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(3)

Circuits and Circuit Elements

Chapter

18

Table of Contents

Section 1 Schematic Diagrams and Circuits Section 2 Resistors in Series or in Parallel Section 3 Complex Resistor Combination

(4)

Section 1 Schematic Diagrams and Circuits

Chapter

18

Objectives

• Interpret and construct circuit diagrams.

• Identify circuits as open or closed.

• Deduce the potential difference across the circuit load, given the potential difference across the

battery’s terminals.

(5)

Chapter

18

Schematic Diagrams

• A schematic diagram is a representation of a circuit that uses lines to represent wires and

different symbols to represent components.

• Some symbols used in schematic diagrams are shown at right.

(6)

Chapter

18

Schematic Diagram and Common Symbols

(7)

Chapter

18

Electric Circuits

• An electric circuit is a set of electrical components connected such that they provide one or more

complete paths for the movement of charges.

• A schematic diagram for a circuit is sometimes called a circuit diagram.

• Any element or group of elements in a circuit that dissipates energy is called a load.

(8)

Chapter

18

Electric Circuits, continued

• A circuit which contains a complete path for electrons to follow is called a closed circuit.

• Without a complete path, there is no charge flow and therefore no current. This situation is called an open circuit.

• A short circuit is a closed circuit that does not contain a load. Short circuits can be hazardous.

(9)

Chapter

18

• The source of potential difference and electrical energy is the circuits emf.

• Any device that transforms nonelectrical energy into electrical energy, such as a battery or a generator, is a source of emf.

• If the internal resistance of a battery is neglected, the emf equals the potential difference across the

source’s two terminals.

(10)

Chapter

18

Internal Resistance, emf, and Terminal Voltage

(11)

Chapter

18

• The terminal voltage is the potential difference across a battery’s positive and negative terminals.

• For conventional current, the terminal voltage is less than the emf.

• The potential difference across a load equals the terminal voltage.

(12)

Chapter

18

Light Bulb

(13)

Section 2 Resistors in Series or in Parallel

Chapter

18

Objectives

• Calculate the equivalent resistance for a circuit of resistors in series, and find the current in and

potential difference across each resistor in the circuit.

• Calculate the equivalent resistance for a circuit of resistors in parallel, and find the current in and

potential difference across each resistor in the circuit.

(14)

Chapter

18

Resistors in Series

• A series circuit describes two or more components of a circuit that provide a single path for current.

• Resistors in series carry the same current.

• The equivalent resistance can be used to find the current in a circuit.

• The equivalent resistance in a series circuit is the sum of the circuit’s resistances.

R_eq = R₁ + R₂ + R₃…

(15)

Chapter

18

Resistors in Series

(16)

Chapter

18

Resistors in Series, continued

• Two or more resistors in the actual circuit have the same effect on the current as one equivalent resistor.

• The total current in a series circuit equals the

potential difference divided by the equivalent

resistance. _



eq

I V

R

(17)

Chapter

18

Sample Problem

Resistors in Series A 9.0 V battery is

connected to four light bulbs, as shown at right.

Find the equivalent

resistance for the circuit and the current in the circuit.

(18)

Chapter

18

Sample Problem, continued

Resistors in Series 1. Define

Given:

∆V = 9.0 V R₁ = 2.0 Ω R₂ = 4.0 Ω R₃ = 5.0 Ω R₄ = 7.0 Ω

Unknown:

R_eq = ? I = ? Diagram:

(19)

Chapter

18

Resistors in Series 2. Plan

Choose an equation or situation: Because the

resistors are connected end to end, they are in series.

Thus, the equivalent resistance can be calculated with the equation for resistors in series.

R_eq = R₁ + R₂ + R₃…

The following equation can be used to calculate the current.

∆V = IR_eq

(20)

Chapter

18

Resistors in Series 2. Plan, continued

Rearrange the equation to isolate the unknown:

No rearrangement is necessary to calculate Req, but

∆V = IR_eq must be rearranged to calculate the current.

 

eq

I V

R

(21)

Chapter

18

Resistors in Series 3. Calculate

Substitute the values into the equation and solve:

   9.0 V 

0.50 A 18.0 Ω

eq

I V

R

Substitute the equivalent resistance value into the equation for current.

eq eq

R = 2.0 Ω + 4.0 Ω + 5.0 Ω + 7.0 Ω R = 18.0 Ω

(22)

Chapter

18

Resistors in Series 4. Evaluate

For resistors connected in series, the equivalent resistance should be greater than the largest

resistance in the circuit.

