subgroups share node b and ground, meaning they are in parallel. Of course, the order of parallel elements in a situation like this does not matter. This orientation, having the source at one end of the circuit, tends to make visualization of current flows a little easier.

Figure 5.4.10 : Circuit for Example 5.4.3 redrawn and with polarities and current directions added.

In any case, the current exits the source and then splits at node $$b$$, some heading through the 10 k$$\Omega$$ with the rest flowing down the 7 k$$\Omega$$ with 2 k$$\Omega$$ series combo. At node $$c$$ the current that entered the 10 k$$\Omega$$ splits again, flowing down into the parallel subgroup of the 12 k$$\Omega$$ and 24 k$$\Omega$$. Knowing the current directions, the voltage polarities are determined directly.

Now that the circuit has been redrawn, several solution paths might come to mind. For starters, the voltage drop across the 1 k$$\Omega$$ is trivial as both its current and resistance are known. From there, a current divider can be used to find the current flowing down into the series 7 k$$\Omega$$ plus 2 k$$\Omega$$ combo. Ohm's law could then be used to find their voltages. Following that, KCL can be used to find the current through the 10 k$$\Omega$$ (and thus its voltage), and then Ohm's law can be applied to find $$V_c$$ by treating the two right-most parallel resistors as a single unit.

Another solution path would be to determine the equivalent resistance that loads the current source. This allows the source voltage to be determined, after which a voltage divider may be applied between the 1 k$$\Omega$$ and the remaining five resistors to find $$V_b$$. From there, two more voltage dividers can be used to find $$V_a$$ and $$V_c$$ which will allow determination of the remaining resistor voltages via KVL. There are other possible paths as well but for lack of a better reason, we shall chase the first path that was outlined.

Before getting too deep, it would be wise to determine the equivalent resistances of a few subgroups. First, the 12 k$$\Omega$$ $$||$$ 24 k$$\Omega$$ combo reduces down to 8 k$$\Omega$$. This is in series with the 10 k$$\Omega$$, leaving 18 k$$\Omega$$ for the three right-most resistors. In the middle branch, the 2 k$$\Omega$$ + 7 k$$\Omega$$ series combo is equivalent to 9 k$$\Omega$$. The parallel combination of 9 k$$\Omega$$ $$||$$ 18 k$$\Omega$$ yields 6 k$$\Omega$$. Finally, this last bit is in series with the 1 k$$\Omega$$ producing 7 k$$\Omega$$ for the equivalent resistance of all resistors.

The source voltage and the voltage across the 1 k$$\Omega$$ resistor are determined via Ohm's law:

$V_{source} = I_{source} \times R_{equivalent} \nonumber$

$V_{source} = 10 mA \times 7k \Omega \nonumber$

$V_{source} = 70 V \nonumber$

$V_{1k} = I_{source} \times R \nonumber$

$V_{1k} = 10mA \times 1k \Omega \nonumber$

$V_{1k} = 10 V \nonumber$

As the source produces 70 volts and the drop on the 1 k$$\Omega$$ is 10 volts, then by KVL, $$V_b$$ must be 70 volts − 10 volts, or 60 volts (which will be useful for a crosscheck in a moment). Using a current divider, the 10 mA source splits between the middle and right branches. The equivalent resistances of those branches are 9 k$$\Omega$$ and 18 k$$\Omega$$, as found earlier.

$I_{middle} = I_{source} \frac{R_{right}}{R_{right}+R_{middle}} \nonumber$

$I_{middle} = 10 mA \frac{18k \Omega}{18 k \Omega +9k \Omega} \nonumber$

$I_{middle} \approx 6.667mA \nonumber$

Via KCL, the right branch current must be 10 mA − 6.667 mA, or approximately 3.333 mA. As a crosscheck, if 6.667 mA passes through a total of 9 k$$\Omega$$, that produces a voltage drop of 60 volts, exactly as expected from the earlier calculation for node $$b$$. At this point, the current through each resistor is known (treating the 12 k$$\Omega$$ $$||$$ 24 k$$\Omega$$ combo as a unit) and therefore Ohm's law may be used to determine their voltage drops. They are summarized below.

$V_{7k} \approx 46.67 V \nonumber$

$V_{2k} \approx 13.33 V \nonumber$

$V_{10k} \approx 33.33 V \nonumber$

$V_{12k} = V_{24k} \approx 26.67 V \nonumber$

Finally, to determine $$V_{ac}$$ we know that $$V_{ac} = V_a − V_c$$, and by observation $$V_a$$ is the drop across the 2 k$$\Omega$$ while $$V_c$$ is the drop across the 12 k$$\Omega$$ $$||$$ 24 k$$\Omega$$ combo.

