693KA8-26-R-00004-Attachment J.1-Sample Problems and Formulas.pdf

Tutoring Support Addendum The following are formulas and example practice problems from Math Refresher, Building Environmental Concepts, and Antennas & Transmission Lines (Basic and Advanced). Note that students will be using the Texas Instruments TI-36X or Casio FX-115ES Plus scientific calculator so tutors should be able to assist with specific calculator operations regarding these models.

Table of Contents (using PDF page numbers)

47027001 Mathematics Refresher Formulas

47027001 Mathematics Refresher Practice Problems

45517001 Building Environmental Concepts Block 1 Formulas

45517001 Building Environmental Concepts Block 2 Formulas

45517001 Building Environmental Concepts Block 3 Formulas

45517001 Building Environmental Concepts Block 1 Practice Problems

45262001 Antennas & Transmission Lines Formulas

45262001 Antennas & Transmission Lines Module 1 Practice Problems

45262001 Antennas & Transmission Lines Module 2 Practice Problems

Attachment J.1-Sample Problems and Formulas

693KA8-26-R-00004

MATHEMATICS REFRESHER

Appendix A: Formulas Version: V.2019-04

FAA47027001

PAGE INTENTIONALLY LEFT BLANK

FAA47027001-AppA / V.2019-04 Page 1

APPENDIX A: MATHEMATICS REFRESHER

FORMULAS

Formula 2-1 c2 = a2+ b2

Formula 2-2 sin = hyp opp opp csc

Formula 2-3 cos = adj adj sec

Formula 2-4 tan = adj opp ; cot = opp adj

Formula 2-5 cos

R

Z

Formula 2-6 cm ssm f E

CosE S

Formula 2-7 Zs = jZ0 tan βs

Formula 2-8 tan =

1S

Formula 3-1 f = T

Formula 3-2 e = EM sin

Formula 3-3 e = EM sin ωt ; e = EM sin 2πf (t) ; e = EM sin 360(f) (t)

Formula 3-4 A = 22 ba tan-1 a b ; A = A cos + j A sin

Formula 4-1

P

P log10dB

Formula 4-2 out in

E

E log20dB

Formula 4-3 Z0 = d

D log r

FAA47027001-AppA / V.2019-04 Page 2

APPENDIX B

Mathematics Refresher Practice Problems

Version: V.2019-04

FAA47027001

FAA47027001-AppB / V.2019-04 Page 1

MATHEMATICS REFRESHER PRACTICE PROBLEMS

LESSON 1. APPLIED ALGEBRA PRACTICE

PROBLEMS

1. Solve:

)2x(

)4x(

8x6x

4x a.

)4x)(2x(

)4x(10 b.

24x3

c. )4x)(2x(

40x6 d.

40x10

2. Simplify:

}b5c10]c4)b5a3(a4ba15[ca10{a3 2

a. bc7a2a3 2

b. b5c10c4b5a3a4ab15ac10a3 2

c. bac10a3 2

d. b5ac3 2

Continued on next page

PROBLEMS (Continued)

FAA47027001-AppB / V.2019-04 Page 2

3. Solve:

)c7aca3)(ca( 22

a. 3223 c7ac8ac4a3

b. 3223 c7ac8ca4a3

c. 3223 c7ac8ca4a3

d. 322 c7ac8ac4a3

4. Solve:

8x4

4x4x

)2x(x4 x4

16x16x4

)2x(4

b. ) )2x)(2x(

)2x(x4

c. 1

2x

2x4 2

FAA47027001-AppB / V.2019-04 Page 3

5. Simplify:

)y3x3(x5)y3x2(x4

a. xy15x15xy12x 22

b. xy3x7 2

c. xy3x 2

d. xy3x7 2

6. Simplify:

c.

FAA47027001-AppB / V.2019-04 Page 4

7. Simplify:

c. 12

d. 4

8. Simplify:

a. 12

b. 11 c.

FAA47027001-AppB / V.2019-04 Page 5

9. Find (x):

x

a. 4

b. 3

c. 12

d. 15

10. Solve for ZL:

OL

OL

ZZ

ZZ

a. ܼ ൌ ೀሺଵା௰ሻ

ଵି௰

b. ܼ ൌ ሺܼ߁ ܼைሻ

c. ܼ ൌ ܼ߁ ைܼ߁

d. ܼ ൌ ைܼ߁

FAA47027001-AppB / V.2019-04 Page 6

11. Simplify:

x4 a6w3 a5w a5w w2a11 c.

a6w3

d. x+4

FAA47027001-AppB / V.2019-04 Page 7

LESSON 2. APPLIED TRIGONOMETRY PRACTICE

PROBLEMS

12.

Find Hypotenuse, given θ = 38°, Adj. = 15.

a. 20.05

b. 19.04

c. 18.18

d. 24.01

13.

Find Adjacent and Opposite, given θ = 62°, Hypotenuse = 35.

a. Adj = 15.1; Opp = 29.1

b. Adj = 16.43; Opp = 30.9

c. Adj = 12.33; Opp = 21.1

d. Adj = 12.2; Opp = 31.1

Continued on next page

FAA47027001-AppB / V.2019-04 Page 8

14.

Find sine, cosine, and tangent functions for θ.

Given Adj. = 10, Opp. = 6

a. Sin = .413; Cos = .858; Tan = .393

b. Sin = .413; Cos = .959; Tan = .6

c. Sin = .515; Cos = .858; Tan = .6

d. Sin = .515; Cos = .959; Tan = .393

15.

Find θ and Adjacent, given Hypotenuse = 1320´ and Opposite = 14.

a. θ = .806°; Adj = 1913.93´

b. θ = .608°; Adj = 1319.93´

c. θ = .806°; Adj = 1319.93´

d. θ = .608°; Adj = 1913.93´

FAA47027001-AppB / V.2019-04 Page 9

16.

