In the preceding discussions, series and parallel dc circuits have been considered separately. The technician will encounter circuits consisting of both series and parallel elements. A circuit of this type is referred to as a COMBINATION CIRCUIT. Solving for the quantities and elements in a combination circuit is simply a matter of applying the laws and rules discussed up to this point.
1. Solving CombinationCircuit Problems
The basic technique used for solving dc combinationcircuit 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 R_{2} and R_{3}.
Given:
Solution:
Now that the equivalent resistance for R_{2} and R_{3} 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 R_{1}, R_{2}, and R_{3}, refer to Figure 1(B). R_{eq1} represents the parallel network of R_{2} and R_{3}. 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 R_{2} (E_{R2}) and also equal to the voltage across R_{3} (E_{R3}).
Given:
Solution:
To find power used by R_{1}:
Given:
Solution:
To find the current through R_{2} and R_{3}, 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 R_{2} and R_{3}, 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 8ohm resistor (R_{1}). In so doing, a voltage drop of 24 volts occurs across resistor R_{1}. At point A, this 3ampere current divides into two currents. Of the total current, 1.8 amps flows through the 20ohm resistor. The remaining current of 1.2 amps flows from point A, down through the 30ohm resistor to point B. This current produces a voltage drop of 36 volts across the 30ohm resistor. (Notice that the voltage drops across the 20 and 30ohm 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.
1.1. Practice Circuit Problem
Figure 2 is a typical combination circuit. To make sure you understand the techniques of solving for the unknown quantities, solve for E_{R1}.
It is not necessary to solve for all the values in the circuit to compute the voltage drop across resistor R_{1} (E_{R1}). First look at the circuit and determine that the values given do not provide enough information to solve for E_{R1} directly.
If the current through R_{1} (I_{R1}) is known, then E_{R1} can be computed by applying the formula:
The following steps will be used to solve the problem.

The total resistance (R_{T}) is calculated by the use of equivalent resistance. Given: Solution: Redraw the circuit as shown in Figure 2(B). Given: Solution: Solution: Redraw the circuit as shown in Figure 2(C). Given: Solution:

The total current (I_{T}) is now computed. Given: Solution:

Solve for the voltage dropped across Req2. This represents the voltage dropped across the network R_{1}, R_{2}, and R_{3} in the original circuit. Given: Solution:

Solve for the current through R_{eq1}. (R_{eq1} represents the network R_{1} and R_{2} in the original circuit.) Since the voltage across each branch of a parallel circuit is equal to the voltage across the equivalent resistor representing the circuit: Given: Solution:

Solve for the voltage dropped across R_{1} (the quantity you were asked to find). Since R_{eq1} represents the series network of R_{1} and R_{2} and total current flows through each resistor in a series circuit, I_{R1} must equal I_{Req1}. Given: Solution:
values indicated: R_{1} = 900Ω, R_{3} = 1kΩ, what is the total power in the circuit? What is E_{R2}?
2. Redrawing Circuits for Clarity
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.

Trace the current paths in the circuit.

Label the junctions in the circuit.

Recognize points which are at the same potential.

Visualize a rearrangement, "stretching" or "shrinking," of connecting wires.

Redraw the circuit into simpler form (through stages if necessary).
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 boxlike 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 boxlike structure and may be easier to work with. Figure 4(A) is a schematic of a circuit shown in a boxlike structure, but may be misleading. This circuit in reality is a seriesparallel 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.
2.1. Redrawing a Complex Circuit
Figure 5(A) shows a complex circuit that may be redrawn for clarification in the following steps.
Note

