**BASIC DC THEORY**

**INTRODUCTION**

The material covered in this lesson contains many new terms that are explained as you progress through the material.

The basic dc circuit is the easiest to understand, so the chapter begins with the basic circuit and from there works into the basic schematic diagram of that circuit.

The schematic diagram is used in all your future work in electricity and electronics. It is very important that you become familiar with the symbols that are used.

This module also explains how to determine the total resistance, current, voltage, and power in a series, parallel, or combination circuit through the use of Ohm’s and Kirchhoff’s laws.

The voltage divider network, series, parallel, and series-parallel practice problem circuits will be used for practical examples of what you have learned.

**THE BASIC ELECTRIC CIRCUIT**

The flashlight is an example of a basic electric circuit.

It contains a source of electrical energy (the dry cells in the flashlight), a load (the bulb) that changes the electrical energy into a more useful form of energy (light), and a switch to control the energy delivered to the load.

Before you study a schematic representation of the flashlight, it is necessary to define certain terms.

The ** LOAD** is any device through which an electrical current flows and which changes this electrical energy into a more useful form.

Some common examples of loads are

- a light bulb, which changes electrical energy to light energy;
- an electric motor, which changes electrical energy into mechanical energy; and
- a speaker, which changes electrical energy into sound.

The ** SOURCE** is the device that furnishes the electrical energy used by the load. It may consist of a simple dry cell (as in a flashlight), a storage battery (as in an automobile), or a power supply (such as a battery charger).

The ** SWITCH**, which permits control of the electrical device, interrupts the current delivered to the load.

**SCHEMATIC REPRESENTATION**

The technician’s main aid in troubleshooting a circuit in a piece of equipment is the SCHEMATIC DIAGRAM. The schematic diagram is a “picture” of the circuit that uses symbols to represent the various circuit components; physically large or complex circuits can be shown on a relatively small diagram.

Before studying the basic schematic, look at ** figure 1**.

This figure shows the symbols that are used in this chapter. These, and others like them, are referred to and used throughout the study of electricity and electronics.

The schematic in **figure 2** represents a flashlight.

View A of the figure shows the flashlight in the off or de-energized state.

- The switch (S1) is open.
- There is no complete path for current (I) through the circuit, and
- The bulb (DS1) does not light.

In figure 2 view B, switch S1 is closed.

- Current flows in the direction of the arrows from the negative terminal of the battery (BAT), through the switch (S1),
- through the lamp (DS1), and
- back to the positive terminal of the battery.

With the switch closed the path for current is complete.

Current will continue to flow until the switch (S1) is moved to the open position or the battery is completely discharged.

**OHM’S LAW**

One of the most important and fundamental laws of electronics is that of ohms law.

Ohm’s Law is made from 3 mathematical equations that show the relationship between *electric voltage*,

**and**

*current,***.**

*resistance***Definition of 1 Volt****–**The unit of electromotive force or electric pressure.**Definition of 1 Amp –**The standard unit of electric current.**Definition of 1 Ohm –**The unit by which electrical resistance is measured.

All three of these: ** voltage**,

**and**

*current,***directly interact in Ohm’s Law. Change any two of them and you affect the third.**

*resistance*Three of the most important electrical properties are elegantly intertwined by way of a simple equation:

**V = I x R**, better known as Ohm’s Law.

*If you need further assistance with Ohms Law please see our Ohms law lesson.*

**GRAPHICAL ANALYSIS OF THE BASIC CIRCUIT**

One of the most valuable methods of analyzing a circuit is by constructing a graph. No other method provides a more convenient or more rapid way to observe the characteristics of an electrical device.

The first step in constructing a graph is to obtain a table of data.

The information in the table can be obtained by taking measurements on the circuit under examination or can be obtained theoretically through a series of Ohm’s law computations. The latter method is used here.

Since there are three variables (E**,** I, and R) to be analyzed, there are three distinct graphs that may be constructed.

- To construct any graph of electrical quantities, it is standard practice to vary one quantity in a specified way and note the changes which occur in a second quantity.

The quantity which is intentionally varied is called the independent variable and is plotted on the horizontal axis.

- The horizontal axis is known as the X-AXIS.

The second quantity, which varies as a result of changes in the first quantity, is called the dependent variable and is plotted on the vertical, or Y-AXIS.

- Any other quantities involved are held constant.

For example, in the circuit shown in **figure 6**, if the resistance was held at **10 ohms **and the voltage was varied, the resulting changes in current could then be graphed.

- The resistance is the constant,
- The voltage is the independent variable, and
- The current is the dependent variable.

This table shows R held constant at 10 ohms as E is varied from 0 to 20 volts in 5-volt steps. Through the use of Ohm’s Law, you can calculate the value of current for each value of voltage shown in the table.

When the table is complete, the information it contains can be used to construct the graph shown in **figure 7**.

For example, when the voltage applied to the 10-ohm resistor is 10 volts, the current is 1 ampere.

These values of current and voltage determine a point on the graph. When all five points have been plotted, a smooth curve is drawn through the points.

Through the use of this curve, the value of current through the resistor can be quickly determined for any value of voltage between 0 and 20 volts.

Since the curve is a straight line, it shows that equal changes of voltage across the resistor produce equal changes in current through the resistor.

This fact illustrates an important characteristic of the Basic Law—the current varies directly with the applied voltage when the resistance is held constant.

When the voltage across a load is held constant, the current depends solely upon the resistance of the load.

Fig-8: Relationship between current and resistance. For example, **figure 8** shows a graph with the voltage held constant at 12 volts.

The independent variable is the resistance which is varied from 2 ohms to 12 ohms.

The current is the dependent variable.

Values for current can be calculated as:

This process can be continued for any value of resistance.

You can see that as the resistance is halved, the current is doubled; when the resistance is doubled, the current is halved.

This illustrates another important characteristic of Ohm’s Law – current varies inversely with resistance when the applied voltage is held constant.

**POWER**

Power, whether electrical or mechanical, pertains to the rate at which work is being done.

- Work is done whenever a force causes motion.
- When a mechanical force is used to lift or move a weight, work is done.

However, the force exerted WITHOUT causing motion, such as the force of a compressed spring acting between two fixed objects, does not constitute work.

Previously, it was shown that voltage is an electrical force and that voltage forces current to flow in a closed circuit.

However, when voltage exists but current does not flow because the circuit is open, no work is done. This is similar to the spring under tension that produced no motion. When voltage causes electrons to move, work is done.

The instantaneous RATE at which this work is done is called the electric power rate and is measured in WATTS. A total amount of work may be done in different lengths of time.

** For example**, a given number of electrons may be moved from one point to another in 1 second or in 1 hour, depending on the RATE at which they are moved.

In both cases, the total work done is the same.

However, when the work is done in a short time, the wattage, or INSTANTANEOUS POWER RATE, is greater than when the same amount of work is done over a longer period of time.

As stated, the basic unit of power is the watt. Power in watts is equal to the voltage across a circuit multiplied by the current through the circuit.

This represents the rate at any given instant at which work is being done.

The symbol (P) indicates electrical power.

Thus, the basic power formula is P = E x I, where E is voltage and I is current in the circuit.

The amount of power changes when either voltage or current, or both voltage and current, are caused to change.

*In practice, the ONLY factors that can be changed are voltage and resistance.*

In explaining the different forms that formulas may take, the current is sometimes presented as a quantity that is changed.

Remember, if the current is changed, it is because either voltage or resistance has been changed.

** Figure 9** shows a basic circuit using a source of power that can be varied from 0 to 8 volts and a graph that indicates the relationship between voltage and power.

The resistance of this circuit is 2 ohms; this value does not change. Voltage (E) is increased (by increasing the voltage source), in steps of 1 volt, from 0 volts to 8 volts.

By applying Ohm’s law, the current (I) is determined for each step of voltage.