Find h, given θ = 2.25°, Adj. = 570´

a. Opp = 24.4´

b. Opp = 26.4´

c. Opp = 22.4´

d. Opp = 18.4´

17. Find the input impedance (Zs) to a shorted transmission line; given Z0 = 50Ω and line length of 65°.

a. Zs = j103.23Ω

b. Zs = j109.23Ω

c. Zs = j107.23Ω

d. Zs = j101.23Ω

Continued on next page

FAA47027001-AppB / V.2019-04 Page 10

18. Find βs if Zs = +j252Ω and Z0 = 75Ω for a shorted transmission line.

a. βs = 33.43°

b. βs = 53.43°

c. βs = 73.43°

d. βs = 93.43°

19. Find the shortest length of transmission line that could be used as a stub.

Given Z0 = 100Ω with SWR of 4:1

How long is the stub in electrical degrees?

a. = 65.31°

b. = 61.31°

c. = 59.31°

d. = 56.31

FAA47027001-AppB / V.2019-04 Page 11

LESSON 3. PERIODIC WAVES AND PHASOR

ANALYSIS PRACTICE PROBLEMS

20. Given a frequency of 113 MHz, find the period of the wave.

a. 9.95 nanoseconds

b. 8.00 nanoseconds

c. 7.55 nanoseconds

d. 9.00 nanoseconds

21. The term “ωt” is equal to:

a. ωt = πr²

b. ωt = 2πr

c. ωt = (2πf)(t)

d. ωt = (180)(f)(t)

22. Find Em if “e” is 27.5 volts and the frequency is 800kHz and the time .7 usec.

a. Em = 74.7 volts

b. Em = -74.7 volts

c. Em = 57.2 volts

d. Em = -57.2 volts

ANALYSIS PRACTICE PROBLEMS (Continued)

FAA47027001-AppB / V.2019-04 Page 12

23. Multiply [1 - j5.196] * [3 - j1.732]

a. -5.99 – j17.29

b. 5.99 + j17.29

c. 18.30 – j109.11

d. -18.30 + j109.11

24. Divide 205 + j36 by 105 + j15

a. 1.96 – j1.83

b. -1.96 + j1.83

c. 1.96 + j.063

d. -1.96 + j.063

ANALYSIS PRACTICE PROBLEMS (Continued)

FAA47027001-AppB / V.2019-04 Page 13

25. Add vectors 30 45° + 100 70° + 50 -30°

a. 98.71 42.41°

b. 133.70 42.41°

c. 133.70 90.17°

d. 98.69 91.17°

FAA47027001-AppB / V.2019-04 Page 14

LESSON 4. APPLIED LOGARITHMS AND DB

PRACTICE PROBLEMS

26. Find the Gain in an amplifier.

a. 15.55 dB

b. 11.76 dB

c. -11.66 dB

d. -15.55 dB

27. Find the input power to an amplifier.

a. Pin = 15 watts

b. Pin = 3.16 watts

c. Pin = 10.33 watts

d. Pin = 11.66 watts

PRACTICE PROBLEMS (Continued)

FAA47027001-AppB / V.2019-04 Page 15

28. Find little “d” if big “D” = 1.625 and Z0 is 50Ω with a dielectric material of polyethylene, where r is 2.25.

Z0 d

D log r

a. d = .955 inches

b. d = .599 inches

c. d = .465 inches

d. d = .564 inches

29. Given a transmission line that is 75 feet long, the attenuation for this type of line is

3.5 dB/100feet. The input power is 100 watts. How much power arrives at the antenna?

a. Pout = 59.64 watts

b. Pout = 44.67 watts

c. Pout = 45.64 watts

d. Pout = 54.64 watts

FAA47027001-AppB / V.2019-04 Page 16

FAA47027001-AppB / V.2019-04 Page 17

LESSON 1. APPLIED ALGEBRA ANSWERS

1. b:

)2x(

)4x(

8x6x

)4x(10

)4x)(2x(

)2x(6

24x3

)4x)(2x(

40x1012x64x

)4x(10)2x(64x

2. a

3ܽଶ ሼ10ܽ ܿ െ ሾ15ܽ ܾ െ 4ܽ െ ሺ3ܽ 5ܾሻ 4ܿሿ 10ܿ െ 5ܾሽ 3ܽଶ ሼ10ܽ ܿ െ ሾ15ܽ ܾ െ 4ܽ െ 3ܽ െ 5ܾ 4ܿሿ 10ܿ െ 5ܾሽ 3ܽଶ ሼ10ܽ ܿ െ 15ܽ െ ܾ 4ܽ 3ܽ 5ܾ െ 4ܿ 10ܿ െ 5ܾሽ 3ܽଶ 10ܽ ܿ െ 15ܽ െ ܾ 4ܽ 3ܽ 5ܾ െ 4ܿ 10ܿ െ 5ܾ 3ܽଶ 2ܽ 7ܿ െ ܾ

LESSON 1. APPLIED ALGEBRA ANSWERS (Continued)