As you redraw the circuit, draw it in simple boxlike form. Each time you reach a junction, a new branch is created by stretching or shrinking the wires. 
Start at the negative terminal of the voltage source. Current flows through R_{1} to a junction and divides into three paths; label this junction (a). Follow one of the paths of current through R_{2} and R_{3} to a junction where the current divides into two more paths. This junction is labeled (b).
The current through one branch of this junction goes through R_{5} and back to the source. (The most direct path.) Now that you have completed a path for current to the source, return to the last junction, (b). Follow current through the other branch from this junction. Current flows from junction (b) through R_{4} to the source. All the paths from junction (b) have been traced. Only one path from junction (a) has been completed. You must now return to junction (a) to complete the other two paths. From junction (a) the current flows through R_{7} back to the source. (There are no additional branches on this path.) Return to junction (a) to trace the third path from this junction. Current flows through R_{6} and R_{8} and comes to a junction. Label this junction (c). From junction (c) one path for current is through R_{9} to the source. The other path for current from junction (c) is through R_{10} to the source. All the junctions in this circuit have now been labeled. The circuit and the junction can be redrawn as shown in Figure 5(C). It is much easier to recognize the series and parallel paths in the redrawn circuit.
3. Effects of Open and Short Circuits
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 R_{3}. There is no path for current through R_{3}. In the circuit, current flows through R_{1} and R_{2} only. Since there is only one path for current flow, R_{1} and R_{2} 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 seriesparallel 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), R_{1} has shorted. R_{1} 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 R_{2} and R_{3}, or 20 ohms. Circuit current has increased to 6 amps. All of this current goes through the parallel network (R_{2}, R_{3}) and this increase in current would most likely damage the components.
In Figure 9(C), R_{3} has shorted. With R_{3} shorted there is a short circuit in parallel with R_{2}. The short circuit routes the current around R_{2}, effectively removing R_{2} from the circuit. Total circuit resistance is now equal to the resistance of R_{1}, or 20 ohms.
As you know, R_{2} and R_{3} form a parallel network. Resistance of the network can be calculated as follows:
Given:
Solution:
The total circuit current with R_{3} shorted is 6 amps. All of this current flows through R_{1} and would most likely damage R_{1}. 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 burnedout 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.
4. Voltage Dividers
Most electrical and electronics equipment use voltages of various levels throughout their circuitry.
One circuit may require a 90volt supply, another a 150volt supply, and still another a 180volt 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 50ohm 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 tapoff 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 R_{1}. (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 ruleofthumb 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.

Determine the load requirement and the available voltage source.

Select bleeder current by applying the 10% ruleofthumb.

Calculate bleeder resistance. The value of R_{1} may be rounded off to 275 ohms:

Calculate the total current (load plus bleeder).

