basic circuit theory

Lesson 2: Basic Circuit Theory

Les

son

2:

B

asic

Cir

cuit

Th

eory

Introduction

This lesson covers basic direct current theory by reviewing the threebasic types types of electrical circuits and the laws that apply to eachtype circuit:

Series Circuits

Parallel Circuits

Series-Parallel Circuits

Objectives

At the completion of this lesson, the student will be able to:

Given examples of series, parallel, and series-parallel circuit, explainthe basic laws of d.c. circuits. Calculate current flow, circuitresistance and voltage drops. Draw and explain equivalent circuitsand their applications.

BASIC CIRCUITS

Unit 2 2-2-2 Electrical FundamentalsLesson 2

Fuse

+ -

Fig. 2.2.1 Series Circuit

Series Circuit

A series circuit is the simplest kind of circuit. In a series circuit,each electrical device is connected to other electrical devices in sucha way that there is only one path for current to flow. In the circuitshown here, current flows from the battery (+) through a fuse(protection device) and a switch (control device) to the lamp (load)and then returns to frame ground. All circuit devices andcomponents are connected in series. The following rules apply to allseries circuits:

At any given point in the circuit the current value is the same.

The total circuit resistance is equal to the sum of all theindividual resistances and is called an equivalent resistance.

The voltage drop across all circuit loads are equal to the appliedsource voltage.

A simple way to express these series circuit rules are:

Voltage is the SUM of all voltage drops.

Current is the SAME at any given point in the circuit.

Resistance is the SUM of all individual resistances.

Applying the Rules


12V

3Ω

6Ω

R1 R2

R3

12V

3Ω

+ - + -


Draw the above circuit on the board. Explain the following circuitparameters.

The circuit is made up of various devices and components, includinga 24 volt power source. Since two of the circuit values are given,solving for the unknown value is simple, if the basic laws of seriescircuits are applied.

The first step in solving the above circuit is to determine the totalcircuit resistance. The following equation is used for determiningtotal resistance: Rt = R1 + R2 + R3, or Rt = 3Ω + 3Ω + 6Ω, or

Rt = 12Ω.

Since the value for the power source was given as 24 volts and thecircuit resistance has been calculated as 12Ω, the only valueremaining to calculate is the current flow. Total circuit current iscalculated by using the Ohm’s Law Circle and writing the followingequation: I = E/R, or I = 24V/12Ω, or I = 2 amperes.

The remaining step is to “plug” the value for current flow into eachof the resistive loads. One of the rules for series circuits stated thatcurrent was the SAME at any given point. Using the equation

E = IxR for each resistor will determine the voltage drop across eachload. The following voltage drops are:

E1 = 2A x 3Ω = 6VE2 = 2A x 3Ω = 6VE3 = 2A x 6Ω = 12V

All of the circuit values have now been calculated. Using the Ohm’sLaw Circle, verify each answer.

A500 mA

12V

24Ω

4Ω

R1 R2R3

R4

____?

12V

12Ω

+ - + -


Unit 2 - 1 - Electrical FundamentalsInstructor Copy Exercise 2.2.1

Series Circuit

Use the Ohm’s Law Circle and the rules for series circuits to calculate the unknown values.In the space below the question, write the equation used to solve the problem and show allmathematical steps.

1. What is the voltage drop across R4? 2 volts

2. What is the total circuit resistance? 48 ohms


4.What is the voltage drop across switch S1? 0 volts

5. What is the current flow through resistor R3? .5 amps

6. A voltmeter placed across R2 reads? 12 volts

Understanding Series Circuits

Name ______________________________

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2.2.

1

Series Circuit

Use the Ohm’s Law Circle and the rules for series circuits to calculate the unknown values.In the space below the question, write the equation used to solve the problem and show allmathematical steps.

1. What is the voltage drop across R4? _______volts

2. What is the total circuit resistance? ________ ohms

3. What is the voltage drop across R1? __________ volts

4.What is the voltage drop across switch S1? ________volts

5. What is the current flow through resistor R3? ______amps

6. A voltmeter placed across resistor R2 reads?________volts

A500 mA

12V

24Ω

4Ω

R1 R2R3

R4

____?

12V

12Ω

+ - + -


Unit 2 - 1 - Electrical FundamentalsStudent Copy Exercise 2.2.1

Understanding Series Circuits

Name ______________________________

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2.1

Fuse

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Parallel Circuit

A parallel circuit is more complex than a series circuit because thereis more than one path for current to flow. Each current path is calleda branch. Because all branches connect to the same positive andnegative terminal, they will all have the same voltage; each branchdrops the same amount of voltage, regardless of resistance within thebranch.

The current flow in each branch can be different, depending on theresistance. Total current in the circuit equals the sum of the branchcurrents.