FAA47027001-AppB / V.2019-04 Page 18

3. c

ሺܽ ܿሻሺ3ܽଶ ܽܿ 7ܿଶሻ 3ܽଶ ܽܿ 7ܿଶ

3ܽଶܿ ܽܿଶ 7ܿଷ

3ܽଷ ܽଶܿ 7ܽܿଶ

3ܽଷ 4ܽଶܿ 8ܽܿଶ 7ܿଷ

4. c

ቆ ଶݔ4 ݔ16 16

ݔ4 ቇ ൬

ݔሺݔ4 2ሻ ଶݔ ݔ4 4

൰ ൬

ݔ4 8 ൰

ቆ 4ሺݔଶ ݔ4 4ሻ

ݔ4 ቇ ൬

ݔሺݔ4 2ሻ ሺݔ 2ሻሺݔ 2ሻ

൰ ൬

4ሺݔ 2ሻ ൰

൬ 4ሺݔ 2ሻሺݔ 2ሻ

ݔ4 ൰ ൬

ݔሺݔ4 2ሻ ሺݔ 2ሻሺݔ 2ሻ

൰ ൬

4ሺݔ 2ሻ ൰

ൌ 1

5. b

െ7ݔଶ ݕݔ3

FAA47027001-AppB / V.2019-04 Page 19

6. a

7. c

ൊ ൬

൰

ൌ 12

8. b

൬

൰ ൊ ൬

൰

ൌ ൬

൰ ൊ ൬

൰

ൌ 11

FAA47027001-AppB / V.2019-04 Page 20

9. c

ݔ

ሺ18ሻ ൌ

ݔ

ሺ18ሻ

ൌ ݔ

ݔ ൌ

ݔ ൌ 12

10. a

߁ ൌ

ሺܼ߁ ܼைሻ ൌ

ሺܼ ܼைሻ

ሺܼ߁ ܼைሻ ൌ ܼ െ ܼை ܼ߁ ைܼ߁ ൌ ܼ െ ܼை ܼை ܼ߁ ைܼ߁ ൌ ܼ െ ܼை ܼை ܼை ܼ߁ ைܼ߁ ൌ ܼ

ܼሺ1 െ ሻ߁ ሺ1 െ ሻ߁

ܼைሺ1 ሻ߁ ሺ1 െ ሻ߁

ܼைሺ1 ሻ߁ ሺ1 െ ሻ߁

FAA47027001-AppB / V.2019-04 Page 21

11. b

ݓ 5ܽ ݔ 4

ݓ3 െ 6ܽ െ4 െ ݔ

ݓ3 െ 6ܽሺെ1ሻ െ4 െ ሺെ1ሻݔ

െ3ݓ 6ܽ

െ2ݓ 11ܽ

11ܽ െ ݓ2

FAA47027001-AppB / V.2019-04 Page 22

LESSON 2. APPLIED TRIGONOMETRY ANSWERS

12. b

ߠݏܥ ൌ ݆݀ܣ ݕܪ

°38ݏܥ ൌ

ݕܪ ൌ

°38ݏܥ

. 788

ൌ 19.04

13. b

ߠݏܥ ൌ

°62ݏܥ ൌ

݆݀ܣ ൌ 35ݔ°62ݏܥ ൌ 16.43

ߠ݊݅ܵ ൌ ݕܪ

ܵ݅݊62° ൌ

ܱ ൌ 35ݔ62°݊݅ܵ ൌ 30.9

14. c

ܿଶ ൌ ܽଶ ܾଶ; ܿ ൌ ඥܽଶ ܾଶ; ܿ ൌ ඥ10ଶ 6ଶ ܿ ൌ √136ܿ ൌ 11.66 ൌ ݕܪ

ߠ݊݅ܵ ൌ ݕܪ

11.66

ൌ .515

ߠݏܥ ൌ ݆݀ܣ ݕܪ

11.66

ൌ .858

ߠ݊ܽܶ ൌ ݆݀ܣ

ൌ .6

(Continued)

FAA47027001-AppB / V.2019-04 Page 23

15. b

ߠ݊݅ܵ ൌ

ݕܪ

; ߠ݊݅ܵ ൌ

; ߠ݊݅ܵ ൌ .010606

ሺܵ݅݊ିଵሻܵ݅݊ߠ ൌ ሺܵ݅݊ିଵሻ.010606 ߠ ൌ .608°

ߠݏܥ ൌ

.݆݀ܣ ൌ 1319.93′

16. c

ߠ݊ܽܶ ൌ

ܶܽ݊2.25° ൌ

ܱ ⥂ൌ 570ݔ2.25°݊ܽܶ ܱ ൌ 22. 4′

(Continued)

FAA47027001-AppB / V.2019-04 Page 24

17. c

Zs = j Zo tan βs

Zs = j50 tan 65°

Zs = j107.23Ω

18. c.

Zs = j Zo tan βs j252Ω = j75Ω tan βs j j = tan βs

3.36 = tan βs

)(tan 1 3.36 = )(tan 1 tan βs βs = 73.43°

19. d tan 1 tan 1

= 5.1)(tan 1

= 56.31°

FAA47027001-AppB / V.2019-04 Page 25

LESSON 3. PERIODIC WAVES AND PHASER

ANALYSIS ANSWERS

20. d

113000000

ܶ ൌ ݏ݀݊ܿ݁ݏ ݊ܽ݊ 9.00

21. c ωt = (2πf)(t) or (360)(f)(t)

22. b

)]t)(f(360sin[ e Em

)]10x7)(.10x8(.360sin[

5.27 E

66m

368.

5.27 E m

volts70.74E m

LESSON 3. PERIODIC WAVES AND PHASER

ANALYSIS ANSWERS (Continued)

FAA47027001-AppB / V.2019-04 Page 26

23. a

1 - j5.196 = 5.29 -79.1°

3 - j1.732 = 3.46 -30°

(5.29)(3.46) = 18.30

-79.1° + -30° = -109.11°

18.30 -109.11° convert to rectangular -5.99 – j17.29

24. c

205 + j36 = 208.1 9.96°

105 + j15 = 106.07 8.13°

(208.1) ÷ (106.07) = 1.96

9.96° - 8.13° = 1.83°

1.96 1.83° convert to rectangular 1.96 + j.063

25. b

3045° = 21.21 + j21.21

10070° = 34.20 + j93.97

50 -30° = 43.30 - j25

98.71 + j90.18 convert to polar 133.7042.41°

FAA47027001-AppB / V.2019-04 Page 27

ANSWERS

26. b dB ൌ 10 ݈݃ ଵܲ

ଶܲ dB ൌ 10 ݈݃ dB ൌ 10ሺ1.176ሻ dB ൌ 11.76

27. b

15dB ൌ 10 ݈݃

10 ݈݃

1.5 ൌ ݈݃

ሺܽ݊݅ݐ ሻ݈݃ 1.5 ൌ ݈݃

ሺܽ݊݅ݐ ሻ݈݃

31.62 ൌ

31.62

ܲ ൌ ݏݐݐܽݓ3.16

ANSWERS (Continued)

FAA47027001-AppB / V.2019-04 Page 28

28. c

50 ൌ

√2.25

1.625

50 ൌ 92 ݈݃

92 ݈݃

1.625

. 54348 ൌ ݈݃

ሺܽ݊݅ݐ ሻ݈݃ . 54348 ൌ ݈݃

ሺܽ݊݅ݐ ሻ݈݃

3.495 ൌ

3.495

݀ ൌ .465 inches

29. d

2.625 ൌ 10 ݈݃

ܲ௨௧

2.625

10 ݈݃

. 2625 ൌ ݈݃

ሺܽ݊݅ݐ ሻ݈݃ . 2625 ൌ ݈݃

ܲ௨௧ ሺܽ݊݅ݐ ሻ݈݃

1.83 ൌ

ܲ௨௧ ൌ

1.83

ܲ௨௧ ൌ ݏݐݐܽݓ54.64

FAA47027001-AppB / V.2019-04 Page 29

NOTES

Course 45517001 1

June 2019 (Rev. 2)

