Chapter 8: Series-Parallel DC Circuits

The basic technique used for solving dc combination-circuit problems is the use of equivalent circuits. To simplify a complex circuit to a simple circuit containing only one load, equivalent circuits are substituted (on paper) for the complex circuit they represent. To demonstrate the method used to solve combination circuit problems, the network shown in Figure 1(A) will be used to calculate various circuit quantities, such as resistance, current, voltage, and power.

Examination of the circuit shows that the only quantity that can be computed with the given information is the equivalent resistance of R₂ and R₃.

Given:

Solution:

Now that the equivalent resistance for R₂ and R₃ has been calculated, the circuit can be redrawn as a series circuit as shown in Figure 1(B).

The equivalent resistance of this circuit (total resistance) can now be calculated.

Given:

Solution:

The original circuit can be redrawn with a single resistor that represents the equivalent resistance of the entire circuit as shown in Figure 1(C).

To find total current in the circuit:

Given:

Solution:

To find total power in the circuit:

Given:

Solution:

To find the voltage dropped across R₁, R₂, and R₃, refer to Figure 1(B). R_eq1 represents the parallel network of R₂ and R₃. Since the voltage across each branch of a parallel circuit is equal, the voltage across R_eq1 (E_eq1) will be equal to the voltage across R₂ (E_R2) and also equal to the voltage across R₃ (E_R3).

Given:

Solution:

To find power used by R₁:

Given:

Solution:

To find the current through R₂ and R₃, refer to the original circuit, Figure 1(A). You know E_R2 and E_R3 from previous calculation.

Given:

Solution:

To find power used by R₂ and R₃, using values from previous calculations:

Given:

Solution:

Now that you have solved for the unknown quantities in this circuit, you can apply what you have learned to any series, parallel, or combination circuit. It is important to remember to first look at the circuit and from observation make your determination of the type of circuit, what is known, and what you are looking for. A minute spent in this manner may save you many unnecessary calculations.

Having computed all the currents and voltages of Figure 1, a complete description of the operation of the circuit can be made. The total current of 3 amps leaves the negative terminal of the battery and flows through the 8-ohm resistor (R₁). In so doing, a voltage drop of 24 volts occurs across resistor R₁. At point A, this 3-ampere current divides into two currents. Of the total current, 1.8 amps flows through the 20-ohm resistor. The remaining current of 1.2 amps flows from point A, down through the 30-ohm resistor to point B. This current produces a voltage drop of 36 volts across the 30-ohm resistor. (Notice that the voltage drops across the 20- and 30-ohm resistors are the same.) The two branch currents of 1.8 and 1.2 amps combine at junction B and the total current of 3 amps flows back to the source. The action of the circuit has been completely described with the exception of power consumed, which could be described using the values previously computed.

It should be pointed out that the combination circuit is not difficult to solve. The key to its solution lies in knowing the order in which the steps of the solution must be accomplished.

You will notice that the schematic diagrams you have been working with have shown parallel circuits drawn as neat square figures, with each branch easily identified.

In actual practice the wired circuits and more complex schematics are rarely laid out in this simple form. For this reason, it is important for you to recognize that circuits can be drawn in a variety of ways, and to learn some of the techniques for redrawing them into their simplified form. When a circuit is redrawn for clarity or to its simplest form, the following steps are used.

To redraw any circuit, start at the source, and trace the path of current flow through the circuit. At points where the current divides, called JUNCTIONS, parallel branches begin. These junctions are key points of reference in any circuit and should be labeled as you find them. The wires in circuit schematics are assumed to have NO RESISTANCE and there is NO VOLTAGE drop along any wire. This means that any unbroken wire is at the same voltage all along its length, until it is interrupted by a resistor, battery, or some other circuit component. In redrawing a circuit, a wire can be "stretched" or "shrunk" as much as you like without changing any electrical characteristic of the circuit.

Figure 3(A) is a schematic of a circuit that is not drawn in the box-like fashion used in previous illustrations. To redraw this circuit, start at the voltage source and trace the path for current to the junction marked (a). At this junction the current divides into three paths. If you were to stretch the wire to show the three current paths, the circuit would appear as shown in Figure 3(B).

While these circuits may appear to be different, the two drawings actually represent the same circuit. The drawing in Figure 3(B) is the familiar box-like structure and may be easier to work with. Figure 4(A) is a schematic of a circuit shown in a box-like structure, but may be misleading. This circuit in reality is a series-parallel circuit that may be redrawn as shown in Figure 4(B). The drawing in part (B) of the figure is a simpler representation of the original circuit and could be reduced to just two resistors in parallel.

Earlier in this chapter the terms open and short circuits were discussed. The following discussion deals with the effects on a circuit when an open or a short occurs.

The major difference between an open in a parallel circuit and an open in a series circuit is that in the parallel circuit the open would not necessarily disable the circuit. If the open condition occurs in a series portion of the circuit, there will be no current because there is no complete path for current flow. If, on the other hand, the open occurs in a parallel path, some current will still flow in the circuit. The parallel branch where the open occurs will be effectively disabled, total resistance of the circuit will INCREASE, and total current will DECREASE.

To clarify these points, Figure 8 illustrates a series parallel circuit. First the effect of an open in the series portion of this circuit will be examined. Figure 8(A) shows the normal circuit, R_T = 40 ohms and I_T = 3 amps. In Figure 8(B) an open is shown in the series portion of the circuit, there is no complete path for current and the resistance of the circuit is considered to be infinite.