18.0 Ω > 7.0 Ω

(23)

Chapter

18

Resistors in Series, continued

• Series circuits require all elements to conduct electricity

• As seen below, a burned out filament in a string of

bulbs has the same effect as an open switch. Because the circuit is no longer complete, there is no current.

(24)

Chapter

18

Comparing Resistors in Series and in Parallel

(25)

Chapter

18

Resistors in Parallel

• A parallel arrangement describes two or more components of a circuit that provide separate conducting paths for current because the components are

connected across common points or junctions

• Lights wired in parallel have more than one path for current. Parallel circuits do not require all elements to conduct.

(26)

Chapter

18

Resistors in Parallel

(27)

Chapter

18

Resistors in Parallel, continued

• Resistors in parallel have the same potential differences across them.

• The sum of currents in parallel resistors equals the total current.

• The equivalent resistance of resistors in parallel can be calculated using a reciprocal relationship

1

R_eq  1

R₁  1

R₂  1 R₃ ...

(28)

Chapter

18

Sample Problem

Resistors in Parallel A 9.0 V battery is connected to four

resistors, as shown at right. Find the

equivalent resistance for the circuit and the total current in the circuit.

(29)

Chapter

18

Resistors in Parallel 1. Define

Given:

∆V = 9.0 V R₁ = 2.0 Ω R₂ = 4.0 Ω R₃ = 5.0 Ω R₄ = 7.0 Ω

Unknown:

R_eq = ? I = ? Diagram:

(30)

Chapter

18

Resistors in Parallel 2. Plan

Choose an equation or situation: Because both sides of each resistor are connected to common points, they are in parallel. Thus, the equivalent resistance can be calculated with the equation for resistors in parallel.

The following equation can be used to calculate the current. ∆V = IR_eq

1

R_eq  1

R₁  1

R₂  1 R₃ ...

(31)

Chapter

18

Resistors in Parallel 2. Plan, continued

Rearrange the equation to isolate the unknown:

No rearrangement is necessary to calculate Req; rearrange ∆V = IR_eq to calculate the total current delivered by the battery.

 

eq

I V

R

(32)

Chapter

18

Resistors in Parallel 3. Calculate

Substitute the values into the equation and solve:



eq

1 1 1 1 1

= + + +

R 2.0 Ω 4.0 Ω 5.0 Ω 7.0 Ω

1 0.50 0.25 0.20 0.14 1.09

= + + +

RΩ Ω Ω Ω Ω

R = 1 Ω = 0.917 Ω 1.09

(33)

Chapter

18

Resistors in Parallel 3. Calculate, continued

Substitute the equivalent resistance value into the equation for current.

  



9.0 V 0.917 Ω 9.8 A

eq

I V

R I

(34)

Chapter

18

Resistors in Parallel 4. Evaluate

For resistors connected in parallel, the equivalent resistance should be less than the smallest

resistance in the circuit.

0.917 Ω < 2.0 Ω

(35)

Chapter

18

Resistors in Series or in Parallel

(36)

Section 3 Complex Resistor Combinations

Chapter

18

Objectives

• Calculate the equivalent resistance for a complex circuit involving both series and parallel portions.

• Calculate the current in and potential difference across individual elements within a complex circuit.

(37)

Chapter

18

Resistors Combined Both in Parallel and in Series

• Many complex circuits can be understood by isolating segments that are in series or in parallel and

simplifying them to their equivalent resistances.

• Work backward to find the current in and potential difference across a part of a circuit.

(38)

Chapter

18

Analysis of Complex Circuits

(39)

Chapter

18

Sample Problem

Equivalent Resistance

Determine the equivalent resistance of the complex circuit shown below.

(40)

Chapter

18

Equivalent Resistance Reasoning

The best approach is to divide the circuit into groups of series and parallel resistors. This way, the

methods presented in Sample Problems A and B can be used to calculate the equivalent resistance for

each group.