The total resistance is always less than the smallest resistance in anybranch.

In the circuit shown in Figure 2.2.4, current flows from the batterythrough a fuse and switch, and then divides into two branches, eachcontaining a lamp. Each branch is connected to frame ground. Thefollowing rules apply to parallel circuits:

The voltage is the same in each parallel branch.

The total current is the sum of each individual branch currents.

The equivalent resistance is equal to the applied voltage dividedby the total current, and is ALWAYS less than the smallestresistance in any one branch.

A simple way to express these parallel rules are:

Voltage is the SAME for all branches.

Current is the SUM of the individual branch currents.

Equivalent resistance is SMALLER than the smallest resistance ofany individual branch.

Fig. 2.2.4 Parallel Circuit

Fuse

24V

4Ω4Ω+ -


The circuit is made up of various devices and components, includinga 24 volt power source. The resistance of each lamp is given alongwith the value of source voltage. Before applying the basic laws ofparallel circuits it will be necessary to determine an equivalentresistance to replace the two 4 ohm parallel branches.

The first step in developing an equivalent circuit is to apply the basicrules for determining the total resistance of the two parallel branches.Remember, the total resistance of the combined branches will besmaller than the smallest resistance of an individual branch. Thecircuit above has two parallel branches, each with a 4 Ω lamp,therefore, the total resistance will be less than 4 Ω.

The following equation is used to solve for total resistance.

Fig. 2.2.5 Circuit

1 Rt

1 R1

1 R2

1 4

1 .50

1 4

1 Rt

1 Rt

+

+

=

=

=

or

.25 + .25 = .50 or

Rt = or Rt = 2 ohms


As stated earlier, one of the rules for parallel circuits states that thevoltage is the SAME in all parallel branches. With 24 volts appliedto each branch, the individual current flow can be calculated usingohms law. The equation I = E/R is used to calculate the current ineach branch as 6 amps. In this particular case, the current flow ineach branch is the same because the resistance values are the same.

Solving current flow in a parallel circuit

A

12V

4ΩR1 R2 R3

12V

4Ω 2Ω+ - + -


Draw the above circuit on the board. Explain the following circuitparameters.

The circuit shown in Fig. 2.2.6 is a typical d.c. circuit with threeparallel branches. The circuit also contains an ammeter connected inseries with the parallel branches (all current flow in the circuit mustpass through the ammeter).

Applying the basic rules for parallel circuits makes solving thisproblem very simple. The source voltage is given (24 volts) and eachbranch resistance is given (R1 = 4Ω; R2= 4Ω; R3 = 2Ω). Applyingthe voltage rule for parallel circuits (voltage is the SAME in allbranches) we can solve the unknown current value in each branch byusing the Ohm’s Law Circle, whereas, I = E/R.

I1 = E1/R1 or I1 = 24/4 or I1 = 6 ampsI2 = E2/R2 or I2 = 24/4 or I2 = 6 ampsI3 = E3/R3 or I3 = 24/2 or I3 = 12 amps

Since current flow in parallel branches is the SUM of all branchcurrents, the equation for total current is It = I1 + I2 + I3 or

6+6+12 = 24 amps. With the source voltage given as 24 volts and thetotal current calculated at 24 amps, the total circuit resistance iscalculated as 1 ohm. (Rt = Et/It).


Understanding Parallel Circuits

12V

3ΩR1 R2 R34Ω 6Ω+ -


Parallel Circuit

Use the Ohm’s Law Circle and the rules for parallel circuits to calculate the unknown values.In the space below the question, write the equation used to solve the problem and show allmathematical steps.


2. What is the total circuit resistance? 1.33 ohms


4. How much current is following through resistor R3? 2 amps

5. What is the total circuit current (It)? 9 amps

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Understanding Parallel Circuits

12V

3ΩR1 R2 R34Ω 6Ω+ -


Parallel Circuit

Use the Ohm’s Law Circle and the rules for parallel circuits to calculate the unknown values.In the space below the question, write the equation used to solve the problem and show allmathematical steps.

1. What is the voltage drop across R3? _________ volts

2. What is the total circuit resistance? __________ ohms

3. What is the voltage drop across R1? _________ volts

4. How much current is following through resistor R3? __________ amps

5. What is the total circuit current (It)?__________ amps

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R2 R312V

R1

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Series-Parallel Circuits

A series-parallel circuit is composed of a series section and a parallelsection. All of the rules previously discussed regarding series andparallel circuits are applicable in solving for unknown circuit values.

Although some series-parallel circuits appear to be very complex,they are solved quite easily using a logical approach. The followingtips will make solving series-parallel circuits less complicated.

Examine the circuit carefully, then determine the path or pathsthat current may flow through the circuit before returning to thesource.