In Figure 8(C) an open is shown in the parallel branch of R₃. There is no path for current through R₃. In the circuit, current flows through R₁ and R₂ only. Since there is only one path for current flow, R₁ and R₂ are effectively in series.

Under these conditions R_T = 120Ω, I_T = 1 amp. As you can see, when an open occurs in a parallel branch, total circuit resistance increases and total circuit current decreases.

A short circuit in a parallel network has an effect similar to a short in a series circuit. In general, the short will cause an increase in current and the possibility of component damage regardless of the type of circuit involved. To illustrate this point, Figure 9 shows a series-parallel network in which shorts are developed. In Figure 9(A) the normal circuit is shown. R_T = 40 ohms and I_T = 3 amps.

In Figure 9(B), R₁ has shorted. R₁ now has zero ohms of resistance. The total of the resistance of the circuit is now equal to the resistance of the parallel network of R₂ and R₃, or 20 ohms. Circuit current has increased to 6 amps. All of this current goes through the parallel network (R₂, R₃) and this increase in current would most likely damage the components.

In Figure 9(C), R₃ has shorted. With R₃ shorted there is a short circuit in parallel with R₂. The short circuit routes the current around R₂, effectively removing R₂ from the circuit. Total circuit resistance is now equal to the resistance of R₁, or 20 ohms.

As you know, R₂ and R₃ form a parallel network. Resistance of the network can be calculated as follows:

Given:

Solution:

The total circuit current with R₃ shorted is 6 amps. All of this current flows through R₁ and would most likely damage R₁. Notice that even though only one portion of the parallel network was shorted, the entire paralleled network was disabled.

Opens and shorts alike, if occurring in a circuit, result in an overall change in the equivalent resistance. This can cause undesirable effects in other parts of the circuit due to the corresponding change in the total current flow. A short usually causes components to fail in a circuit which is not properly fused or otherwise protected. The failure may take the form of a burned-out resistor, damaged source, or a fire in the circuit components and wiring.

Fuses and other circuit protection devices are installed in equipment circuits to prevent damage caused by increases in current. These circuit protection devices are designed to open if current increases to a predetermined value. Circuit protection devices are connected in series with the circuit or portion of the circuit that the device is protecting. When the circuit protection device opens, current flow ceases in the circuit.

A more thorough explanation of fuses and other circuit protection devices is presented in Module 3, Introduction to Circuit Protection, Control, and Measurement.

Most electrical and electronics equipment use voltages of various levels throughout their circuitry.

One circuit may require a 90-volt supply, another a 150-volt supply, and still another a 180-volt supply. These voltage requirements could be supplied by three individual power sources. This method is expensive and requires a considerable amount of room. The most common method of supplying these voltages is to use a single voltage source and a VOLTAGE DIVIDER. Before voltage dividers are explained, a review of what was discussed earlier concerning voltage references may be of help.

As you know, some circuits are designed to supply both positive and negative voltages. Perhaps now you wonder if a negative voltage has any less potential than a positive voltage. The answer is that 100 volts is 100 volts. Whether it is negative or positive does not affect the feeling you get when you are shocked.

Voltage polarities are considered as being positive or negative in respect to a reference point, usually ground. Figure 10 will help to illustrate this point.

Figure 10(A) shows a series circuit with a voltage source of 100 volts and four 50-ohm resistors connected in series. The ground, or reference point, is connected to one end of resistor R 1. The current in this circuit determined by Ohm’s law is .5 amp. Each resistor develops (drops) 25 volts. The five tap-off points indicated in the schematic are points at which the voltage can be measured. As indicated on the schematic, the voltage measured at each of the points from point A to point E starts at zero volts and becomes more positive in 25 volt steps to a value of positive 100 volts.

In Figure 10(B), the ground, or reference point has been moved to point B. The current in the circuit is still .5 amp and each resistor still develops 25 volts. The total voltage developed in the circuit remains at 100 volts, but because the reference point has been changed, the voltage at point A is negative 25 volts. Point E, which was at positive 100 volts in Figure 10(A), now has a voltage of positive 75 volts. As you can see the voltage at any point in the circuit is dependent on three factors; the current through the resistor, the ohmic value of the resistor, and the reference point in the circuit.

A typical voltage divider consists of two or more resistors connected in series across a source voltage (E_s). The source voltage must be as high or higher than any voltage developed by the voltage divider. As the source voltage is dropped in successive steps through the series resistors, any desired portion of the source voltage may be "tapped off" to supply individual voltage requirements. The values of the series resistors used in the voltage divider are determined by the voltage and current requirements of the loads.

Figure 11 is used to illustrate the development of a simple voltage divider. The requirement for this voltage divider is to provide a voltage of 25 volts and a current of 910 milliamps to the load from a source voltage of 100 volts. Figure 11(A) provides a circuit in which 25 volts is available at point B. If the load was connected between point B and ground, you might think that the load would be supplied with 25 volts. This is not true since the load connected between point B and ground forms a parallel network of the load and resistor R₁. (Remember that the value of resistance of a parallel network is always less than the value of the smallest resistor in the network.)

Since the resistance of the network would now be less than 25 ohms, the voltage at point B would be less than 25 volts. This would not satisfy the requirement of the load.

To determine the size of resistor used in the voltage divider, a rule-of-thumb is used. The current in the divider resistor should equal approximately 10 percent of the load current. This current, which does not flow through any of the load devices, is called bleeder current.

Given this information, the voltage divider can be designed using the following steps.