circuit analisys and ohms law

8/3/2019 Circuit Analisys and Ohms Law

1/15

Excerpt from

Circuit Analysis andOhms LawBy

Robert Cecci

Show Bookma


2/15

he following is a sample excerpt from a study unit converted into the Adobe Acrobat formatsample online exam is available for this excerpt.

cuit analysis, one of the most fundamental jobs of an electrician or electronics technician.ith the knowledge of how voltage, current, and resistance are related in a circuit, a person csily learn to troubleshoot any circuit quickly and efficiently. This excerpt of the study unitcuses on circuit resistance.

ter reading through the following material, feel free to take the sample exam based on thcerpt.

Preview

he sample text, which is from the Basic Electronics program, introduces the concept of
http://enrollonline/SampleTest.jhtmlhttp://enrollonline/SampleTest.jhtml


3/15

CALCULATING CIRCUIT RESISTANCE . . . . . . . . . . . . . . . . . . . .

Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1

Circuit Basics. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2

Calculating Resistance in Series Circuits . . . . . . . . . . . . . . . . . . 3

Calculating Resistance in Parallel Circuits . . . . . . . . . . . . . . . . . 4Calculating Resistance in Series-Parallel Circuits . . . . . . . . . . . . . 10

Practical Example . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11

EXAMINATION

Contents
http://enrollonline/SampleTest.jhtmlhttp://enrollonline/SampleTest.jhtml


4/15

ALCULATING CIRCUIT RESISTANCE

troductionCircuit analysis is one of the most important tasks of an electricianelectronics technician. In order to troubleshoot circuits and electricdevices, you must clearly understand how voltage, current, and restance are related in circuits. With this knowledge, you can performcalculations that will provide you with important information abougiven circuit. Then, youll be able to quickly and easily locate thesource of trouble in a faulty circuit.

Lets take a moment now to review some important topics. Remem ber that current is the flow of electric charge in a circuit. The amouof electric current flowing through a circuit is called amperage.Oampere of current is equal to the charge of 6,240,000,000,000,000electrons flowing past a given point in a circuit per second. Note thin this definition of ampere, it doesnt matter which direction theelectrons are flowing in.

There are two theories of electric current flow that are used to de-scribe the movement of electrons in a circuit. You should be awareof both theories, since youll see both of them referred to in textboand technical manuals. In the electron theoryof current flow, elecare said to flow from the negative point of a circuit to the positive

point. In contrast, in the conventional theoryof current flow, curresaid to flow from the positive point of a circuit to the negative poin

Why are there two theories of current flow, and which is correct?The answer is that both theories are used in the study and applicatof electricity. The conventional theory is older and is based on theassumption that electric current flows from the positive point of acircuit to the negative point. In later years, scientists determined thelectrons actually flow from the negative point of a circuit to the ptive. Thus, the newer electron theory is the correct way to explaincurrent flow.

Circuit Analysis and Ohms Law


5/15

also used to explain the rules of magnetism and the electrical forceproduced in motors and generators.

One good example of the conventional theory in practice is aneveryday car battery. In a car battery, the positive battery terminal considered to be the hot terminal and is therefore painted red.According to the electron theory, the negative terminal is actually source of electron flow. However, it would be too confusing to chathe standards of battery production to adapt to a new theory of current flow! Therefore, the conventional theory continues to be usedwhen referring to batteries.

For most practical purposes, all you need to know is that currentflows out of a power source, flows through a circuit load, and then

turns to the power source. It doesnt really matter in which directiothe current flows. However, you should be aware of the two theoriof current flow and what they mean. For our purposes in this studyunit, when we talk about circuits, well use the electron theory to dscribe current flow. In addition, all our electrical diagrams are drawusing the electron or negative-to-positive standard. This is becausemost of the actual circuit diagrams youll see in your work are draaccording to the electron theory.

ircuit BasicsMost DC circuits contain three basic parts: a power source, conductorand a resistance. Resistance in a circuit may come from heaters, motorelay coils, lights, or any other electrical devices. However, for instructional purposes, all the circuit examples in this text will contain resisto

Now, lets briefly review the difference between a series and a parcircuit. In a series circuit, components are connected end-to-end.

typical series circuit is shown in Figure 1. In this circuit, three re(R1, R2, and R3) are connected end-to-end, and power is supplied b battery. Currentleaving the negat battery terminalflow through R1and R3 , and backthe positive batteterminal. Becaus

components in thi i

Circuit Analysis and Ohm's

Fit Width


6/15

In contrast, a parallel circuit is connected much differently. In a palel circuit (Figure 2), the components are connected across each oCurrent leaves the negative battery terminal and then splits into thrdifferent circuit paths. Some of the current flows through R1, sothrough R2, and the remainder through R3. The current then flowthrough the conductors back to the positive battery terminal. In a pallel circuit, if one circuit path is broken or opened, current continuto flow through the other paths, and part of the circuit keeps worki

In a parallel circuit, the amount of current that flows through each sistor is related to the values of the resistors. If all three resistors hexactly the same value, then the current through each resistor will the same. If one of the resistors is higher in value than the other twresistors, less current will flow through that resistor.

alculating Resistance in Series CircuitsSeries circuits are very simple circuits. Therefore, the process of clating the total resistance of a series circuit is very simple. To calcuthe total resistance of a series circuit, simply add the values of all tresistors in the circuit.