Block 1 Formula Sheet

Ohms Law Formulas and Power Formulas

Series:

...321 IIIIT

...321 EEEET

...321 RRRRT

...321 PPPPT

Parallel:

...321 EEEET

...321 IIIIT

T

T

TT

I

E R

RR

RR

RRR

RT

...321 PPPPT

Bridge:

R

R

Wheatstone Bridge:

X

R

R

1 a

Inductors:

R

L T

Capacitors:

CxRT

Inductors in Series: same as resistors

Inductors in Parallel: same as resistors

Capacitors in Series: opposite of resistors

Capacitors in Parallel: opposite of resistors

Course 45517001 2

June 2019 (Rev. 2)

1 Horsepower = 746 Watts

Conductance:

𝐺 =

𝑅

Time Constants:

TC#1 = 63.2%

TC#2 = 86.5%

TC#3 = 95.0%

TC#4 = 98.2%

TC#5 = 99.3%

(sec) ohmsR henrysL

CRT

Resistor Color Coding

COLOR

SIGNIFICANT

FIGURE

DECIMAL

MULTIPLIER

TOLERANCE

PERCENT

RELIABILITY

LEVEL PER

1,000 HRS

BLACK 0 1 - -

BROWN 1 10 1 1.0%

RED 2 100 2 0.1%

ORANGE 3 1,000 - 0.01%

YELLOW 4 10,000 - 0.001%

GREEN 5 100,000 - -

BLUE 6 1,000,000 - -

VIOLET 7 10,000,000 - -

GRAY 8 100,000,000 - -

WHITE 9 1,000,000,000 - -

GOLD - .1 5 -

SILVER - .01 10 -

NO COLOR - - 20 -

May 2019 (Rev. 1)

Block 2 Formula Sheet

Formulas Instantaneous Voltage: SinxEe

MAX

VPP: 2xVV

PPP

VP:

PP

P

V V or

707.

RMS

P

V V or 414.1xVV

RMSP

or

636.

avg

P

V

VRMS: 707.xVV

PRMS

Vavg: 636.xVV Pavg

Period (T):

F

Frequency (F):

F

EG Frequency (f):

RPMxP f

EG RPM:

fx

RPM

Formulas Purely resistive:

RIE

Series AC circuit:

321321321 .,., RRRRetcIIIIetcEEEE

TTT

Parallel AC circuits:

T

321T

321T

R

R

R

R

.etc,IIII etc,EEEE

AC Power:

EIP

Applies to Engine Generators

2 Course 45517

Capacitive

General Capacitive Reactance:

FC

X

C

Ohm’s Law:

C

XIE

C

X

E I

I

E X

Series RC

Circuits

321 IIIIT

Applied Voltage: 22 )()(

CR

EEE

Impedance:

A

C I

E ZorXRZ

True Power: IEPor

E PorRIP

Reactive Power:

VARCVARCVAR

X

E PorXIPorEIP

Apparent Power:

)(Re)( ractivepoweTruepowerVA orVARWVAorZIVA

Power Factor:

ampsvolt watts or powerapparent powertrue

PF

A

R

E

E PFor

Z

PF

Phase Angle:

Hyp

Opp or

Adj

Opp or

Hyp

Adj 111 sintancos

Continued on next page.

Course 45517001 3

Parallel RC

.,321 etcEEEET

C

C

R

R X

E Ior

R

I

A

C

I

E Zor

XR

Current:

CRT

III

True Power:

E

C

X

E

VAR

powerreactivepowertrueVAorVARWVA

EIV

TA

ampsvolt watts powerapparent powertrue

PF

Hyp

Adj1cos

4 Course 45517

Inductive

Inductive

Reactance fLX

L

Series RL

Ohm’s Law:

I

E X

X

E

IXIE

L

L

Series Current: etcIIII

I A

Voltage:

LLR

XIERIE

LR

EEEA

L

T

A XRZor I

True Power: WattsEIP

Reactive Power: VARSEIP LR

VAEIP

AA

or

RTA

PPP or 22 VARSWattsP

A

Adj

Opp or

Hyp

Adj or

Hyp

Opp 111 tancossin

Power Factor: cos PFor AmpsVolt

Watts

PowerApparent

PowerTrue

PF

A

R

E

E PFor

Z

R

PF

Continued on next page.

Course 45517001 5

Parallel RL

Voltage:

LRA

EEE

E Z

Z

IZIE

XLRT

III

XRZ

True Power:

E

PRIPEIP RRR

22 werreactivepotruepowerPA

XLXLR

EIP

Phase Shift:

Adj

Opp

Hyp

Adj

Hyp

Opp 111 tancossin werapparentpo truepower PForPF cos

6 Course 45517

RCL

Series RCL

Impedance:

When XL is larger - 22

CL

XXRZ

When XC is larger - 22

LC

XXRZ

CLRT

IIII

Z

E I A

Voltage: RIE RR

LLL

CCC

XIE

XIE

XLXCRT

EEEE

Power:

True power - RRT EIP

Apparent Power -

AA

EIP or 22

CLT PPPPA

Reactive Power -

CR

EIP (cap.) and LR EIP (ind.)

PF or cosPF

R cos or

A

R

E cos

Resonant Frequency:

F

Continued on next page.

Course 45517001 7

Parallel RCL

Circuits Impedance:

CL XXRZ

or

T

A

I

Voltage:

CLRA

EEEE

E

IR

LC

RT

C

C

L

L

IIIIor X

E I

X

Power:

True power - RRT EIP

Apparent Power -

ATA

EIP or 22

CLTA PPPP

Reactive Power -

CCCR

EIP (cap.) and

LLLR

EIP (ind.)