(41)

Chapter

18

1. Redraw the circuit as a group of resistors along one side of the circuit.

Because bends in a wire do not affect the circuit, they do not need to be represented in a schematic

diagram. Redraw the circuit without the corners,

keeping the arrangement of the circuit elements the same.

TIP: For now, disregard the emf source, and work only with the

resistances.

(42)

Chapter

18

Equivalent Resistance 2. Identify components

in series, and calcu- late their equivalent resistance.

Resistors in group (a) and (b) are in series.

For group (a):

R_eq = 3.0 Ω + 6.0 Ω = 9.0 Ω For group (b):

R_eq = 6.0 Ω + 2.0 Ω = 8.0 Ω

(43)

Chapter

18

3. Identify components in parallel, and calculate their equivalent resis- tance.

Resistors in group (c) are in parallel.

    



1 1 1 0.12 0.25 0.37

8.0Ω 4.0Ω 1Ω 1Ω 1Ω

2.7 Ω

eq eq

R R

(44)

Chapter

18

4. Repeat steps 2 and 3 until the resistors in the circuit are reduced to a single equivalent resistance.The remainder of the resistors, group (d), are in series.

  



9.0Ω 2.7Ω For group (d

1.0Ω ):

12.7Ω

eq eq

R R

(45)

Chapter

18

Sample Problem

Current in and Potential Difference Across a Resistor Determine the current in and potential difference

across the 2.0 Ω resistor highlighted in the figure below.

(46)

Chapter

18

Current in and Potential Difference Across a Resistor Reasoning

First determine the total circuit current by reducing the resistors to a single equivalent resistance. Then rebuild the circuit in steps, calculating the current and potential difference for the equivalent resistance of each group until the current in and potential difference across the 2.0 Ω resistor are known.

(47)

Chapter

18

Current in and Potential Difference Across a Resistor 1. Determine the equivalent resistance of the circuit.

The equivalent resistance of the circuit is 12.7 Ω, as calculated in the previous Sample Problem.

(48)

Chapter

18

Current in and Potential Difference Across a Resistor 2. Calculate the total current in the circuit.

Substitute the potential difference and equivalent

resistance in ∆V = IR, and rearrange the equation to find the current delivered by the battery.

   9.0 V 

0.71 A 12.7 Ω

eq

I V

R

(49)

Chapter

18

3. Determine a path from the

equivalent resistance found in step 1 to the 2.0 Ω resistor.

Review the path taken to find the equivalent resistance in the figure at right, and work backward through this path. The equivalent resistance for the entire circuit is the same as the equivalent resistance for group (d).

The center resistor in group (d) in turn is the equivalent resistance for group (c). The top resistor in group (c) is the equivalent resistance for group (b), and the right resistor in group (b) is the 2.0 Ω resistor.

(50)

Chapter

18

Current in and Potential Difference Across a Resistor 4. Follow the path determined in step 3, and calculate

the current in and potential difference across each equivalent resistance. Repeat this process until the desired values are found.

(51)

Chapter

18

4. A. Regroup, evaluate, and calculate.

Replace the circuit’s equivalent resistance with group (d). The resistors in group (d) are in series; therefore, the current in each resistor is the same as the current in the equivalent resistance, which equals 0.71 A.

The potential difference across the 2.7 Ω resistor in group (d) can be calculated using ∆V = IR.

Given: I = 0.71 A R = 2.7 Ω Unknown: ∆V = ?

∆V = IR = (0.71 A)(2.7 Ω) = 1.9 V

(52)

Chapter

18

4. B. Regroup, evaluate, and calculate.

Replace the center resistor with group (c).

The resistors in group (c) are in parallel; therefore, the potential difference across each resistor is the same as the potential difference across the 2.7 Ω equivalent resistance, which equals 1.9 V. The current in the 8.0 Ω resistor in group (c) can be calculated using ∆V = IR.

Given: ∆V = 1.9 V R = 8.0 Ω Unknown: I = ?

   1.9 V 

0.24 A 8.0 Ω

I V

R

(53)

Chapter

18

4. C. Regroup, evaluate, and calculate.

Replace the 8.0 Ω resistor with group (b).

The resistors in group (b) are in series; therefore, the current in each resistor is the same as the current in the 8.0 Ω equivalent resistance, which equals 0.24 A.

The potential difference

across the 2.0 Ω resistor can be calculated using ∆V = IR.