For instance, when E is 1 volt, the current is:

- I = E/R
- I = (1 volt)/(2 ohms)
- I = 0.5 ampere

Power (P), in watts, is determined by applying the basic power formula

- P = E x I
- P = 1 volt x 0.5 amperes = P = 0.5 watts

You should notice that when the voltage was increased to 2 volts, the power increased from .5 watts to 2 watts or 4 times. When the voltage increased to 3 volts, the power increased to 4.5 watts or 9 times. This shows that if the resistance in a circuit is held constant, the power varies directly with the SQUARE OF THE VOLTAGE.

Another way of proving that power varies as the square of the voltage when resistance is held constant is:

Thus, another formula for power, with current and resistance as its factors, is P = I** _{2R}**.

*This can be proved by:*

Up to this point, four of the most important electrical quantities have been discussed. These are **voltage (E), current (I), resistance (R), and power (P)**. You must understand the relationships which exist among these quantities because they are used throughout your study of electricity.

In the preceding paragraphs,** P** was expressed in terms of alternate pairs of the other three basic quantities **E**, **I**, and **R**. In practice, you should be able to express any one of these quantities in terms of any two of the others.

** Figure 11** is a summary of

**12 basic formulas**you should know.

The four, **E**, **I**, **R**, and **P** are at the center of the figure. Adjacent to each quantity are three segments.

Note that in each segment, the basic quantity is expressed in terms of two other basic quantities, and no two segments are alike.

For example, the formula wheel in **figure 11** could be used to find the formula to solve the following problem:

*A circuit has a voltage source that delivers 6 volts and the circuit uses 3 watts of power. **What is the resistance of the load?*

Since R is the quantity you have been asked to find, look in the section of the wheel that has R in the center.

**POWER RATING**

Electrical components are often given a power rating. The power rating, in watts, indicates the rate at which the device converts electrical energy into another form of energy, such as light, heat, or motion.

An example of such a rating is noted when comparing a 150-watt lamp to a 100-watt lamp. The higher wattage rating of the 150-watt lamp indicates it is capable of converting more electrical energy into light energy than the lamp with the lower rating.

Other common examples of devices with power ratings are soldering irons and small electric motors. In some electrical devices, the wattage rating indicates the maximum power the device is designed to use rather than the normal operating power.

A 150-watt lamp, for example, uses 150 watts when operated at the specified voltage printed on the bulb. In contrast, a device such as a resistor is not normally given a voltage or a current rating.

A resistor is given a power rating in watts and can be operated at any combination of voltage and current as long as the power rating is not exceeded. In most circuits, the actual power used by a resistor is considerably less than the power rating of the resistor because a 50% safety factor is used.

For example, if a resistor normally used 2 watts of power, a resistor with a power rating of 3 watts would be used.

Resistors of the same resistance value are available in different wattage values.

Carbon resistors, for example, are commonly made in wattage ratings of 1/8, 1/4, 1/2, 1, and 2 watts. The larger the physical size of a carbon resistor the higher the wattage rating. This is true because a larger surface area of material radiates a greater amount of heat more easily.

When resistors with wattage ratings greater than 5 watts are needed, wirewound resistors are used. Wirewound resistors are made in values between 5 and 200 watts. Special types of wirewound resistors are used for power in excess of 200 watts.

As with other electrical quantities, prefixes may be attached to the word watt when expressing very large or very small amounts of power. Some of the more common of these are the kilowatt (1,000 watts), the megawatt (1,000,000 watts), and the milliwatt (1/1,000 of a watt).

**POWER CONVERSION AND EFFICIENCY**

The term power consumption is common in the electrical field. It is applied to the use of power in the same sense that gasoline consumption is applied to the use of fuel in an automobile.

Another common term is power conversion. Power is used by electrical devices and is converted from one form of energy to another.

- An electrical motor converts electrical energy to mechanical energy.
- An electric light bulb converts electrical energy into light energy and
- An electric range converts electrical energy into heat energy.

Power used by electrical devices is measured in energy. This practical unit of electrical energy is equal to 1 watt of power used continuously for 1 hour.

The term kilowatt-hour (kWh) is used more extensively on a daily basis and is equal to 1,000 watt-hours.

The EFFICIENCY of an electrical device is the ratio of power converted to useful energy divided by the power consumed by the device. This number will always be less than one (1.00) because of the losses in any electrical device. If a device has an efficiency rating of .95, it effectively transforms 95 watts into useful energy for every 100 watts of input power. The other 5 watts are lost to heat or other losses which cannot be used.

Calculating the amount of power converted by an electrical device is a simple matter. You need to know the length of time the device is operated and the input power or horsepower rating.

Horsepower, a unit of work, is often found as a rating on electrical motors. One horsepower is equal to 746 watts.

Example: A 3/4-hp motor operates 8 hours a day.

How much power is converted by the motor per month?

How many kWh does this represent?

*Given:*

T = 8 hrs x 30 days

P = 3/4 hp

*Solution:*

*Convert horsepower to watts*

P = hp x 746 watts

P = 3/4 x 746 watts

P = 559 watts

Convert watts to watt-hours

P = work x time

P = 559 watts x 8 x 30

P = 134,000 watt-hours per month

(NOTE: These figures are rounded to the nearest 1000.)

If the motor actually uses 137 kWh per month, what is the efficiency of the motor?

*Given:*

Power converted = 134 kWh per month

Power used = 137 kWh per month

*Solution:*

**SERIES DC CIRCUITS**

When two unequal charges are connected by a conductor, a complete pathway for current exists. An electric circuit is a complete conducting pathway. It consists not only of the conductor but also includes the path through the voltage source. Inside the voltage source, current flows from the positive terminal, through the source, emerging at the negative terminal.

**SERIES CIRCUIT CHARACTERISTICS**

A SERIES CIRCUIT is defined as a circuit that contains only ONE PATH for current flow.

To compare the basic circuit that has been discussed and a more complex series circuit.

Figure 13 shows two circuits.

- The basic circuit has only one lamp and
- The series circuit has three lamps connected in series.

**Resistance in a Series Circuit**

Referring to figure 13, the current in a series circuit must flow through each lamp to complete the electrical path in the circuit.

Each additional lamp offers added resistance.

In a series circuit, THE TOTAL CIRCUIT RESISTANCE (RT) IS EQUAL TO THE SUM OF THE INDIVIDUAL RESISTANCES.

**As an equation RT = R1 + R2 + R3 + …R _{n}**

NOTE: The subscript _{n} denotes any number of additional resistances that might be in the equation.

** Example:** In figure 14 a series circuit consisting of three resistors:

- one resistor is 10 ohms,
- one resistor is 15 ohms, and
- one resistor is 30 ohms, is shown.
- A voltage source provides 110 volts.

What is the total resistance?

In some circuit applications, the total resistance is known and the value of one of the circuit resistors has to be determined.

The equation R_{T} = R_{1} + R_{2} + R_{3} can be transposed to solve for the value of the unknown resistance.

**Current in a Series Circuit**

Since there is only one path for current in a series circuit, the same current must flow through each component of the circuit.

To determine the current in a series circuit, only the current through one of the components needs to be known.

The fact that the same current flows through each component of a series circuit can be verified by inserting meters into the circuit at various points, as shown in **figure 16**.

If this were done, each meter would be found to indicate the same value of current.

**Voltage in a Series Circuit**

The voltage dropped across the resistor in a circuit consisting of a single resistor and a voltage source is the total voltage across the circuit and is equal to the applied voltage.

The total voltage across a series circuit that consists of more than one resistor is also equal to the applied voltage but consists of the sum of the individual resistor voltage drops.

In any series circuit, the SUM of the resistor voltage drops must equal the source voltage. This statement can be proven by an examination of the circuit shown in **figure 17**.