Total Reactive Power -

ACLR

EIIP

R

I

I

H

A 1cos or

T

X

I

I

H

O 1sin or R

X

I

I

A

O 1tan

A

T

P

PF or cosPF

Resonant Frequency:

F

8 Course 45517

Transformers

Formulas Turns Ratio:

P

S

N

Voltage Ratio:

S

P

S

N

N

E

E

P

P

S

S E

N

N E , S s p

P E N

Power Ratio:

SP

PP

Current Ratio:

S

S

P

N

N

I

I

S

P

S

P I

N

N I , P

S

P S I

N

Impedance Ratio:

P

S

P

S

P

Z

Z

N

N or

N

N

ZS

ZP

Impedance matching:

S

P

P N

Z

Efficiency ratio:

powerinput poweroutput eff

Block 3 Formula Sheet mhos

Igm

GS

D =

414.1•= RMSP VV

414.1÷= PRMS VV

( )3.7. ordropvoltagediodeVE PPEAK −=

( )3.7. ordropvoltagediodeEV eakPP +=

LPPEAK REI ÷=

½ wave - ( )318.•= PEAKAVG EE

½ wave - ( )318.÷= AVGPEAK EE

½ wave - ( ) 414.1/7.3.318.0/ orEV AVGRMS +=

Full wave - ( )637.•= PEAKAVG EE

Op-Amp Gain:

Inverting-

IN

F

RA =

Non-Inverting - 1+=

IN

F

RA

Diff Amp - ( ) ( )INF RREEOutput /21 •−=

Practice Problems Block 1, Lesson 3 Resistors and Conductors

3-36 Course 45517001

Figure 3-1 Practice Problem

Practice ANSWERS RT =2,100 Ω == 2.1 kΩ

R1 = BROWN BLACK BROWN

R2 = GRAY BLACK BROWN

R3 = BROWN RED RED

CALCULATIONS

RT

RT = R1 + R2 + R3

RT = 100 Ω + 800 Ω + 1200 Ω

RT = 2100 Ω or 2.1 kΩ

Color Code

R1 = BROWN BLACK BROWN

R2 = GRAY BLACK BROWN

R3 = BROWN RED RED

Course 45517001 3-37

Figure 3-2 Practice Problem 2

RT = 47 kΩ == 47,000 Ω

R1 = BROWN RED RED

R2 = ORANGE BLUE ORANGE

R3 = WHITE GRAY RED

CALCULATIONS

RT

RT = R1 + R2 + R3

RT = 1.2 kΩ + 36 kΩ + 9.8 kΩ = 47 kΩ = 47,000 Ω

Color Code

R1 = BROWN RED RED

R2 = ORANGE BLUE ORANGE

R3 = WHITE GRAY RED

3-38 Course 45517001

Figure 3-2 Practice Problem 3

RT = 1,980 kΩ ==1.98 MΩ == 1,980,000 Ω

R1 = YELLOW GREEN YELLOW

R2 = BLUE VIOLET YELLOW

R3 = GRAY BLUE YELLOW

CALCULATIONS

RT

RT = R1 + R2 + R3

RT = 450 kΩ + 670 kΩ + 860 kΩ = 1980 kΩ = 1.98 MΩ = 1,980,000 Ω

Color Code

R1 = YELLOW GREEN YELLOW

R2 = BLUE VIOLET YELLOW

R3 = GRAY BLUE YELLOW

Course 45517001 3-39

Figure 3-4 Practice Problem 4

RT = 9,420 kΩ == 9.42 MΩ == 9,420,000 Ω

R1 = ORANGE WHITE GREEN

R2 = YELLOW VIOLET GREEN

R3 =GRAY RED YELLOW

CALCULATIONS

RT

Convert all R values to kΩ R1 = 3.9 MΩ == 3,900 kΩ R2 = 4.7 MΩ == 4,700 kΩ R3 = 820 kΩ

RT = R1 + R2 + R3

RT = 3900 kΩ + 4700 kΩ + 820 kΩ = 9,420 kΩ RT = 9,420 kΩ = 9.42 MΩ = 9,420,000 Ω

Color Code

R1 = ORANGE WHITE GREEN

R2 = YELLOW VIOLET GREEN

R3 = GRAY RED YELLOW

3-40 Course 45517001

Figure 3-5 Practice Problem 5

RT = 9,060 Ω == 9.06 kΩ

R1 = RED ORANGE RED

R2 = GREEN GRAY RED

R3 = WHITE BLUE BROWN

R1 = 115 Ω R2 = 290 Ω R3 = 48 Ω

CALCULATIONS

RT

Convert all values to Ω R1 = 2.3 kΩ == 2,300 Ω R2 = 5.8 kΩ == 5,800 Ω R3 = 960 Ω

RT = R1 + R2 + R3

RT = 2,300 Ω + 5,800 Ω + 960 Ω = 9,060 Ω = 9.06 kΩ

Continued on next page.

Course 45517001 3-41

Practice (Cont’d)

Color Code

R1 = RED ORANGE RED

R2 = GREENGRAY RED

R3 = WHITE BLUE BROWN

Tolerance 5% 5% = 0.05 or 5

R1 = 2,300 Ω × 0.05 = 115 Ω R2 = 5,800 Ω × 0.05 = 290 Ω R3 = 960 Ω × 0.05 = 48 Ω

3-42 Course 45517001

Figure 3-6 Practice Problem 6

RT = 2,360 kΩ == 2.36 MΩ == 2,360,000 Ω

R1 = ORANGE WHITE YELLOW

R2 = YELLOW VIOLET YELLOW

R3 = BROWN GREEN GREEN

10% TOLERANCE

R1 = 39,000 Ω == 39 kΩ R2 = 47,000 Ω ==47 kΩ R3 =150,000 Ω = 150 kΩ

TOLERANCE RANGE

R1 = 351 – 429 kΩ R2 = 423 – 517 kΩ R3 = 1,350 – 1,650 kΩ

Course 45517001 3-43

CALCULATIONS

RT

Convert all values to kΩ R1 = 390 kΩ R2 = 0.47 MΩ = 470 kΩ R3 = 1.5 MΩ = 1,500 kΩ

RT = R1 + R2 = R3

RT = 390 kΩ + 470 kΩ + 1,550 kΩ =2,360 kΩ=2.36 MΩ=2,360,000 Ω

Color Code

R1 = ORANGE WHITE YELLOW

R2 = YELLOW VIOLET YELLOW

R3 = BROWN GREEN GREEN

10% Tolerance 10% = 0.10 or 10

R1 = 390,000 Ω × 0.10 = 39,000 Ω = 39 kΩ R2 = 470,000 Ω × 0.10 = 47,000 Ω = 47 kΩ R3 = 1,500,000 Ω × 0.10 = 150,000 Ω = 150 kΩ