Calculate the resistance of the other divider resistor(s). The voltage divider circuit can now be drawn as shown in Figure 11(B).
4.1. MultipleLoad Voltage Dividers
A multipleload voltage divider is shown in Figure 12. An important point that was not emphasized before is that when using the 10% ruleofthumb to calculate the bleeder current, you must take 10% of the total load current.
Given the information shown in Figure 12, you can calculate the values for the resistors needed in the voltagedivider circuits. The same steps will be followed as in the previous voltage divider problem.
Given:
The bleeder current should be 10% of the total load current.
Solution:
Since the voltage across R_{1} (E_{R1}) is equal to the voltage requirement for load 1, Ohm’s law can be used to calculate the value for R_{1}.
Solution:
The current through R_{2} (I_{R2}) is equal to the current through R_{1} plus the current through load 1.
Solution:
The voltage across R_{2} (E_{R2}) is equal to the difference between the voltage requirements of load 1 and load 2.
Ohm’s law can now be used to solve for the value of R_{2}.
Solution:
The current through R_{3} (I_{R3}) is equal to the current through R_{2} plus the current through load 2.
The voltage across R_{3} (E_{R3}) equals the difference between the voltage requirement of load 3 and load 2.
Ohm’s law can now be used to solve for the value of R_{3}.
Solution:
The current through R_{4} (I_{R4}) is equal to the current through R_{3} plus the current through load 3. IR_{4} is equal to total circuit current (I_{T}).
The voltage across R_{4} (E_{R4}) equals the difference between the source voltage and the voltage requirement of load 3.
Ohm’s law can now be used to solve for the value of R_{4}.
Solution:
With the calculations just explained, the values of the resistors used in the voltage divider are as follows:
4.2. Power in the Voltage Divider
Power in the voltage divider is an extremely important quantity. The power dissipated by the resistors in the voltage divider should be calculated to determine the power handling requirements of the resistors. Total power of the circuit is needed to determine the power requirement of the source.
The power for the circuit shown in Figure 12 is calculated as follows:
Given:
Solution:
The power in each resistor is calculated just as for R 1. When the calculations are performed, the following results are obtained:
To calculate the power for load 1:
Given:
Solution:
The power in each load is calculated just as for load 1. When the calculations are performed, the following results are obtained.
Total power is calculated by summing the power consumed by the loads and the power dissipated by the divider resistors. The total power in the circuit is 15.675 watts.
The power used by the loads and divider resistors is supplied by the source. This applies to all electrical circuits; power for all components is supplied by the source.
Since power is the product of voltage and current, the power supplied by the source is equal to the source voltage multiplied by the total circuit current (E_{s} x I_{T}).
In the circuit of Figure 12, the total power can be calculated by:
Given:
Solution:
4.3. Voltage Divider With Positive and Negative Voltage Requirements
In many cases the load for a voltage divider requires both positive and negative voltages. Positive and negative voltages can be supplied from a single source voltage by connecting the ground (reference point) between two of the divider resistors. The exact point in the circuit at which the reference point is placed depends upon the voltages required by the loads.
For example, a voltage divider can be designed to provide the voltage and current to three loads from a given source voltage.
Given:
The circuit is drawn as shown in Figure 13. Notice the placement of the ground reference point. The values for resistors R_{1}, R_{3}, and R_{4} are computed exactly as was done in the last example. IR_{1} is the bleeder current and can be calculated as follows:
Solution:
Calculate the value of R_{1}.
Solution:
Calculate the current through R_{2} using Kirchhoff’s current law.
At point A:
(or 195mA leaving point A)
Since E_{R2} = E load 2, you can calculate the value of R_{2}.
Solution:
Calculate the current through R_{3}.
The voltage across R_{3} (E_{R3}) equals the difference between the voltage requirements of loads 3 and 2.
Solution:
Calculate the value of R_{3}.
Solution:
Calculate the current through R_{4}.
The voltage across E_{R4} equals the source voltage (E_{s}) minus the voltage requirement of load 3 and the voltage requirement of load 1. Remember Kirchhoff’s voltage law which states that the sum of the voltage drops and emfs around any closed loop is equal to zero.
Solution:
Calculate the value of R_{4}.
Solution:
With the calculations just explained, the values of the resistors used in the voltage /divider are as follows:
From the information just calculated, any other circuit quantity, such as power, total current, or resistance of the load, could be calculated.
4.4. Practical Application of Voltage Dividers
In actual practice the computed value of the bleeder resistor does not always come out to an even value. Since the ruleofthumb for bleeder current is only an estimated value, the bleeder resistor can be of a value close to the computed value. (If the computed value of the resistance were 510 ohms, a 500 ohm resistor could be used.) Once the actual value of the bleeder resistor is selected, the bleeder current must be recomputed. The voltage developed by the bleeder resistor must be equal to the voltage requirement of the load in parallel with the bleeder resistor.
The value of the remaining resistors in the voltage divider is computed from the current through the remaining resistors and the voltage across them. These values must be used to provide the required voltage and current to the loads.
If the computed values for the divider resistors are not even values; series, parallel, or seriesparallel networks can be used to provide the required resistance.
Example: A voltage divider is required to supply two loads from a 190.5 volts source. Load 1 requires +45 volts and 210 milliamps; load 2 requires +165 volts and 100 milliamps.
Calculate the bleeder current using the ruleofthumb.
Given:
Solution:
Calculate the ohmic value of the bleeder resistor.
Given:
Solution:
Since it would be difficult to find a resistor of 1451.6 ohms, a practical choice for R_{1} is 1500 ohms.
Calculate the actual bleeder current using the selected value for R_{1}.
Given:
Solution:
Using this value for I_{R1}, calculate the resistance needed for the next divider resistor. The current (I_{R2}) is equal to the bleeder current plus the current used by load 1.
Given:
Solution:
The voltage across R_{2} (E_{R2}) is equal to the difference between the voltage requirements of loads 2 and 1, or 120 volts.
Calculate the value of R_{2}.
Given:
Solution:
The value of the final divider resistor is calculated with I_{R3} (I_{R2} + I load 2) equal to 340 mA and E_{R3} (E_{s}  E load 2) equal to 25.5V.
Given:
Solution:
A 75ohm resistor may not be easily obtainable, so a network of resistors equal to 75 ohms can be used in place of R_{3}.
Any combination of resistor values adding up to 75 ohms could be placed in series to develop the required network. For example, if you had two 37.5ohm resistors, you could connect them in series to get a network of 75 ohms. One 50ohm and one 25ohm resistor or seven 10ohm and one 5ohm resistor could also be used.
A parallel network could be constructed from two 150ohm resistors or three 225ohm resistors. Either of these parallel networks would also be a network of 75 ohms.
The network used in this example will be a seriesparallel network using three 50ohm resistors.
With the information given, you should be able to draw this voltage divider network.
Once the values for the various divider resistors have been selected, you can compute the power used by each resistor using the methods previously explained. When the power used by each resistor is known, the wattage rating required of each resistor determines the physical size and type needed for the circuit. This circuit is shown in Figure 14.