In this circuit, a source potential (E_{T}) of 20 volts is dropped across a series circuit consisting of two 5-ohm resistors.

The total resistance of the circuit (R_{T}) is equal to the sum of the two individual resistances, or 10 ohms.

Using Ohm’s law the circuit current may be calculated as follows:

Since the value of the resistors is known to be 5 ohms each, and the current through the resistors is known to be 2 amperes, the voltage drops across the resistors can be calculated.

The voltage (E_{1}) across R_{1} is, therefore: By inspecting the circuit, you can see that R_{2} is the same ohmic value as R_{1} and carries the same current.

The voltage drop across R_{2} is therefore also equal to 10 volts. Adding these two 10-volts drops together gives a total drop of 20 volts, exactly equal to the applied voltage.

For a series circuit then:

**E _{T} = E_{1} = E_{2} + E_{3} = . . . E_{n}**

Example: A series circuit consists of three resistors having values of 20 ohms, 30 ohms, and 50 ohms, respectively.

Find the applied voltage if the current through the 30-ohm resistor is 2 amps. To solve the problem, a circuit diagram is first drawn & labeled in figure 18.

*Substituting:*

E** _{T}** = (R

_{1}+I

_{1}) + (R

_{2}+I

_{2}) + (R

_{3}+I

_{3})

E** _{T}** = (20 ohms x 2 amps) + (30 ohms x 2 amps)+ (50 ohms x 2 amps)

E** _{T}** = 40 volts + 60 volts + 100 volts

E** _{T}** = 200 volts

NOTE: When you use Ohm’s law, the quantities for the equation MUST be taken from the SAME part of the circuit. In the above example, the voltage across R_{2} was computed using the current through R_{2} and the resistance of R_{2}. The value of the voltage dropped by a resistor is determined by the applied voltage and is in proportion to the circuit resistances.

The voltage drops that occur in a series circuit are in direct proportion to the resistance. This is the result of having the same current flow through each resistor – the larger the ohmic value of the resistor, the larger the voltage drop across it.

**Power in a Series Circuit**

Each of the resistors in a series circuit consumes power which is dissipated in the form of heat. Since this power must come from the source, the total power must be equal to the power consumed by the circuit resistances.

In a series circuit, the total power is equal to the SUM of the power dissipated by the individual resistors.

Total power (PT) is equal to:

** P _{T} = P_{1} + P_{2} + P_{3} … P_{n}**

Example: A series circuit consists of three resistors having values of 5 ohms, 10 ohms, and 15 ohms.

Find the total power when 120 volts is applied to the circuit. (See fig. 19.)

To check the answer, the total power delivered by the source can be calculated.

P source = I source x E source

P source = 4 amps x 120 volts

P source = 480 watts

The total power is equal to the sum of the power used by the individual resistors.

**SUMMARY OF CHARACTERISTICS**

The important factors governing the operation of a series circuit are listed below.

These factors have been set up as a group of rules so that they may be easily studied.

These rules must be completely understood before the study of more advanced circuit theory is undertaken.

**Rules for Series DC Circuits**

- The same current flows through each part of a series circuit.
- The total resistance of a series circuit is equal to the sum of the individual resistances.
- The total voltage across a series circuit is equal to the sum of the individual voltage drops.
- The voltage drop across a resistor in a series circuit is proportional to the ohmic value of the resistor.
- The total power in a series circuit is equal to the sum of the individual powers used by each circuit component.

**SERIES CIRCUIT ANALYSIS**

To establish a procedure for solving series circuits, the following sample problems will be solved.

Fig-20: solving for various values in a series circuit ** Example:** Three resistors of 5 ohms, 10 ohms, and 15 ohms are connected in series with a power source of 90 volts as shown in figure 3-20.

Find the total resistance, circuit current, voltage drop of each resistor, power of each resistor, and total power of the circuit.

In solving the circuit the total resistance will be found first. Next, the circuit current will be calculated. Once the current is known, the voltage drops and power dissipation can be calculated.

** Example:** Four resistors, R

**= 10 ohms, R**

_{1}**= 10 ohms, R**

_{2}**= 50 ohms, and**

_{3}R** _{4}** = 30 ohms are connected in series with a power source as shown in figure 21.

The current through the circuit is ½ ampere.

- What is the battery voltage?
- What is the voltage across each resistor?
- What is the power expended in each resistor?
- What is the total power?

An important fact to keep in mind when applying Ohm’s law to a series circuit is to consider whether the values used are component values or total values.

When the information available enables the use of Ohm’s law to find total resistance, total voltage, and total current, total values must be inserted into the formula.

** NOTE:** I

**is equal to I in a series circuit.**

_{T}However, the distinction between I** _{T}** and I in the formula should be noted. The reason for this is that future circuits may have several currents, and it will be necessary to differentiate between I

**and other currents.**

_{T}To compute any quantity (E, I, R, or P) associated with a single given resistor; the values used in the formula must be obtained from that particular resistor.

For example, to find the value of an unknown resistance, the voltage across and the current through that particular resistor must be used.

**KIRCHHOFF’S VOLTAGE LAW**

In 1847, G. R. Kirchhoff extended the use of Ohm’s law by developing a simple concept concerning the voltages contained in a series circuit loop. Kirchhoff’s voltage law states:

*“The algebraic sum of the voltage drops in any closed path in a circuit and the electromotive forces in that path is equal to zero.”*

To state Kirchhoff’s law another way, the voltage drops and voltage sources in a circuit are equal at any given moment in time.

If the voltage sources are assumed to have one sign (positive or negative) at that instant and the voltage drops are assumed to have the opposite sign, the result of adding the voltage sources and voltage drops will be zero.

NOTE: The terms electromotive force and emf are used when explaining Kirchhoff’s law because Kirchhoff’s law is used in alternating current circuits (covered in a future Module).

In applying Kirchhoff’s law to direct current circuits, the terms electromotive force and emf apply to voltage sources such as batteries or power supplies.

Through the use of Kirchhoff’s law, circuit problems can be solved which would be difficult, and often impossible, with knowledge of Ohm’s law alone.

When Kirchhoff’s law is properly applied, an equation can be set up for a closed loop and the unknown circuit values can be calculated.

Kirchhoff’s Voltage Law – is one of two fundamental laws in electronics and electrical engineering, the other being Kirchhoff’s Current Law.

- Kirchhoff’s Voltage Law is a fundamental law, as fundamental as Conservation of Energy in mechanics, for example, because Kirchhoff’s Voltage Law is really conservation of electrical energy.
- Kirchhoff’s Voltage Law and KCL are the starting point for analysis of any circuit.
- Kirchhoff’s Current Law and Kirchhoff’s Voltage Law always hold and are usually the most useful piece of information you will have about a circuit after the circuit itself.

**POLARITY OF VOLTAGE**

To apply Kirchhoff’s voltage law, the meaning of voltage polarity must be understood. In the circuit shown in figure 22, the current is shown flowing in a counterclockwise direction.

Notice that the end of resistor R** _{1}**, into which the current flows, is marked NEGATIVE (-)

The end of R** _{1}** at which the current leaves is marked POSITIVE (+).

These polarity markings are used to show that the end of R_{1} into which the current flows is at a higher negative potential than the end of the resistor at which the current leaves.

Point C, which is at the same potential as point B, is labeled negative.

This is to indicate that point C is more negative than point D.

To say a point is positive (or negative) without stating what the polarity is based upon has no meaning. In working with Kirchhoff’s law, positive and negative polarities are assigned in the direction of current flow.

**APPLICATION OF KIRCHHOFF’S VOLTAGE LAW**

Kirchhoff’s voltage law can be written as an equation, as shown below:

**E _{a} + E_{b} + E_{c} + . . . E_{n} = 0**

Where E_{a}, E_{b}, etc., are the voltage drops or emf’s around any closed circuit loop.