Tolerance Range R1 (351 kΩ – 429 kΩ) 𝐋𝐋𝐋𝐋𝐋𝐋𝐋𝐋𝐋𝐋 𝐋𝐋𝐋𝐋𝐋𝐋𝐋𝐋𝐋𝐋 390,000 Ω − 39,000 Ω = 351,000 Ω = 351 kΩ 𝐔𝐔𝐔𝐔𝐔𝐔𝐔𝐔𝐔𝐔 𝐋𝐋𝐋𝐋𝐋𝐋𝐋𝐋𝐋𝐋 390,000 Ω + 39,000 Ω = 429,000 Ω = 429 kΩ

R2 (423 kΩ – 517 kΩ) 𝐋𝐋𝐋𝐋𝐋𝐋𝐋𝐋𝐋𝐋 𝐋𝐋𝐋𝐋𝐋𝐋𝐋𝐋𝐋𝐋 470,000 Ω − 47,000 Ω = 423,000 Ω = 423 kΩ 𝐔𝐔𝐔𝐔𝐔𝐔𝐔𝐔𝐔𝐔 𝐋𝐋𝐋𝐋𝐋𝐋𝐋𝐋𝐋𝐋 470,000 Ω + 47,000 Ω = 517,000 Ω = 517 kΩ

R3 (1,350 kΩ – 1,650 KΩ) 𝐋𝐋𝐋𝐋𝐋𝐋𝐋𝐋𝐋𝐋 𝐋𝐋𝐋𝐋𝐋𝐋𝐋𝐋𝐋𝐋 1,500,000 Ω − 150,000 Ω = 1,350,000 Ω = 1,350 kΩ 𝐔𝐔𝐔𝐔𝐔𝐔𝐔𝐔𝐔𝐔 𝐋𝐋𝐋𝐋𝐋𝐋𝐋𝐋𝐋𝐋 1,500,000 Ω + 150,000 Ω = 1,650,000 Ω = 1,650 kΩ