Redraw a complex circuit to simplify its appearance.

When simplifying a series-parallel circuit, begin at the farthestpoint from the voltage source. Replace series and parallel resistorcombinations one step at a time.

A correctly redrawn series-parallel (equivalent) circuit willcontain only ONE series resistor in the end.

Apply the simple series rules for determining the unknownvalues.

Return to the original circuit and “plug” in the known values.Use Ohm’s Law to solve the remaining values.

Fig. 2.2.8 Series-parallel Circuit


Solving a series-parallel problem

The series-parallel circuit as shown in Fig. 2.2.8 shows a 2Ω resistorin series with a parallel branch containing a 6Ω resistor and a 3Ωresistor. To solve this problem it is necessary to determine theequivalent resistance for the parallel branch. Using the followingequation, solve for the parallel equivalent (Re) resistance.

Redraw Fig. 2.2.9 inserting the equivalent resistance for the parallelbranch. Then solve circuit totals using simple Ohm’s Law rules forseries circuits.

R2 R312V

R1

+ -

2Ω

6Ω 3Ω

Fig. 2.2.9 Series-parallel Circuit

1 Re

1 R2

1 R3

= +

1 Re

1 Re

1 Re

1 6

1 3=

=

=

+ or

.1666 + .3333 = .50 or

1 .50

or Re = 2 ohms

R2R3

12V

R1

+ -

2Ω

Re 2Ω

EquivalentResistance


Fig. 2.2.10 Equivalent Series Circuit

R2 R312V

R1

+ -

2Ω

6Ω 3Ω

6V3A

6V1A

6V2A

Fig. 2.2.11 Circuit

Circuit calculations indicate that the total current flow in the circuit is3 amps. Since all current flow that leaves the source must return weknow that the 3 amps must flow through R1. It is now possible tocalculate the voltage drop across R1 by using the equation E = I x R,or E = 3A x 2W, or E1 = 6 volts.

If 6 volts is consumed by resistor R1, the remaining source voltage(6V) is applied to both parallel branches. Using Ohm’s Law for theparallel branch reveals that 1 amp flows through R2 and 2 amps flowthrough R3 before combining into the total circuit current of 3 ampsreturning to the negative side of the power source.

Using the rules for series circuits, the total circuit resistance cannow be calculated using the equation Rt = R1 + Re or Rt = 2+2 or 4

ohms.

The remaining value unknown is current. Again using Ohm's LawCircle, current can be calculated by the equation I = E/R, or I = 12/4, or I = 3 amps.

Consult the original series-parallel circuit and put in the knownvalues.


Other methods and tips for solving complex series-parallelcircuits

As stated earlier, complex circuits can be easily solved by carefullyexamining the path for current flow and then re-drawing the circuit.No matter how complex a circuit appears, drawing an equivalentcircuit and reducing the circuit to its lowest form (series circuit) willprovide the necessary information to “plug” into the original circuit.

Draw the following example of a series-parallel circuit on the boardand explain the following steps for reducing the circuit to a simpleseries circuit.

R2

R3

R1

+ -

TP1TP2

Fig. 2.2.12 Complex Series-parallel Circuits

Step #1:Trace current flow from the (+) side of the battery to the (-) side ofthe battery. All the current leaving the source is available at TP1 (testpoint 1). At TP1 the current is divided among the two parallelbranches and then re-combined at TP2 before flowing through theseries resistor R3 and returning to ground. Now that you haveidentified the path of current flow, the next step is drawing anequivalent circuit for the parallel branches.

Step #2:Using Ohm’s Law calculate the equivalent resistance for the parallelbranch. There are two methods (equations) available for solvingparallel branch resistances. They are:

1 Re

1 R1

1 R2=

=

+

ReR1 x R2R1 + R2

(called product over sum method) (for combiningtwo parallel resistances)

or


When the circuit contains only two branches the “product over summethod” is the easiest equation.

Step #3:Redraw the circuit substituting the Re value to represent the

equivalent resistance. The circuit now has two resistors in series,shown as Re and R3. Further reduce the circuit by combining Re and

R3 as a single resistance called Rt. The following drawings reflect

those steps.

R3

+ -

Re

R3

+ -

Re

Rt = Re + R3

Rt

+ -

Fig. 2.2.12a

Fig. 2.2.12b

Fig. 2.2.12c


Series-Parallel Circuit

Use the Ohm’s Law Circle and the rules for series and parallel circuits to calculate the unknownvalues. In the space below the question, write the equation used to solve the problem and show allmathematical steps.