To set up the equation for an actual circuit, the following procedure is used.

- Assume a direction of current through the circuit. (The correct direction is desirable but not necessary.)
- Using the assumed direction of the current, assign polarities to all resistors through which the current flows.
- Place the correct polarities on any sources included in the circuit.
- Starting at any point in the circuit, trace around the circuit, writing down the amount and polarity of the voltage across each component in succession.

The polarity used is the sign AFTER the assumed current has passed through the component.

Stop when the point at which the trace was started is reached.

- Place these voltages, with their polarities, into the equation and solve for the desired quantity.

Example: Three resistors are connected across a 50-volt source.

What is the voltage across the third resistor if the voltage drops across the first two resistors are 25 volts and 15 volts?

Solution: First, a diagram, such as the one shown in figure 23, is drawn. Next, a direction of current is assumed (as shown).

Using this current, the polarity markings are placed at each end of each resistor and also on the terminals of the source.

Starting at point A, trace around the circuit in the direction of current flow, recording the voltage and polarity of each component.

Starting at point A and using the components from the circuit:

Substituting values from the circuit:

The unknown voltage (E_{X}) is found to be **10 volts**

Using the same idea as above, you can solve a problem in which the current is the unknown quantity.

Example: A circuit having a source voltage of 60 volts contains three resistors of 5 ohms, 10 ohms, and 15 ohms.

Find the circuit current.

Solution: Draw and label the circuit (fig. 24).

Establish a direction of current flow and assign polarities.

Next, starting at any point – point A will be used in this example – write out the loop equation.

Since the current obtained in the above calculations is a positive 2 amps, the assumed direction of the current was correct.

To show what happens if the incorrect direction of current is assumed, the problem will be solved as before, but with the opposite direction of the current.

The circuit is redrawn showing the new direction of current and new polarities in figure 25.

Starting at **point** A the loop equation is:

E** _{3}** + E

**+E**

_{A}**+E**

_{1}**= 0**

_{2}(I x R** _{3}**) + (E

**) + (I x R**

_{A}**) + (I x R**

_{1}**) = 0**

_{2}*Substitution Values*

(I x 15 ohms) + 60 volts + (I x5 ohms) + (I x 10 ohms) = 0

*Combining like terms*

(I x 30 ohms) + (60 volts) = 0

(I x 30 ohms) + -60 volts

Notice that the AMOUNT of current is the same as before. The polarity, however, is NEGATIVE.

The negative polarity simply indicates the wrong direction of current was assumed.

Should it be necessary to use this current in further calculations on the circuit using Kirchhoff’s law, the negative polarity should be retained in the calculations.

**Series Aiding and Opposing Sources **

In many practical applications, a circuit may contain more than one source of emf.

Sources of emf that cause current to flow in the same direction are considered to be SERIES AIDING and the voltages are added.

Sources of emf that would tend to force current in opposite directions are said to be SERIES OPPOSING, and the effective source voltage is the difference between the opposing voltages.

When two opposing sources are inserted into a circuit current flow would be in a direction determined by the larger source.

Examples of series aiding and opposing sources are shown in figure 26.

A simple solution may be obtained for a multiple-source circuit through the use of Kirchhoff’s voltage law. In applying this method, the same procedure is used for the multiple-source circuit as was used above for the single-source circuit. This is demonstrated by the following example.

Example: Using Kirchhoff’s voltage equation, find the amount of current in the circuit shown in fig 27.

Solution: As before, a direction of current flow is assumed and polarity signs are placed on the drawing.

The loop equation will be started at point A.

**CIRCUIT TERMS AND CHARACTERISTICS**

Before you learn about the types of circuits other than the series circuit, you should become familiar with some of the terms and characteristics used in electrical circuits. These terms and characteristics will be used throughout your study of electricity and electronics.

**REFERENCE POINT**

A reference point is an arbitrarily chosen point to which all other points in the circuit are compared. In series circuits, any point can be chosen as a reference and the electrical potential at all other points can be determined in reference to that point.

In figure 28 point A shall be considered the reference point.

- Each series resistor in the illustrated circuit is of equal value.
- The applied voltage is equally distributed across each resistor.
- The potential at point B is 25 volts more positive than at point A.
- Points C and D are 50 volts and 75 volts more positive than point A respectively.

When point B is used as the reference, as in figure 29, point D would be positive 50 volts in respect to the new reference point.

The former reference point, A, is 25 volts negative in respect to point B.

As in the previous circuit illustration, the reference point of a circuit is always considered to be at zero potential. Since the earth (ground) is said to be at a zero potential, the term GROUND is used to denote a common electrical point of zero potential.

In figure 30, point A is the zero reference or ground, and the symbol for ground is shown connected to point A.

Point C is 75 volts positive in respect to ground.

In most electrical equipment, the metal chassis is the common ground for many electrical circuits. When each electrical circuit is completed, common points of a circuit at zero potential are connected directly to the metal chassis, thereby eliminating a large amount of connecting wire. The electrons pass through the metal chassis (a conductor) to reach other points of the circuit.

An example of a chassis grounded circuit is illustrated in figure 31.

Most voltage measurements used to check proper circuit operation in electrical equipment are taken in respect to ground. One meter lead is attached to a grounded point and the other meter lead is moved to various test points.

**OPEN CIRCUIT**

A circuit is said to be OPEN when a break exists in a complete conducting pathway. Although an open occurs when a switch is used to de-energize a circuit, an open may also develop accidentally.

To restore a circuit to proper operation, the open must be located, its cause determined, and repairs made.

Sometimes an open can be located visually by a close inspection of the circuit components. Defective components, such as burned-out resistors, can usually be discovered by this method. Others, such as a break-in wire covered by insulation or the melted element of an enclosed fuse, are not visible to the eye.

Under such conditions, the understanding of the effect an open has on circuit conditions enables a technician to make use of test equipment to locate the open component.

In figure 32, the series circuit consists of two resistors and a fuse.

Notice the effects on circuit conditions when the fuse opens. Current ceases to flow; therefore, there is no longer a voltage drop across the resistors.

Each end of the open conducting path becomes an extension of the battery terminals and the voltage felt across the open is equal to the applied voltage (E_{A}).

An open circuit has INFINITE resistance. INFINITY represents a quantity so large it cannot be measured.

The symbol for infinity is ∞

In an open circuit: R_{T} = ∞

**SHORT CIRCUIT**

A short circuit is an accidental path of low resistance which passes an abnormally high amount of current.

A short circuit exists whenever the resistance of a circuit or the resistance of a part of a circuit drops in value to almost zero ohms.

A short often occurs as a result of improper wiring or broken insulation.

In figure 33, a short is caused by improper wiring. Note the effect on current flow.

Since the resistor has in effect been replaced with a piece of wire, practically all the current flows through the short and very little current flows through the resistor. Electrons flow through the short (a path of almost zero resistance) and the remainder of the circuit by passing through the 10-ohm resistor and the battery. The amount of current flow increases greatly because its resistive path has decreased from 10,010 ohms to 10 ohms.

Due to the excessive current flow, the 10-ohm resistor becomes heated. As it attempts to dissipate this heat, the resistor will probably be destroyed.

Figure 34 shows a pictorial wiring diagram, rather than a schematic diagram, to indicate how broken insulation might cause a short circuit.

**SOURCE RESISTANCE**

A meter connected across the terminals of a good 1.5-volt battery reads about 1.5 volts. When the same battery is inserted into a complete circuit, the meter reading decreases to something less than 1.5 volts.

This difference in terminal voltage is caused by the INTERNAL RESISTANCE of the battery (the opposition to current offered by the electrolyte in the battery).

All sources of electromotive force have some form of internal resistance which causes a drop in terminal voltage as current flows through the source.

This principle is illustrated in figure 35, where the internal resistance of a battery is shown as R_{1}.