Lets begin by looking at a simple two-resistor circuit (Figure 3)this figure, a 300 (ohm) resistor ( R1) and a 500 resistor (R2) connected in series. The total value of resistance ( RT ) is equal tosum of the resistors. The resistance is calculated as follows:

rcuit Analysis and Ohm's Law

FIGURE 2The current inthis parallel circuit (I)flows from the batteryand then separatesonto three circuit paths( I1 , I 2 and I3 ).

Fit Width


7/15

This same addition rule applies no matter how many resistors are

connected in series. For example, look at the circuit in Figure 4.five resistors are connected in series. The total resistance in this cirwould be calculated as follows:

RT = R1 + R2 + R3 + R4 + R5 Set up the problem.

RT = 1,000 + 2,000 + 100 + 100 + 60 Add the resistor values (1,000 + 2,000 + 100 + 60 = 3,260).

RT =

3,260

Answer: The total resistance is 3,260

In summary, note that in a series circuit, the total circuit resistancewill always be greater than the value of any one resistor.


FIGURE 3To calculatethe total resistance in this series circuit, simply addthe values of the two resistors.

FIGURE 4In thisfive-resistor circuit, thetotal resistance can befound by adding thevalues of the five resistors.

Fit Width

Fit Width


8/15

First, if a parallel circuit contains resistors of equal value, the totalcuit resistance will be equal to a fraction of the value of one resistif a parallel circuit contains two equal resistors, then the total circuresistance is equal to one-half the value of either resistor. If the circontains three equal resistors, the total circuit resistance is equal toone-third of the value of one resistor. If the circuit contains four eqresistors, the total circuit resistance is equal to one-fourth the valueone resistor, and so on.

Figure 5 shows a circuit containing two 1,000 resistors connecparallel. The total resistance in this circuit is equal to one-half of 1(that is, 1,000 2). Therefore, the total circuit resistance is 500

In another example, if a circuit contains four 1,000 resistors conected in parallel, the total circuit resistance is equal to one-fourth 1,000 (1,000 4). Therefore, the total circuit resistance is 250 .

Now, lets look at a different type of problem. If two resistors withdifferent valuesare connected in parallel, youll need to use a differmethod to calculate the total circuit resistance. To find the total restance in such a circuit, use the following formula:

RR RR RT

=

+1 2

1 2

Figure 6 shows a 60 resistor (R1) and a 20 resistor (R2) connin parallel. Lets find the total resistance in this circuit.


FIGURE 5Since both parallel resistors are of the same value, the total resistance is equal toone-half the value of either resistor.

Fit Width


9/15

RT =

+

60 2060 20

Multiply (60 20 = 1,200).Add (60 + 20 = 80).

RT =1 200

80

, Divide (1,200 80 = 15).

RT = 15 Answer: The total resistance of this ciris 15 .

Note that the total resistance of a parallel circuit is always lower ththe lowest value of any one resistor in the circuit.

Now, lets look at another example problem. Suppose a circuitcontains a 1,000 resistor (R1) and a 10 resistor (R2) connecteparallel. Use the formula to calculate the total resistance of the circ

R R RR RT

=

+1 2

1 2

Write the formula.

RT =

+

1 000 101 000 10 , ,

Substitute 1,000 for R1 and 10 for

RT =

+

1 000 101 000 10 , ,

Multiply (1,000 10 = 10,000).Add (1,000 + 10 = 1,010).

10 000 Di id (10 000 1 010 9 9)


FIGURE 6This parallelcircuit contains two resistors with different values. You must use theformula shown tocalculate the total resistance.

Fit Width


10/15

Note that the resistor with the lowest value in this circuit was R2 (1and the total resistance of the circuit is less than 10 .

In this circuit, most of the current will flow through R2 since it hsuch a low resistance value. Very little current will flow throughsince its resistance is so large compared to that of R2.

Now, lets look at a parallel circuit containing three resistors withdifferent values (Figure 7). There are two methods of calculating total resistance in such a circuit.

As shown, the values of the three resistors are 100 , 200 , and 30To calculate the total resistance, first use the formula to calculate ttotal resistance of R1 and R2. Then, use that value and the value oto calculate the total circuit resistance.

R R RR RP1

1 2

1 2

=

+Write the formula. The abbreviation Rstands for the total resistance of the firsresistors.

R P1 =

+

100 200100 200

Substitute 100 for R1 and 200 for

R P1 =

+

100 200

100 200

Multiply (100 200 = 20,000).Add (100 + 200 = 300).


FIGURE 7This circuit contains these resistorsconnected in parallel.