2. What is the total circuit resistance? 8 ohms


4. How much current is following through resistor R3? 3 amps

5. What is the total circuit current (It)? 3 amps

Understanding Series-Parallel Circuits

R2

R3

R1

+ -

6Ω

3Ω

6Ω

24VFig. 2.2.13 Series-parallel Circuits

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Series-Parallel Circuit

Use the Ohm’s Law Circle and the rules for series and parallel circuits to calculate the unknownvalues. In the space below the question, write the equation used to solve the problem and show allmathematical steps.

1. What is the voltage drop across R3? ________ volts

2. What is the total circuit resistance? _________ ohms

3. What is the voltage drop across R1? __________ volts

4. How much current is following through resistor R3? _______ amps

5. What is the total circuit current (It)? ________ amps

Understanding Series-Parallel Circuits

R2

R3

R1

+ -

6Ω

3Ω

6Ω

24VFig. 2.2.13 Series-parallel Circuits

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.2.3

Name ___________________________________

Name _________________________________

ELECTRICAL CIRCUITS QUIZ ANSWERS

Read each question carefully and fill in the blanks.

1. If the resistance of a typical electrical circuit equaled 6 ohms, and the power source provided 12volts, select the correct equation for solving for current flow.

a. 6 ohms divided by 12 volts = 500 milliampsb. 12 volts divided by 6 ohms = 2 ampsc. 12 volts X 6 ohms = 72 ampsd. None of the above

2. In a circuit, what could cause an increase in current if voltage stayed the same?

a. A decrease in resistanceb. An increase in resistancec. An opend. A short to power

3. In the Ohm's Law Circle, "E" stands for _________.

a. currentb. resistancec. voltaged. watts

4. In the Ohm's Law Circle, "I" stands for ________.


5. If the voltage applied to a resistor is increased and the resistance stayed the same, the current will_____________.

a. decreaseb. increasec. stay the samed. vary

Unit 2 - 1 - Electrical FundamentalsInstructor Copy Quiz 2.2.1

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6. An electrical terminal in a dome lamp circuit is seriously corroded. The dome lamp is very dim.The overall resistance of the circuit has _________.

a. decreasedb. increasedc. remained the samed. no effect on dome lamp brightness.

7. The basic unit of measurement for power is the ________.

a. ampereb. voltc. wattd. ohm

8. A _________ circuit has only one path for current.

a. seriesb. parallelc. series-paralleld. all of the above

9. In a series circuit with more than one resistor, the total resistance of the circuit is ________ theresistance of any single resistor.

a. greater thanb. less thanc. equal tod. smaller than the smallest value

10. In a parallel circuit with more than one resistor, the total resistance of the circuit is ________ theresistance of any single resistor.

a. greater thanb. less than c. equal tod. the square root of

Unit 2 - 2 - Electrical FundamentalsInstructor Copy Quiz 2.2.1

Name _________________________________

ELECTRICAL CIRCUITS QUIZ

Read each question carefully and fill in the blanks.

1. If the resistance of a typical electrical circuit equaled 6 ohms, and the power source provided 12volts, select the correct equation for solving for current flow.

a. 6 ohms divided by 12 volts = 500 milliampsb. 12 volts divided by 6 ohms = 2 ampsc. 12 volts X 6 ohms = 72 ampsd. None of the above

2. In a circuit, what could cause an increase in current if voltage stayed the same?

a. A decrease in resistanceb. An increase in resistancec. An opend. A short to power

3. In the Ohm's Law Circle, "E" stands for _________.


4. In the Ohm's Law Circle, "I" stands for ________.


5. If the voltage applied to a resistor is increased and the resistance stayed the same, the current will_____________.

a. decreaseb. increasec. stay the samed. vary

Unit 2 - 1 - Electrical FundamentalsStudent Copy Quiz 2.2.1

Stu

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Qu

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.2.1

6. An electrical terminal in a dome lamp circuit is seriously corroded. The dome lamp is very dim.The overall resistance of the circuit has _________.

a. decreasedb. increasedc. remained the samed. no effect on dome lamp brightness.

7. The basic unit of measurement for power is the ________.

a. ampereb. voltc. wattd. ohm

8. A _________ circuit has only one path for current.

a. seriesb. parallelc. series-paralleld. all of the above

9. In a series circuit with more than one resistor, the total resistance of the circuit is ________ theresistance of any single resistor.

a. greater thanb. less thanc. equal tod. smaller than the smallest value

10. In a parallel circuit with more than one resistor, the total resistance of the circuit is ________ theresistance of any single resistor.

a. greater thanb. less than c. equal tod. the square root of

Unit 2 - 2 - Electrical FundamentalsStudent Copy Quiz 2.2.1

basic circuit theory

Documents

series circuit rules

circuit loads

circuit values

eachtype circuit

total circuit resistance

basic circuit theorylesson

circuit devices andcomponents

series circuituse