In the schematic, the internal resistance is indicated by an additional resistor in series with the battery. The battery, with its internal resistance, is enclosed within the dotted lines of the schematic diagram.

With the switch open, the voltage across the battery terminals reads 15 volts. When the switch is closed, the current flow causes voltage drops around the circuit.

The circuit current of 2 amperes causes a voltage drop of 2 volts across R** _{i}**. The 1-ohm internal battery resistance thereby drops the battery terminal voltage to 13 volts.

Internal resistance cannot be measured directly with a meter. An attempt to do this would damage the meter.

The effect of the source resistance on the power output of a dc source may be shown by an analysis of the circuit in ** figure 36**.

When the variable load resistor (R** _{L}**) is set at the zero-ohm position (equivalent to a short circuit), current (I) is calculated using the following formula:

This is the maximum current that may be drawn from the source. The terminal voltage across the short circuit is zero volts and all the voltage is across the resistance within the source.

If the load resistance (R** _{L}**) were

*increased (the internal resistance remaining the same)*, the current drawn from the source would decrease. Consequently, the voltage drop across the internal resistance would decrease.

At the same time, the terminal voltage applied across the load would increase and approach a maximum as the current approaches zero amps.

**POWER TRANSFER AND EFFICIENCY**

Maximum power is transferred from the source to the load when the resistance of the load is equal to the internal resistance of the source. This theory is illustrated in the table and the graph of figure 36 When the load resistance is 5 ohms, matching the source resistance, the maximum power of 500 watts is developed in the load.

The efficiency of power transfer (ratio of output power to input power) from the source to the load increases as the load resistance is increased. The efficiency approaches 100 percent as the load resistance approaches a relatively large value compared with that of the source since less power is lost in the source.

The efficiency of power transfer is only 50 percent at the maximum power transfer point (when the load resistance equals the internal resistance of the source).

The efficiency of power transfer approaches zero efficiency when the load resistance is relatively small compared with the internal resistance of the source. This is also shown on the chart of figure 36.

The problem of a desire for both high efficiency and maximum power transfer is resolved by a compromise between maximum power transfer and high efficiency.

Where the amounts of power involved are large and the efficiency is important, the load resistance is made large relative to the source resistance so that the losses are kept small. In this case, the efficiency is high.

Where the problem of matching a source to a load is important, as, in communications circuits, a strong signal may be more important than a high percentage of efficiency. In such cases, the efficiency of power transfer should be only about 50 percent; however, the power transfer would be the maximum which the source is capable of supplying.

You should now understand the basic concepts of series circuits. The principles that have been presented are of lasting importance. Once equipped with a firm understanding of series circuits, you hold the key to an understanding of the parallel circuits to be presented next.

**PARALLEL DC CIRCUITS**

The discussion of electrical circuits presented up to this point has been concerned with series circuits in which there is only one path for current.

There is another basic type of circuit known as the PARALLEL CIRCUIT with which you must become familiar. Where the series circuit has only one path for current, the parallel circuit has more than one path for current.

Ohm’s law and Kirchhoff’s law apply to all electrical circuits, but the characteristics of a parallel dc circuit are different than those of a series dc circuit.

*PARALLEL CIRCUIT CHARACTERISTICS*

A PARALLEL CIRCUIT is defined as one having more than one current path connected to a common voltage source. Parallel circuits, therefore, must contain two or more resistances that are not connected in series. An example of a basic parallel circuit is shown in figure 37.

Start at the voltage source (E** _{S}**) and trace counterclockwise around the circuit. Two complete and separate paths can be identified in which current can flow. One path is traced from the source, through resistance R

**, and back to the source. The other path is from the source, through resistance R**

_{1}**, and back to the source.**

_{2}**Voltage in a Parallel Circuit**

You have seen that the source voltage in a series circuit divides proportionately across each resistor in the circuit.

IN A PARALLEL CIRCUIT, THE SAME VOLTAGE IS PRESENT IN EACH BRANCH. (A branch is a section of a circuit that has a complete path for current.) In figure 3-37 this voltage is equal to the applied voltage (E** _{S}**).

This can be expressed in equation form as:

**E _{S} = E_{R1} = E_{R2}**

Voltage measurements taken across the resistors of a parallel circuit, as illustrated by figure 38 verify this equation.

Each meter indicates the same amount of voltage. Notice that the voltage across each resistor is the same as the applied voltage.

** Example:** Assume that the current through a resistor of a parallel circuit is known to be 4.5 milliamperes (4.5 mA) & the value of the resistor is 30,000 ohms (30kΩ). Determine the source voltage. The circuit is shown in figure 39.

*Given:*** **

R** _{2}** = 30,000 ohms (30kΩ).

I** _{R2}** = 4.5 milliamps (4.5 mA or 00.45 amps)

*Solution:*

E + IR

E** _{R2}** = 00.45 amps x 30,000 ohms

E** _{R2}** = 135 volts

Since the source voltage is equal to the voltage of a branch:

E** _{S}** = ER2

E** _{S}** = 135 volts

To simplify the math operation, the values can be expressed in powers of ten as follows:

If you are not familiar with the use of the powers of 10 or would like to brush up on it, the Mathematics, modules will be of great help to you.

**Current in a Parallel Circuit**

Ohm’s law states that the current in a circuit is inversely proportional to the circuit resistance. This fact is true in both series and parallel circuits. There is a single path for current in a series circuit.

The amount of current is determined by the total resistance of the circuit and the applied voltage.

In a parallel circuit, the source current divides among the available paths.

The behavior of current in parallel circuits will be shown by a series of illustrations using example circuits with different values of resistance for a given value of applied voltage.

** Part (A) of figure 40** shows a basic series circuit.

Here, the total current must pass through the single resistor. The amount of current can be determined.

**Part (B) of figure 40** shows the same resistor (R** _{1}**) with a second resistor (R2) of equal value connected in parallel across the voltage source.

When Ohm’s law is applied, the current flow through each resistor is found to be the same as the current through the single resistor in part (A).

It is apparent that if there are 5 amperes of current through each of the two resistors, there must be a TOTAL CURRENT of 10 amperes drawn from the source. The total current of 10 amperes, as illustrated in figure 40(B), leaves the negative terminal of the battery and flows to point a.

Since point A is a connecting point for the two resistors, it is called a JUNCTION. At junction A, the total current divides into two currents of 5 amperes each. These two currents flow through their respective resistors and rejoin at junction B. The total current then flows from junction b back to the positive terminal of the source.

The source supplies a total current of 10 amperes and each of the two equal resistors carries one-half the total current. Each individual current path in the circuit of figure 40(B) is referred to as a BRANCH. Each branch carries a current that is a portion of the total current. Two or more branches form a NETWORK.

From the previous explanation, the characteristics of current in a parallel circuit can be expressed in terms of the following general equation:

**I _{T} = I_{1} + I_{2} + . . . I_{n}**

Compare part (A) of figure 41 with part (B) of the circuit in figure 40.

Notice that doubling the value of the second branch resistor (R** _{2}**) has no effect on the current in the first branch (I

**), but does reduce the second branch current (I**

_{R1}**) to one-half its original value. The total circuit current drops to a value equal to the sum of the branch currents.**

_{R2}These facts are verified by the following equations.

The amount of current flow in the branch circuits and the total current in the circuit shown in figure 41(B) is determined by the following computations.

Notice that the sum of the ohmic values in each circuit shown in figure 41 is equal (30 ohms) and that the applied voltage is the same (50 volts).

However, the total current in 41(B) (15 amps) is twice the amount in 41(A) (7.5 amps).

It is apparent, therefore, that the manner in which resistors are connected in a circuit, as well as their actual ohmic values, affect the total current. The division of current in a parallel network follows a definite pattern.