$V_{ac} = V_a−V_c \nonumber$

$V_{ac} \approx 13.33V−26.67 V \nonumber$

$V_{ac} \approx −13.33 V \nonumber$

Do not let the negative sign be bothersome. All it indicates is that node $$a$$ is at a lower potential than node $$c$$, i.e., that node $$a$$ is 13.33 volts below node $$c$$. It is also purely coincidental that $$V_{ac}$$ and $$V_a$$ have the same magnitude. In the lab, if a voltmeter is connected such that the red (positive) lead is attached to node $$a$$ and the black (negative or reference) lead is attached to node $$c$$, the meter will show a negative value. If the leads are connected in reverse, thus measuring $$V_{ca}$$, the voltage will show up as positive (meaning that node $$c$$ is higher in potential than node $$a$$, the reverse way of looking at the situation)1.

## Any Route Works

In the example just completed it was noted that, by definition, $$V_{ac} = V_a − V_c$$, and this dictated the method employed to find said voltage. It turns out that this is just one of many ways of determining a voltage from one point to another. In general:

$\text{To find some voltage } V_{XY}, \text{start at point X and proceed along any convenient route to Y, subtracting voltage rises (polarities of − to +) and adding voltage drops (polarities of + to −) along the way.} \nonumber$

In the prior example, an obvious route for $$V_{ac}$$ would be to go down the 2 k$$\Omega$$ and then up the 12 k$$\Omega$$. That gives a drop of approximately13.33 volts (positive). Going up the 12 k$$\Omega$$ produces a rise of approximately 26.67 volts (negative). The result is approximately −13.33 volts, as previously calculated. The technique states that any route may be used, no matter how inefficient or wacky it might be. To verify this, one odd route might be to go up the 7 k$$\Omega$$, down through the 1 k$$\Omega$$ and source, and then up through the 24 k$$\Omega$$. The summation would be − 46.67 − 10 + 70 − 26.67, or approximately −13.33 volts once again. An easy way to remember whether the potential is added or subtracted is to just look at the polarity sign where the component is entered along your chosen route.

As promised earlier in the chapter, it's time to look at a ladder network. While ladders can use any resistance values desired, certain ratios turn out to be eminently practical. Of particular note is the R-2R ladder, an example of which is shown in Figure 5.4.11 , and which is being driven by a current source. In this configuration, only two different resistor values are used; some initial value and a second value of precisely twice that size. As drawn in Figure 5.4.11 , all horizontal resistors use value R (6 k$$\Omega$$ here) while all of the vertical resistors use 2R (12 k$$\Omega$$ here) with the exception of the very last, or terminating, resistor which uses R.

Figure 5.4.11 : An R-2R ladder network.

It turns out that this arrangement offers something very unique: a halving of successive node voltages and branch currents. This is extremely useful in digital to analog conversion circuits; the sort of hardware that would turn the digital bits of an MP3 or WAV file into listenable analog signals to feed loudspeakers or headphones.

Here is how it works. Starting at the far right end, it should be obvious that the last two 6 k$$\Omega$$ resistors comprise a 50% voltage divider, and thus $$V_e$$ will half of $$V_d$$. The combination of the final three resistors works out to 12 k$$\Omega$$ $$||$$ (6 k$$\Omega$$ + 6 k$$\Omega$$), or 6 k$$\Omega$$ again. Thus, the divider from $$c$$ to $$d$$ is also 50%. This process repeats as we move to the left, each equivalent working out to R and resulting in a 50% voltage divider. It will not matter what value is chosen for R. As long as 2R is used for the other set of resistors, the result will always be successive 50% voltage dividers.

##### Example 5.4.5

Determine all lettered node voltages in the circuit of Figure 5.4.11 .

Continuing the prior analysis, the entire resistor network presents an equivalent resistance of 12 k$$\Omega$$. The node $$a$$ voltage is determined by Ohm's law:

$V_a = I \times R \nonumber$

$V_a = 24mA \times 12 k \Omega \nonumber$

$V_a = 288 V \nonumber$

The prior analysis indicated that the subsequent node voltages should be, from left to right, 144 volts, 72 volts, 36 volts and 18 volts. A quick check follows. The source of 24 mA flows into the left-most 6 k$$\Omega$$, producing a drop of 144 volts by Ohm's law. This can be subtracted from $$V_a$$ (288 V) to arrive at $$V_b$$, or 144 volts, or half of the prior node voltage, as expected. The current through the left-most vertical resistor must be $$V_b$$/12 k$$\Omega$$, or 12 mA. That's half of the entering current, as expected. KCL indicates that the remaining current, 24 mA − 12 mA, or 12 mA, must be flowing into the second 6 k$$\Omega$$ between nodes $$b$$ and $$c$$. This will produce a voltage drop of 72 volts across that resistor. In other words, $$V_c$$ must be 72 volts less than $$V_b$$ (144 volts), or 72 volts. This is half of $$V_b$$, as expected. Likewise, the current flowing down through the second 12 k$$\Omega$$ must be $$V_c$$/12 k$$\Omega$$, or 6 mA, once again half of the preceding current, as expected. The process continues as we move to the right, each subsequent node seeing half the voltage of the prior node and each current getting cut in half as well.