Fit Width


11/15

We can now alter Figure 7 to include this value (66.67 ) in placR1 and R2. Figure 8 shows the updated circuit diagram. The circuinow contains two resistors connected in parallel.

Now, use the value 66.67 and the value of the third resistor to fthe total circuit resistance.

RR RR RT

P

P

=

+1

1

3

3

Write the formula.

RT =

+

66 67 30066 67 300

.

.Substitute 66.67 for RP1 and 300 f

RT =

+

66 67 30066 67 300

.

.Multiply (66.67 300 = 20,001).Add (66.67 + 300 = 366.67).

RT =20 001366 67

,.

Divide (20,001 366.67 = 54.55).

RT = 54.55 Answer: The total resistance of this cir54.55 (rounded).

Note that the total resistance of this circuit (54.55 ) is less than

lowest value of any one resistor (100 ).


FIGURE 8In thisupdated version of thecircuit diagram in Figure7, RP1 is used in place of R1 and R2 .

Fit Width


12/15

To find the total resistance of a parallel circuit using the reciprocalmethod, first find the reciprocal of each resistor value. Then, add athe reciprocal values together. Finally, find the reciprocal of that vto get the total resistance.

This isnt as difficult as it sounds! Just use the following reciprocaformula to make your calculations:

1 1 1 1R R R RT

= + +

1 2 3

Lets look at an example problem. The circuit shown in Figure 9contains three resistors with values of 20 , 40 , and 50 . Usereciprocal formula to find the total circuit resistance.

1 1 1 1R R R RT

= + +

1 2 3

Write the formula.

1 120

140

150RT

=

+

+

Substitute the values of R1, R2, and R3Divide to simplify each of the fractions(1 20 = 0.05; 1 40 = 0.025; 1 50 =

1 0 05 0 025 0 02RT

= + +. . . Add (0.05 + 0.025 + 0.02 = 0.095).

1 0 095RT

= . The reciprocal of 0.095 is 10095. .

RT =1

0 095.Divide (1 0.095 = 10.53).

RT = 10.53 Answer: The total resistance of this cir10.53 (rounded).


Page Width


13/15

The reciprocal method can be used to calculate the total resistancecircuit no matter how many resistors are connected in parallel.

alculating Resistance in Series-Parallel CircuitsWhen youre working with real-life circuits, youll find that theyrnot always neatly laid out in series or parallel. Often, circuits conta both series and parallel connections. This type of circuit is called acombination or series-parallel circuit. The circuit in Figure 10 coa total of four resistors joined in both series and parallel connectio

Now, lets calculate the total resistance of the circuit shown in FiguThe first step is to find the resistance of the parallel portion of thcircuit. This value can easily be found by using the resistance formu

RR RR RP

=

+2 3

2 3

Write the formula.

R P =

+

20 4020 40

Substitute the values of R2 and R3.

R P =

+

20 4020 40

Multiply (20 40 = 800).Add (20 + 40 = 60).


FIGURE 10This circuit contains four resistors

connected in both seriesand parallelarrangements.

Fit Width


14/15

Now, lets substitute this value for the parallel section in our circuidiagram. The revised diagram is shown in Figure 11. With this stution, the circuit diagram in Figure 11 is now a simple series cir

Since you now have a series circuit, your calculations are easy to cplete. Simply add the remaining resistor values together to get the tal resistance for this circuit.

RT = R1 + RP + R4 Set up the problem.

RT = 10 + 13.33 + 30 Substitute the values for R1 , RP , and R(10 + 13.33 + 30 = 53.33).

RT = 53.33 Answer: The total resistance of this cir53.33 .

ractical ExampleThe ability to calculate total circuit resistance in series and parallecuits is an important skill for the industrial electrician or electronictechnician. This knowledge can often be used to troubleshoot a faumachine quickly and efficiently.

For example, suppose youre working on a press that has two solenoid valves connected in parallel The valves are activated by a pu


FIGURE 11In thisupdated version of Figure 10, the values of R2 and R 3 have been replaced by the valueRP .

Fit Width


15/15

Now, suppose that the machine is malfunctioning. When the operapresses the button, the locator retracts, but the shield doesnt lift. Tproblem could be the result of an open solenoid coil or a stuck solenoid valve. How can you figure out what the problem is?

To troubleshoot this machine, use a meter to measure the resistancthe two parallel coils.

Since the two solenoids have the same resistance value, the total ccuit resistance should be one-half the resistance value of one solencoil.

R P =96

2Divide the value of one resistance by 2

RT = 48 Answer: The total circuit resistance is

The total circuit resistance is 48 , so if you read 48 on your mthe solenoid valve is stuck. If you read 96 , one of the two coilsone for the shield, has opened causing the failure of the solenoidvalve.


FIGURE 12In this circuit,two solenoid coils areconnected in parallel.

Fit Width
http://enrollonline/SampleTest.jhtml

circuit analisys and ohms law

Documents