This pattern is described by KIRCHHOFF’S CURRENT LAW which states:

*“The algebraic sum of the currents entering and leaving any junction of conductors is equal to zero.”*

This law can be stated mathematically as **I _{a} + l_{b} + . . . I_{n} + 0**

** where:** Ia, Ib, etc., are the currents entering and leaving the junction.

Currents ENTERING the junction are considered to be POSITIVE and currents LEAVING the junction are considered to be NEGATIVE.

When solving a problem using Kirchhoff’s current law, the currents must be placed into the equation WITH THE PROPER POLARITY SIGNS ATTACHED.

Example: Solve for the value of I_{3} in Figure 42

Given:

I** _{1}** = 10 amps

I** _{2}** = 3 amps

I** _{4}** = 5 amps

Solution: Place the currents into the equation with the proper signs

I** _{3}** has a value of 2 amperes, and the negative sign shows it to be a current LEAVING the junction.

** Example:** Using figure 43, solve for the magnitude and direction of I

_{3}.

I_{3} is 2 amperes and its positive sign shows it to be a current entering the junction.

**Resistance in a Parallel Circuit**

In the example diagram, figure 44, there are two resistors connected in parallel across a 5-volt battery. Each has a resistance value of 10 ohms. A complete circuit consisting of two parallel paths is formed and current flows as shown.

Computing the individual currents shows that there is one-half of an ampere of current through each resistance.

The total current flowing from the battery to the junction of the resistors, and returning from the resistors to the battery, is equal to 1 ampere. The total resistance of the circuit can be calculated by using the values of total voltage (E** _{T}**) and total current (I

**).**

_{T}This computation shows the total resistance to be 5 ohms; one-half the value of either of the two resistors.

Since the total resistance of a parallel circuit is smaller than any of the individual resistors, the __total resistance__ of a parallel circuit is not the sum of the individual resistor values as was the case in a series circuit.

The total resistance of resistors in parallel is also referred to as EQUIVALENT RESISTANCE (R** _{eq}**). The terms total resistance and equivalent resistance are used interchangeably.

There are several methods used to determine the equivalent resistance of parallel circuits. The best method for a given circuit depends on the number and value of the resistors.

For the circuit described above, where all resistors have the __same value__, the following simple equation is used:

This equation is valid for any number of parallel resistors of EQUAL VALUE.

Example: Four 40-ohm resistors are connected in parallel. What is their equivalent resistance?

** Figure 45** shows two resistors of unequal value in parallel.

Since the total current is shown, the equivalent resistance can be calculated.

The equivalent resistance of the circuit shown in figure 45 is smaller than either of the two resistors (R_{1}, R_{2}).

An important point to remember is that the equivalent resistance of a parallel circuit is always less than the resistance of any branch. Equivalent resistance can be found if you know the individual resistance values and the source voltage.

By calculating each branch current, adding the branch currents to calculate the total current, and dividing the source voltage by the total current, the total can be found. This method, while effective, is somewhat lengthy.

A quicker method of finding equivalent resistance is to use the general formula for resistors in parallel:

If you apply the general formula to the circuit shown in figure 45 you will get the same value for equivalent resistance (2) as obtained in the previous calculation that used source voltage and total current.

Convert the fractions to a common denominator.

Since both sides are reciprocals (divided into one), disregard the reciprocal function.

The formula you were given for equal resistors in parallel.

is a simplification of the general formula for resistors in parallel.

There are other simplifications of the general formula for resistors in parallel which can be used to calculate the total or equivalent resistance in a parallel circuit.

**RECIPROCAL METHOD** This method is based upon taking the reciprocal of each side of the equation.

This presents the general formula for resistors in parallel as:

This formula is used to solve for the equivalent resistance of a number of unequal parallel resistors. You must find the lowest common denominator in solving these problems. If you are a little hazy on finding the lowest common denominator, brush up on it in Mathematics Module 1

** Example:** Three resistors are connected in parallel as shown in figure 46.

The resistor values are: R1 = 20 ohms, R2 = 30 ohms, R3 = 40 ohms. What is the equivalent resistance? (Use the reciprocal method.)

**PRODUCT OVER THE SUM METHOD:** A convenient method for finding the equivalent, or total, the resistance of two parallel resistors is by using the following formula.

This equation, called the product over the sum formula, is used so frequently it should be committed to memory.

**Example:** What is the equivalent resistance of a 20-ohm and a 30-ohm resistor connected in parallel, as in figure 47?

**Power in a Parallel Circuit**

Power computations in a parallel circuit are essentially the same as those used for the series circuit.

Since power dissipation in resistors consists of a heat loss, power dissipations are additive regardless of how the resistors are connected in the circuit. The total power is equal to the sum of the power dissipated by the individual resistors.

Like the series circuit, the total power consumed by the parallel circuit is:

**Example:** Find the total power consumed by the circuit in figure 48.

Since the total current and source voltage are known, the total power can also be computed by:

**Equivalent Circuits**

In the study of electricity, it is often necessary to reduce a complex circuit into a simpler form.

Any complex circuit consisting of resistances can be redrawn (reduced) to a basic equivalent circuit containing the voltage source and a single resistor representing total resistance.

This process is called reduction to an EQUIVALENT CIRCUIT. The parallel circuit shown in Figure 49 part A shows the original circuit.

To create the equivalent circuit, you must first calculate the equivalent resistance.

Once the equivalent resistance is known, a new circuit is drawn consisting of a single resistor (to represent the equivalent resistance) and the voltage source, as shown in Figure 49 part B.

**Rules for Parallel DC Circuits**

- The same voltage exists across each branch of a parallel circuit and is equal to the source voltage.
- The current through a branch of a parallel network is inversely proportional to the amount of resistance of the branch.
- The total current of a parallel circuit is equal to the sum of the individual branch currents of the circuit.
- The total resistance of a parallel circuit is found by the general formula:

or one of the formulas derived from this general formula.

- The total power consumed in a parallel circuit is equal to the sum of the power consumption of the individual resistances.

**SOLVING PARALLEL CIRCUIT PROBLEMS**

Problems involving the determination of resistance, voltage, current, and power in a parallel circuit are solved as simply as in a series circuit.

The procedure is the same

- (1) draw the circuit diagram,
- (2) state the values given and the values to be found,
- (3) select the equations to be used in solving for the unknown quantities based upon the known quantities, and
- (4) substitute the known values in the equation you have selected and solve for the unknown value.

Example: A parallel circuit consists of five resistors.

The value of each resistor is known and the current through R** _{1}** is known.

You are asked to calculate the value for total resistance, total power, total current, source voltage, the power used by each resistor, and the current through resistors R** _{2}**, R

**, R**

_{3}**, and R**

_{4}**.**

_{5}**Rules for Parallel DC Circuits**

The same voltage exists across each branch of a parallel circuit and is equal to the source voltage.

Once the equivalent resistance is known, a new circuit is drawn consisting of a single resistor (to represent the equivalent resistance) and the voltage source, as shown above in Figure 49B.

This may appear to be a large amount of mathematical manipulation. However, if you use the step-by-step approach, the circuit will fall apart quite easily.

The first step in solving this problem is for you to draw the circuit and indicate the known values as shown in figure 50.

There are several ways to approach this problem.

With the values you have been given, you could first solve for R** _{T}**, the power used by R

**, or the voltage across R**

_{1}**, which you know is equal to the source voltage and the voltage across each of the other resistors.**

_{1}Solving for R** _{T}** or the power used by R

**will not help in solving for the other unknown values. Once the voltage across R**

_{1}**is known, this value will help you calculate other unknowns. Therefore the logical unknown to solve for is the source voltage (the voltage across R**

_{1 }**).**

_{1}Now that source voltage is known, you can solve for current in each branch.