Given: I = 0.24 A R = 2.0 Ω Unknown: ∆V = ?

  

 

(0.24 A)(2.0 Ω) 0.48 V

V IR V

I  0.24 A

(54)

Multiple Choice

1. Which of the following is the correct term for a circuit that does not have a closed-loop path for electron flow?

A. closed circuit B. dead circuit C. open circuit D. short circuit

Chapter

18

(55)

Multiple Choice, continued

1. Which of the following is the correct term for a circuit that does not have a closed-loop path for electron flow?

A. closed circuit B. dead circuit C. open circuit D. short circuit

Chapter

18

(56)

2. Which of the following is the correct term for a circuit in which the load has been unintentionally bypassed?

F. closed circuit G. dead circuit H. open circuit J. short circuit

Chapter

18

(57)

2. Which of the following is the correct term for a circuit in which the load has been unintentionally bypassed?

F. closed circuit G. dead circuit H. open circuit J. short circuit

Chapter

18

(58)

Use the diagram below to answer questions 3–5.

Chapter

18

3. Which of the circuit elements contribute to the load of the circuit?

A. Only A

B. A and B, but not C C. Only C

D. A, B, and C

(59)

Chapter

18

3. Which of the circuit elements contribute to the load of the circuit?

A. Only A

B. A and B, but not C C. Only C

D. A, B, and C

(60)

Chapter

18

4. Which of the following is the correct equation for the equivalent resis-

tance of the circuit?

 

 

  

.

1 1 1

. .

1 1 1 1

.

eq A B

eq

eq A B C

R R R

R I V

R R R R

F G H

J

(61)

Chapter

18

 

 

 



1 1 1

. . .

1 1 1 1

.

e

eq A B

eq

q A B

eq A B C

R R R

R I V

R R

R R R

R R

G F

H J

(62)

Chapter

18

5. Which of the following is the correct equation for the current in the

resistor?

  

 

 

 

A B C

B

eq

B total A

B

I I I I I V

R

I I I I V

R A.

B.

C.

D.

(63)

Chapter

18

5. Which of the following is the correct equation for the current in the

resistor?







 





A B C

B B

eq

total A

B

I V

R

I I I I

I I I I V

R A.

C.

D.

B.

(64)

Chapter

18

  

 

 

    

 

–1

.

1 1 1 1

. .

1 1

.

eq A B C

eq

eq A

B C

R R R R

R I V

R R

F G H

J

(65)

Chapter

18



  



    

 



1 1 –1

. .

1 1 1 1

. .

eq A B C

B C

eq

eq A

R R

R

R R R R

R I V

R F

G H

J

(66)

Chapter

18

7. Which of the following is the correct equation for the current in resistor B?

  

 

 

 

A B C

B

eq

B total A

B B

B

I I I I I V

R

I I I I V

R A.

B.

C.

D.

(67)

Chapter

18

7. Which of the following is the correct equation for the current in resistor B?

  

 

 

 

A B C

B

eq B total

B B

B

A

I I I I

I V

R I V

R

I I I A.

B.

C.

D.

(68)

8. Three 2.0 Ω resistors are connected in series to a 12 V battery. What is the potential difference across

each resistor?

F. 2.0 V G. 4.0 V H. 12 V J. 36 V

Chapter

18

(69)

8. Three 2.0 Ω resistors are connected in series to a 12 V battery. What is the potential difference across

each resistor?

F. 2.0 V G. 4.0 V H. 12 V J. 36 V

Chapter

18

(70)

Use the following passage to answer questions 9–11.

Six light bulbs are

connected in parallel to a 9.0 V battery. Each bulb has a resistance of 3.0 Ω.

Chapter

18

9. What is the potential difference across each bulb?

A. 1.5 V B. 3.0 V C. 9.0 V D. 27 V

(71)

Six light bulbs are

Chapter

18

9. What is the potential difference across each bulb?

A. 1.5 V B. 3.0 V C. 9.0 V D. 27 V

(72)

Six light bulbs are

Chapter

18

10. What is the current in each bulb?

F. 0.5 A G. 3.0 A H. 4.5 A J. 18 A

(73)

Six light bulbs are

Chapter

18

10. What is the current in each bulb?

F. 0.5 A G. 3.0 A H. 4.5 A J. 18 A

(74)

Six light bulbs are

Chapter

18

11. What is the total current in the circuit?

A. 0.5 A B. 3.0 A C. 4.5 A D. 18 A

(75)

Six light bulbs are

Chapter

18

11. What is the total current in the circuit?

A. 0.5 A B. 3.0 A C. 4.5 A D. 18 A

(76)

Short Response

12. Which is greater, a battery’s terminal voltage or the same battery’s emf? Explain why these two quantities are not equal.