## Computer Simulation

In order to verify the results of Example 5.4.4 , the R-2R ladder circuit is entered into a simulator, as shown in Figure 5.4.12 . The numbered nodes 1 through 5 in the simulation schematic correspond to the lettered nodes of the original circuit, $$a$$ through $$e$$, respectively. This is one example where the use of virtual instruments would result in a very cluttered display with five multimeters. Consequently, the simulation uses a DC operating point analysis with a single output window, as shown in Figure 5.4.13 .

Although the node numbers are not in ascending order, the voltages are a perfect match to the results computed previously. Each successive node sees half the voltage of the prior node. Although not explicitly listed here, it should be obvious that the currents will also be cut in half when progressing from left to right, via a quick application of Ohm's law.

Figure 5.4.12 : The circuit of Example 5.4.3 in a simulator.

Figure 5.4.13 : Simulation results of the circuit of Example 5.4.4 .

## Bridges

A bridge is a particular series-parallel configuration consisting of two pairs of series connected elements placed in parallel. An example of a resistive bridge being driven by a voltage source is shown in Figure 5.4.14 . In this circuit, $$R_1$$ and $$R_2$$ create one series connection while $$R_2$$ and $$R_3$$ create the other. The two pairs are then placed in parallel. While this circuit is shown with a voltage source, it could also be driven by a current source. Other elements might also be added.

Figure 5.4.14 : A resistive bridge network.

A typical use for a bridge is the measurement of some environmental quantity. As stated in Chapter 2, there are certain resistive devices that are sensitive to environmental change. Examples include photoresistors, whose resistance is a function of light level; and thermistors, whose resistance is a function of temperature.

To understand the operation, the voltages $$V_b$$ and $$V_c$$ are established by the voltage dividers comprised of $$R_1$$ and $$R_2$$; and $$R_3$$ and $$R_4$$, respectively. With ordinary resistors, these voltages would be unchanging and therefore $$V_{bc}$$ would also be a fixed value. We shall set it up such that $$V_b$$ and $$V_c$$ are the same voltage, and thus $$V_{bc}$$ will be zero. Now assume that we replace $$R_1$$ with a photoresistor. As the light level increases, its resistance decreases. This will cause $$V_b$$ to rise which will cause $$V_{bc}$$ to go positive. If the light level decreases, the resistance of $$R_1$$ will increase, forcing $$V_b$$ to drop which will cause $$V_{bc}$$ to go negative. The greater the change in light, the larger $$V_{bc}$$ will be, and its sign indicates whether the light level has increased or decreased compared to the original set point. By simply placing some manner of voltmeter between points $$b$$ and $$c$$, we can create a display that indicates light levels.

If we want to flip the sign, we would place the photoresistor in the position of $$R_2$$ instead of $$R_1$$. Further, it is possible to increase the sensitivity by using sensors at opposite corners (e.g., $$R_1$$ and $$R_4$$), and comparative or differential measurements are possible by using both sides (e.g., $$R_1$$ with $$R_3$$).

##### Example 5.4.6

Determine $$V_{bc}$$ in the circuit of Figure 5.4.15 .

Figure 5.4.15 : Circuit for Example 5.4.5 .

Perhaps the most straightforward way to do this is to treat each half as a voltage divider and then subtract $$V_c$$ from $$V_b$$.

$V_b = E \frac{R_2}{R_2+R_1} \nonumber$

$V_b = 12V \frac{2k \Omega}{2k \Omega +1k \Omega} \nonumber$

$V_b = 8V \nonumber$

$V_c = E \frac{R_4}{R_4+R_3} \nonumber$

$V_c = 12 V \frac{4k \Omega}{ 4k \Omega +3k \Omega} \nonumber$

$V_c \approx 6.857V \nonumber$

Thus, $$V_{bc}$$ is 8 volts minus 6.857 volts, or 1.143 volts.

## Computer Simulation

In order to verify the results of Example 5.4.5 , the bridge network of Figure 5.4.15 is entered into a simulator, as depicted in Figure 5.4.16 .

Figure 5.4.16 : The circuit of Example 5.4.5 in a simulator.

A DC operating point analysis is run and the results are shown in Figure 5.4.17 . The voltages are precisely as expected.

Figure 5.4.17 : Simulation results for the bridge circuit of Example 5.4.5 .