Given:

E** _{S}** = 180V

R** _{2}** = 30Ω

R** _{3}** = 18Ω

R** _{4}** = 18Ω

R** _{5}** = 18Ω

Since R** _{3}** = R

**= R**

_{4}**and the voltage across each branch is the same:**

_{5}I** _{R4}** = 10A

I** _{R5}** = 10A

**Solving for total resistance.**

An alternate method for solving R** _{T}** can be used. By observation, you can see that R

_{3}_{,}R

_{4}_{,}and R

**are of equal ohmic value. Therefore an equivalent resistor can be substituted for these three resistors in solving for total resistance.**

_{5}The circuit can now be redrawn using a resistor labeled R** _{eq1} **in place of R

**, R**

_{3}**, and R**

_{4}**as shown in figure 51.**

_{5}An equivalent resistor can be calculated and substituted for R** _{1}** and R

**by use of the product over the sum formula.**

_{2}

The circuit is now redrawn again using a resistor labeled R** _{eq2}** in place of R

**and R**

_{1}**as shown in figure 52.**

_{2}You are now left with two resistors in parallel. The product over the sum method can now be used to solve for total resistance.

This agrees with the solution found by using the general formula for solving for resistors in parallel. The circuit can now be redrawn as shown in figure 53 and the total current can be calculated.

This solution can be checked by using the values already calculated for the branch currents.

Solution:

I** _{T}** + I

**+ I**

_{R1}**+…I**

_{R2}

_{RN}I** _{T}** = 9A + 6A +10A +10A +10A

IT = 45A

Now that total current is known, the next step is to find total power.

**Solving for the power in each branch.**

The previous calculation for total power can now be checked.

**SERIES-PARALLEL DC CIRCUITS**

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.

**SOLVING COMBINATION-CIRCUIT PROBLEMS**

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 3-54(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}Now that the equivalent resistance for R** _{2}** and R

**has been calculated, the circuit can be redrawn as a series circuit as shown in figure 54(B).**

_{3}Since the voltage across each branch of a parallel circuit is equal, the voltage across R** _{eq1 }**(E

**) will be equal to the voltage across R**

_{eq1}**(E**

_{2}**) and also equal to the voltage across R**

_{R2}**(E**

_{3}**).**

_{R3}

*To find power used by R _{1}:*

*To find the current through ***R _{2 }**

**and R**_{3}

*: (Known from previous calculations)**To find power used by R _{2} and R_{3}, using values from previous calculations:*

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 54, 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 (R_{1}).

In so doing, a voltage drop of 24 volts occurs across resistor R_{1}. At point A, this 3-ampere current divides into two currents. Of the total current, 1.8 amps flow 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.

**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 that 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 56(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 56(B).

While these circuits may appear to be different, the two drawings actually represent the same circuit. The drawing in figure 56(B) is a familiar box-like structure and may be easier to work with.

Figure 3-57(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 3-57(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.

Redrawing a Complex Circuit Figure 58(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 box-like 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 R4 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 R7 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 R6 and R8 and comes to a junction. Label this junction (c). From junction (c) one path for current is through R9 to the source. The other path for current from junction (c) is through R10 to the source. All the junctions in this circuit have now been labeled.

**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, the total resistance of the circuit will INCREASE, and the total current will DECREASE. To clarify these points, figure 3-61 illustrates a series-parallel circuit. First, the effect of an open in the series portion of this circuit will be examined.

In figure 3-61(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 3-61(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 and 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 62 shows a series-parallel network in which shorts are developed.

In figure 62 (A) the normal circuit is shown. R_{T} = 40 ohms and I_{T} = 3 amps.

In figure 62 (B), R1 has shorted. R_{1} now has zero ohms of resistance.

The total resistance of the circuit is now equal to the resistance of the parallel network of R_{2} and R_{3}, or 20 ohms. The 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 62 (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.

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

**. Notice that even though only one portion of the parallel network was shorted, the entire paralleled network was disabled.**

_{1}Opens and shorts alike, if occurring in a circuit, resulting 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 that 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.

**VOLTAGE DIVIDERS**

Most electrical and electronic 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 are 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 63 will help to illustrate this point.

Figure 3-63(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 fig 63(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 63 (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 (Es).

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 64 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 64(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 the 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.

**1. Determine the load requirement and the available voltage source.**

**2. Select bleeder current by applying the 10% rule-of-thumb.**

**3. Calculate bleeder resistance.**

The value of R1 may be rounded off to 275 ohms:

**4. Calculate the total current (load plus bleeder).**

**5. Calculate the resistance of the other divider resistor(s).**

The voltage divider circuit can now be drawn as shown in figure 3-64(B).

**MULTIPLE-LOAD VOLTAGE DIVIDERS**

A multiple-load voltage divider is shown in figure 3-65. An important point that was not emphasized before is that when using the 10% rule-of-thumb to calculate the bleeder current, you must take 10% of the total load current.

Given the information shown in figure 65, you can calculate the values for the resistors needed in the voltage-divider circuits. The same steps will be followed as in the previous voltage divider problem.

The bleeder current should be 10% of the total load current.

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}.

The current through R_{2} (I_{R2}) is equal to the current through R_{1} plus the current through load 1.

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}.*

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}.*

The current through R_{4} (I_{R4}) is equal to the current through R_{3} plus the current through load 3.

I_{R4} 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}.*

With the calculations just explained, the values of the resistors used in the voltage divider are as follows.

**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.

The total power of the circuit is needed to determine the power requirement of the source. The power for the circuit shown in figure 65 is calculated as follows:

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:

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 65, the total power can be calculated by:

**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.

The circuit is drawn as shown in figure 66. 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. I_{R1} is the bleeder current and can be calculated as follows:

Calculate the value of R_{1}.

*Calculate the current through R _{2} using Kirchhoff’s current law.*

At point A:

(or 195 mA leaving point A)

Since E_{R2} = E load 2, you can calculate the value of R_{2}.

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.

Calculate the value of R_{3}.

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.

Calculate the value of R_{4}.

With the calculations just explained, the values of the resistors used in the voltage/divider are as follows:

R** _{1}** = 566Ω

R** _{2}** = 816Ω

R** _{3}** = 566Ω

R** _{4}** = 101Ω

From the information just calculated, any other circuit quantity, such as power, total current, or resistance of the load, could be calculated.

**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 rule-of-thumb 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 series-parallel 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 rule-of-thumb.

Calculate the ohmic value of the bleeder resistor.

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}

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.

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}.

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.

A 75-ohm 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 37.5-ohm resistors, you could connect them in series to get a network of 75 ohms. One 50-ohm and one 25-ohm resistor or seven 10-ohm and one 5-ohm resistor could also be used.

A parallel network could be constructed from two 150-ohm resistors or three 225-ohm resistors. Either of these parallel networks would also be a network of 75 ohms.

The network used in this example will be a series-parallel network using three 50-ohm 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 67.

**QUIVALENT CIRCUIT TECHNIQUES**

The circuit solutions that you have studied up to this point have been obtained mainly through the use of formulas derived from Ohm’s law. As in many other fields of science, electricity has its share of special shortcut methods.

Some of the special circuit analysis techniques are THEVENIN’S THEOREM, which uses a process of circuit reduction to Thevenin’s equivalent circuit; and NORTON’S THEOREM, which is the reduction of a circuit to Norton’s equivalent.

Another method is called LOOP ANALYSIS. This uses Kirchhoff’s voltage law to simultaneously solve problems in parallel branches of a circuit.

The use of these methods should be reserved until you have become thoroughly familiar with the methods covered thus far in this chapter. You may want to explore some of the special techniques later in your career.

**ELECTRICAL SAFETY**

Safety precautions must always be observed by persons working around electric circuits and equipment to avoid injury from electric shock.

The danger of shock from a 450-volt ac electrical service system is well recognized by operating personnel. This is shown by the relatively low number of reports of serious shock received from this voltage, despite its widespread use.