Chapter

18

(77)

Short Response, continued

12. Which is greater, a battery’s terminal voltage or the same battery’s emf? Explain why these two quantities are not equal.

Answer: A battery’s emf is slightly greater than its terminal voltage. The difference is due to the battery’s internal resistance.

Chapter

18

(78)

13. Describe how a short circuit could lead to a fire.

Chapter

18

(79)

13. Describe how a short circuit could lead to a fire.

Answer: In a short circuit, the equivalent resistance of the circuit drops very low, causing the current to be very high. The higher current can cause wires still in the circuit to overheat, which may in turn cause a fire in materials contacting the wires.

Chapter

18

(80)

14. Explain the advantage of wiring the bulbs in a string of decorative lights in parallel rather than in series.

Chapter

18

(81)

14. Explain the advantage of wiring the bulbs in a string of decorative lights in parallel rather than in series.

Answer: If one bulb is removed, the other bulbs will still carry current.

Chapter

18

(82)

Extended Response

15. Using standard symbols for circuit elements, draw a diagram of a circuit that contains a battery, an open switch, and a light bulb in parallel with a resistor. Add an arrow to indicate the direction of current if the

switch were closed.

Chapter

18

(83)

Extended Response, continued

15. Using standard symbols for circuit elements, draw a diagram of a circuit that contains a battery, an open switch, and a light bulb in parallel with a resistor. Add an arrow to indicate the direction of current if the

switch were closed.

Answer:

Chapter

18

(84)

Chapter

18

16. For the circuit shown, calculate the following:

a. the equivalent

resistance of the circuit b. the current in the light

bulb.

Show all your work for both calculations.

(85)

Chapter

18

16. For the circuit shown, calculate the following:

a. the equivalent

resistance of the circuit b. the current in the light

bulb.

Show all your work for both calculations.

Answer: a. 4.2 Ω b. 2.9 A

(86)

Chapter

18

17. After a period of time, the 6.0 Ω resistor fails and breaks. Describe what happens to the brightness of the bulb.

Support your answer.

(87)

Chapter

18

17. Answer: The bulb will grow dim. The loss of the 6.0 Ω resistor

causes the equivalent resistance of the circuit to increase to 4.5 Ω. As a result, the current in the bulb drops to 2.7 A, and the brightness of the bulb decreases.

(88)

Chapter

18

18. Find the current in and potential difference across each of the resistors in the following circuits:

a. a 4.0 Ω and a 12.0 Ω resistor wired in series with a 4.0 V source.

b. a 4.0 Ω and a 12.0 Ω resistor wired in parallel with a 4.0 V source.

Show all your work for each calculation.

(89)

a. a 4.0 Ω and a 12.0 Ω resistor wired in series with a 4.0 V source.

b. a 4.0 Ω and a 12.0 Ω resistor wired in parallel with a 4.0 V source.

Answers: a. 4.0 Ω: 0.25 A, 1.0 V 12.0 Ω: 0.25 A, 3.0 V

b. 4.0 Ω: 1.0 A, 4.0 V 12.0 Ω: 0.33 A, 4.0 V

Chapter

18

(90)

Chapter

18

a. a 150 Ω and a 180 Ω resistor wired in series with a 12 V source.

b. a 150 Ω and a 180 Ω resistor wired in parallel with a 12 V source.

(91)

Chapter

18

Answer: a.150 Ω: 0.036 A, 5.4 V 180 Ω: 0.036 A, 6.5 V

b. 150 Ω: 0.080 A, 12 V 180 Ω: 0.067 A, 12 V

a. a 150 Ω and a 180 Ω resistor wired in series with a 12 V source.

b. a 150 Ω and a 180 Ω resistor wired in parallel with a 12 V source.

(92)

Diagram Symbols

Chapter