3-44 Course 45517001

Figure 3-7 Practice Problem 7

RT = 500 Ω

CALCULATIONS

Convert all R values to Ω R1 = 1 kΩ = 1,000 Ω R2 = 1 kΩ = 1,000 Ω

RT =

R1 × R2 R1 + R2

1,000 Ω × 1,000 Ω 1,000 Ω + 1,000 Ω

1,000,000 Ω2

2,000 Ω

Course 45517001 3-45

Figure 3-8 Practice Problem 8

RT = 277 Ω

CALCULATIONS

RT =

R1 × R2 R1 + R2

900 Ω × 400 Ω 900 Ω + 400 Ω

360,000 Ω2

1,300 kΩ

RT = 276.92 Ω ≈ 277 Ω

3-46 Course 45517001

Figure 3-9 Practice Problem 9

RT = 4,421 Ω == 4.42 k Ω

CALCULATIONS

Convert all R values to Ω R1 = 9.1 kΩ = 9,100 Ω R2 = 8.6 kΩ = 8,600 Ω

R1 + R2

9,100 Ω × 8,600 Ω 9,100 Ω + 8,600 Ω

78,260,000 Ω2

17,700 Ω

RT = 4,421.47Ω ≈ 4.42 kΩ

Course 45517001 3-47

Figure 3-10 Practice Problem 10

RT = 24.8 kΩ

CALCULATIONS

R1 + R2

39 kΩ × 68 kΩ 39 kΩ + 68 kΩ

2,652(kΩ)2

107 kΩ

RT = 24.785 kΩ ≈ 24.8 kΩ

3-48 Course 45517001

Figure 3-11 Practice Problem 11

RT = 360 kΩ == 0.360 MΩ == 360,000 Ω

CALCULATIONS

Convert all R values to kΩ R1 = 4.7 MΩ = 4,700 kΩ R2 = 390 kΩ

R1 + R2

4,700 kΩ × 390 kΩ 4,700 kΩ + 390 kΩ

1,833,000(kΩ)2

5,090 kΩ

RT = 360.118 kΩ ≈ 360 kΩ

Course 45517001 3-49

Figure 3-12 Practice Problem 12

RT = 1.79 MΩ == 1,790 kΩ == 1,790,000 Ω

R1 + R2

2.9 MΩ × 4.7 MΩ

2.9 MΩ + 4.7 MΩ

13.63(MΩ)2

7.6 MΩ

RT = 1.79 MΩ

3-50 Course 45517001

Figure 3-13 Practice Problem 13

RT = 33.33 Ω

R1 + 1

R2 + 1 R3 + 1

R4

100 Ω + 1 100 Ω + 1

200 Ω + 1 200 Ω

0.01 Ω + 0.01 Ω + 0.005 Ω + 0.005 Ω

0.03 Ω

= 33.33 Ω

Course 45517001 3-51

Figure 3-14 Practice Problem 14

RT = 0.562 kΩ == 562 Ω

CALCULATIONS

RT =

R1 + 1

R2 + 1 R3 + 1

R4

1 kΩ + 1 5 kΩ + 1

3 kΩ + 1 4 kΩ

1 kΩ + 0.2 kΩ + 0.33 kΩ + 0.25 kΩ

1.78 kΩ

RT = 0.5618 kΩ = 561.8Ω

3-52 Course 45517001

Figure 3-15 Practice Problem 15

RT = 10 kΩ

R1 + 1

R2 + 1 R3 + 1

R4

20 kΩ + 1 60 kΩ + 1

45 kΩ + 1 90 kΩ

0.05 kΩ + 0.0167 kΩ + 0.0222 kΩ + 0.0111 kΩ

0.997 kΩ

Course 45517001 3-53

Figure 3-16 Practice Problem 16

RT = 0.2 MΩ == 200 kΩ

CALCULATIONS

Convert all R values to MΩ R1 = 600 kΩ = 0.6 MΩ R2 = 600kΩ = 0.6 MΩ

R3 = 1.2 MΩ

R4 = 1.2 MΩ

R1 + 1

R2 + 1 R3 + 1

R4

0.6 MΩ + 1

0.6 MΩ + 1

1.2 MΩ + 1

1.2 MΩ

1.67 MΩ + 1.67 MΩ + 0.83 MΩ + 0.83 MΩ

5 MΩ

RT = 0.2 MΩ = 200 kΩ

3-54 Course 45517001

Figure 3-17 Practice Problem 17

RT = 18.59 Ω

CALCULATIONS

RT =

R1 + 1

R2 + 1 R3 + 1

R4

100 Ω + 1 95 Ω + 1

75 Ω + 1 50 Ω

0.01 Ω + 0.0105 Ω + 0.0133 Ω + 0.02 Ω

0.0538 Ω

Course 45517001 3-55

Figure 3-18 Practice Problem 18

RT = 2.095 kΩ ≈ 2.1 kΩ

CALCULATIONS

RT =

R1 + 1

R2 + 1 R3 + 1

R4

16 kΩ + 1

4.7 kΩ + 1

9.8 kΩ + 1 10 kΩ

0.0625 kΩ + 0.2128 kΩ + 0.1020 kΩ + 0.1 kΩ

0.4773 kΩ

3-56 Course 45517001

Figure 3-19 Practice Problem 19

REQ = 50 Ω

RT = 250 Ω

CALCULATIONS

Solve REQ (for R2 and R3)

REQ =

R1 × R2 R1 + R2 or R2 × R3 R2 + R3

100 Ω × 100 Ω 100 Ω + 100 Ω

10,000 Ω2

Solve RT using REQ

RT = R1 + REQ + R4

RT = 100 Ω + 50 Ω + 100 Ω = 250 Ω

Course 45517001 3-57 May 2019 (Rev. 1)

Figure 3-20 Practice Problem 20

Practice ANSWERS REQ = 1,000 Ω == 1 kΩ RT = 2,000 Ω == 2 kΩ

CALCULATIONS

Convert all R values to Ω R1 = 500 Ω R2 = 3 kΩ = 3,000 Ω R3 = 1.5 kΩ = 1,500 Ω R4 = 500 Ω

Solve REQ (for R2 and R3)

REQ =

R1 × R2 R1 + R2 or R2 × R3 R2 + R3

3,000 Ω × 1,500 Ω 3,000 Ω + 1,500 Ω

4,500,000 Ω2

4,500 Ω

REQ = 1,000 Ω = 1 kΩ

Solve RT using REQ

RT = R1 + REQ + R4

RT = 500 Ω + 1,000 Ω + 500 Ω = 2,000 Ω = 2 kΩ

3-58 Course 45517001

Figure 3-21 Practice Problem 21

REQ = 4,150 Ω == 4.15 kΩ RT = 15.75 kΩ

Solve REQ (for R2 and R3)

REQ =

R1 + R2 or R2 × R3 R2 + R3

7.5 kΩ × 9.3 kΩ

7.5 kΩ + 9.3 kΩ

69.75(kΩ)2

16.8 kΩ

REQ = 4.15 kΩ

RT = 6.8 kΩ + 4.15 kΩ + 4.8 kΩ = 15.75 kΩ

Course 45517001 3-59

Figure 3-22 Practice Problem 22

REQ = 0.290 MΩ == 290 kΩ

RT = 2.66 MΩ

CALCULATIONS

Convert all values to MΩ

R1 = 1.9 MΩ

R2 = 1.5 MΩ

R3 = 360 kΩ = 0.36 MΩ R4 = 470 kΩ = 0.47 MΩ

Solve REQ (for R2 and R3)

REQ =

R1 × R2 R1 + R2 or R2 × R3 R2 + R3

1.5 MΩ × 0.36 MΩ

1.5 MΩ + 0.36 MΩ

0.54 (MΩ)2

1.86 MΩ

REQ = 0.290 MΩ = 290 kΩ

RT = 1.9 MΩ + 0.29 MΩ + 0.47 MΩ = 2.66 MΩ

3-60 Course 45517001

Figure 3-23 Practice Problem 23

REQ = 0.825 kΩ = 825 Ω RT = 8.825 kΩ

CALCULATIONS

Convert all R values to kΩ R1 = 4.7 kΩ R2 = 900 Ω = 0.9 kΩ R3 = 10 kΩ R4 = 3.3 kΩ

R1 + R2 or R2 × R3 R2 + R3

0.9 kΩ × 10 kΩ

0.9 kΩ + 10 kΩ

9 (kΩ)2

10.9 kΩ

REQ = 0.825 kΩ = 825 Ω

RT = 4.7 kΩ + 0.825 kΩ + 3.3 kΩ = 8.825 kΩ

Course 45517001 3-61

Figure 3-24 Practice Problem 24

REQ = 214.35 kΩ RT = 267.15 kΩ

REQ =

R1 × R2 R1 + R2 or R2 × R3 R2 + R3

340 kΩ × 580 kΩ 340 kΩ + 580 kΩ

197,200 (kΩ)2

920 kΩ

REQ = 214.35 kΩ

RT = 6.8 kΩ + 214.35 kΩ + 47 kΩ = 267.15 kΩ

3-62 Course 45517001

Figure 3-25 Practice Problem 25

REQA = 200 Ω

REQB = 67 Ω

RT = 267 Ω

Solve REQA (for R2 and R3)

REQA = R1 + R2

REQA = 100 Ω + 100 Ω = 200 Ω

Solve REQB (for REQA and R4)

REQB =

R1 + R2 or

REQA × R4

REQA + R4

200 Ω × 100 Ω

200 kΩ + 100 Ω

20,000 Ω2

300 Ω

REQ = 66.67 Ω ≈ 67 Ω

Solve for RT using REQB

RT = R1 + REQB + R5

RT = 100 Ω + 67 Ω + 100 Ω = 267 Ω

Course 45517001 3-63

Figure 3-26 Practice Problem 26

REQA = 1,160 Ω == 1.16 kΩ

REQB = 513 Ω

RT = 2,093 Ω == 2.093 kΩ

CALCULATIONS

Solve REQA (for R2 and R3)