On the other hand, a number of fatalities have been reported due to contact with low-voltage circuits.

Despite a fairly widespread, but totally unfounded, popular belief to the contrary, low-voltage circuits (115 volts and below) are very dangerous and can cause death when the resistance of the body is lowered.

Fundamentally, current, rather than voltage, is the measure of shock intensity. The passage of even a very small current through a vital part of the human body can cause DEATH.

The voltage necessary to produce the fatal current is dependent upon the resistance of the body, contact conditions, the path through the body, etc.

For example, when a 60-hertz alternating current, is passed through a human body from hand to hand or from hand to foot, and the current is gradually increased, it will cause the following effects:

At about 1 milliampere (0.001 ampere), the shock can be felt; at about 10 milliamperes (0.01 ampere), the shock is of sufficient intensity to prevent voluntary control of the muscles; and at about 100 milliamperes (0.1 ampere) the shock is fatal if it lasts for 1 second or more.

The above figures are the results of numerous investigations and are approximate because individuals differ in their resistance to electrical shock.

It is most important to recognize that the resistance of the human body cannot be relied upon to prevent a fatal shock from 115 volts or less—

FATALITIES FROM VOLTAGES AS LOW AS 30 VOLTS HAVE BEEN RECORDED.

Tests have shown that body resistance under unfavorable conditions may be as low as 300 ohms, and possibly as low as 100 ohms from temple to temple if the skin is broken.

Short circuits can be caused by accidentally placing or dropping a metal tool, rule, flashlight case, or other conducting article across an energized line.

The arc and fire which result, even on relatively low-voltage circuits, may cause extensive damage to equipment and serious injury to personnel.

*ALL LIVE ELECTRIC CIRCUITS SHALL BE TREATED AS POTENTIAL HAZARDS AT ALL TIMES.*

**DANGER SIGNALS**

Personnel should constantly be on the alert for any signs which might indicate a malfunction of electrical equipment.

Besides the more obvious visual signs, the reaction of other senses, such as hearing, smell, and touch, should also make you aware of possible electrical malfunctions.

Examples of signs which you must be alert for are:

- fire, smoke, sparks, arcing, or an unusual sound from an electric motor;
- frayed and damaged cords or plugs; receptacles, plugs, and cords which feel warm to the touch; slight shocks felt when handling electrical equipment;
- unusually hot running electric motors and other electrical equipment;
- an odor of burning or overheated insulation;
- electrical equipment which either fails to operate or operates irregularly; and
- electrical equipment which produces excessive vibrations is also an indication of malfunctions.

When any of the above signs are noted, they are to be reported immediately to a qualified technician, DO NOT DELAY,

*Do not operate faulty equipment.**Above all, do not attempt to make any repairs yourself – unless you are qualified to do so.*

**Stand clear of any suspected hazard and instruct others to do likewise.**

**Warning Signs**—they have been placed for your protection. To disregard them is to invite personal injury as well as possible damage to equipment.

Switches and receptacles with a temporary warning tag, indicating work is being performed, are not to be touched.

**Working Near Electrical Equipment**—When work must be performed in the immediate vicinity of electrical equipment, check with the technician responsible for the maintenance of the equipment so you can avoid any potential hazards of which you may not be immediately aware.

**Authorized Personnel Only**—Because of the danger of fire, damage to equipment, and injury to personnel, all repair, and maintenance work on electrical equipment shall be done only by authorized persons.

Keep your hands off of all equipment that you have not been specifically authorized to handle.

Particularly stay clear of electrical equipment opened for inspection, testing, or servicing.

**Circuit Breakers and Fuses**—Covers for all fuse boxes, junction boxes, switch boxes, and wiring accessories should be kept closed.

Any cover which is not closed or is missing should be reported to the technician responsible for its maintenance.

Failure to do so may result in injury to personnel or damage to equipment in the event accidental contact is made with exposed live circuits.

**ELECTRICAL FIRES**

Carbon dioxide (CO2) is used in fighting electrical fires. It is non-conductive and, therefore, the safest to use in terms of electrical safety. It also offers the least likelihood of damaging equipment.

However, if the discharge horn of a CO2 extinguisher is allowed to accidentally touch an energized circuit, the horn may transmit a shock to the person handling the extinguisher.

The very qualities which cause CO2 to be a valuable extinguishing agent also make it dangerous to life. When it replaces oxygen in the air to the extent that combustion cannot be sustained, respiration also cannot be sustained. Exposure of a person to an atmosphere of high concentration of CO2 will cause suffocation.

** ****FIRST AID FOR ELECTRIC SHOCK**

A person who has stopped breathing is not necessarily dead but is in immediate danger.

Life is dependent upon oxygen, which is breathed into the lungs and then carried by the blood to every body cell.

Since body cells cannot store oxygen, and since the blood can hold only a limited amount (and that only for a short time), death will surely result from a continued lack of breathing. However, the heart may continue to beat for some time after breathing has stopped, and the blood may still be circulated to the body cells.

Since the blood will, for a short time, contain a small supply of oxygen a person’s life may be saved.

The process by which a person who has stopped breathing can be saved is called ARTIFICIAL VENTILATION (RESPIRATION). The purpose of artificial ventilation is to force air out of the lungs and into the lungs, in rhythmic alternation, until natural breathing is reestablished.

Artificial ventilation should be given only when natural breathing has stopped; it should NOT be given to any person who is breathing naturally.

You should not assume that an individual who is unconscious due to electrical shock has stopped breathing. To tell if someone suffering from an electrical shock is breathing, place your hands on the person’s sides, at the level of the lowest ribs. If the victim is breathing, you will usually be able to feel the movement.

*Remember: DO NOT GIVE ARTIFICIAL VENTILATION TO A PERSON WHO IS BREATHING NATURALLY.*

Records show that seven out of ten victims of electric shock were revived when artificial respiration was started in less than 3 minutes.

After 3 minutes, the chances of revival decrease rapidly.

Once it has been determined that breathing has stopped, the person nearest the victim should start the artificial ventilation without delay and send others for assistance and medical aid.

The only logical, permissible delay is that required to free the victim from contact with the electricity in the quickest, safest way.

This step, while it must be taken quickly, must be done with great care; otherwise, there may be two victims instead of one.

In the case of portable electric tools, lights, appliances, equipment, or portable outlet extensions, this should be done by turning off the supply switch or by removing the plug from its receptacle.

If the switch or receptacle cannot be quickly located, the suspected electrical device may be pulled free of the victim.

Other persons arriving on the scene must be clearly warned not to touch the suspected equipment until it is de-energized.

Aid should be enlisted to unplug the device as soon as possible.

The injured person should be pulled free of contact with stationary equipment (such as a bus bar) if the equipment cannot be quickly de-energized, or if considerations of military operation or unit survival prevent the immediate shutdown of the circuits.

This can be done quickly and safely by carefully applying the following procedures:

- Protect yourself with dry insulating material.
- Use a dry board, belt, clothing, or other available nonconductive material to free the victim from electrical contact.

*DO NOT TOUCH THE VICTIM UNTIL THE SOURCE OF ELECTRICITY HAS BEEN REMOVED.*

Once the victim has been removed from the electrical source, it should be determined, if the person is breathing.

If the person is not breathing, a method of artificial ventilation is used.

Sometimes victims of electrical shock suffer cardiac arrest (heart stoppage) as well as loss of breathing.

Artificial ventilation alone is not enough in cases where the heart has stopped.

A technique known as Cardiopulmonary Resuscitation (CPR) has been developed to provide aid to a person who has stopped breathing and suffered a cardiac arrest.

Because you most likely will be working in the field of electricity, the risk of electrical shock is higher than in most other occupations.

You should, at your earliest opportunity, learn the technique of CPR.

CPR is relatively easy to learn and is taught in courses available from the American Red Cross