REQA = R1 + R2

REQA = 480 Ω + 680 Ω = 1,160 Ω = 1.16 kΩ

R1 + R2 or

REQA × R4

REQA + R4

1,160 Ω × 920 Ω 1,160 Ω + 920 Ω

1,067,200 Ω2

2,080 Ω

REQ = 513 Ω

RT = 860 Ω + 513 Ω + 720 Ω = 2,093 Ω = 2.093 kΩ

3-64 Course 45517001

Figure 3-27 Practice Problem 27

REQA = 6.7 kΩ REQB = 3.47 kΩ RT = 13.97 kΩ

CALCULATIONS

Solve REQA (for R2 and R3)

REQA = R1 R2

REQA = 4.9 kΩ + 1.8 kΩ = 6.7 kΩ

R1 + R2 or

REQA × R4

REQA + R4

6.7 kΩ × 7.2 kΩ

6.7 kΩ + 7.2 kΩ

48.24 (kΩ)2

13.9 kΩ

REQ = 3.47 kΩ

RT = 3.3 kΩ + 3.47 kΩ + 7.2 kΩ = 13.97 kΩ

Course 45517001 3-65

Figure 3-28 Practice Problem 28

REQA = 18.9 kΩ REQB = 10.16 kΩ == 10,160 Ω RT = 88.17 kΩ == 88,170 Ω

CALCULATIONS

Solve REQA (for R2 and R3)

REQA = R1 + R2

REQA = 15 kΩ + 3.9 kΩ = 18.9 kΩ

R1 + R2 or

REQA × R4

REQA + R4

18.9 kΩ × 22 kΩ

18.9 kΩ + 22 kΩ

415.8 (kΩ)2

40.9 kΩ

REQ = 10.17 kΩ

RT = 56 kΩ + 10.17 kΩ + 22 kΩ = 88.17 kΩ = 88,170 Ω

3-66 Course 45517001

Figure 3-29 Practice Problem 29

REQA = 5.6 kΩ REQB = 4.46 kΩ RT = 16.36 kΩ

CALCULATIONS

Convert all values to kΩ R2 = 900 Ω = 0.900 kΩ R3 = 4.7 kΩ

Solve REQA (for R2 and R3)

REQA = R1 + R2

REQA = 0.900 kΩ + 4.7 kΩ = 5.6 kΩ

Solve REQB (for REQA and R4)

REQ =

R1 × R2 R1 + R2 or R2 × R3 R2 + R3

5.6 kΩ × 22 kΩ

5.6 kΩ + 22 kΩ

123.2(kΩ)2

27.6 kΩ

REQ = 4.46 kΩ

RT = R1 + REQB + R5

RT = 1.9 kΩ + 4.46 kΩ + 10 kΩ = 16.36 kΩ

Course 45517001 3-67

Figure 3-30 Practice Problem 30

REQA = 2.62 MΩ

REQB = 0.832 MΩ = 823 kΩ RT = 1,223 kΩ == 1.223 MΩ

3-68 Course 45517001

CALCULATIONS

Convert all values to MΩ R2 = 720 kΩ = 0.720 MΩ

R3 = 1.9 MΩ

Solve REQA (for R2 and R3)

REQA = R1 + R2

REQA = 0.720 MΩ + 1.9 MΩ = 2.62 MΩ

Solve REQB (for REQA and R4)

REQ =

R1 × R2 R1 + R2 or

REQA × R4

REQA + R4

2.62 MΩ × 1.2 MΩ

2.62 MΩ + 1.2 MΩ

3.144 (MΩ)2

3.82 MΩ

REQ = 0.823 MΩ = 823 kΩ

RT = R1 + REQB + R5

RT = 100 kΩ + 823 kΩ + 300 kΩ = 1,223 kΩ = 1.223 MΩ

Practice Problems Block 1, Lesson 5

Ohm’s Law and Power

Course 45517001 5-29

Figure 5-1 Practice Problem

I = 0.04 A = 40 mA

I = E

R

5 V

125 Ω = 0.04 A = 40 mA

5-30 Course 45517001

Figure 5-2 Practice Problem 2

E = 375 V

Convert all values to standard units

30 mA = 0.03 A

12.5 kΩ = 12,500 Ω

E = I × R = 0.03 A × 12,000 Ω = 375 V

Course 45517001 5-31

Figure 5-3 Practice Problem 3

R = 2,400 Ω = 2.4 kΩ

RED YELLOW RED

Convert all values to standard units

20 mA = 0.020 A

R =

I

48 V

0.020 A

= 2,400 Ω = 2.4 kΩ

Color Code

RED YELLOW RED

2 4 20

5-32 Course 45517001

Figure 5-4 Practice Problem 4

R1 = 11,936 W

P = 23.9 A

1 hp = 746 W

R1 = 16 × 746 W = 11,936 W

P = I × E and I =

E

11,936 W

500 V = 23.87 A ≈ 23.9 A

Course 45517001 5-33

Figure 5-5 Practice Problem 5

I = 0.0694 A = 69.4 mA

Since I is 69 mA, reducing I to 32 mA means that R must increase.

NOTE: Since I is approximately halved, R would approximately double.

E

R

50 V

720 Ω = 0.0694 A = 69.4 mA

5-34 Course 45517001

Figure 5-6 Practice Problem 6

9.2 kW, 480 V

R = 25.0 Ω

I = 19.17 A ≈ 19.2 A

8.3 kW, 480 V

27.76 Ω ≈ 27.8 Ω

Convert kW to W

9.2 kW = 9,200 W

E2

P

(480 V)2

9,200 W

230,400 V2

9,200 W = 25.0 Ω

E

9,200 W

480 V = 19.17 A ≈ 19.2 A

P

(480 V)2

8,300 W

230,400 V2

8,300 W = 27.76 Ω ≈ 27.8 Ω

Course 45517001 5-35

Figure 5-7 Practice Problem 7

E = 664.5 V

I = 0.1384 A = 138.4 mA

Convert kΩ to Ω

4.8 kΩ = 4,800 Ω

E = √P × R = √92 W × 4,800 Ω = 664.5 V

I = √

R

92 W

4,800 Ω = 0.1384 A = 138.4 mA

5-36 Course 45517001

Figure 5-8 Practice Problem 8

E = 148.15 V ≈ 148.2 V

R 5.49 Ω ≈ 5.5 Ω

50% E = 74.075 V ≈ 74.1 V P = 2000.7 W ≈ 2 kW R1 hp = 5.36 hp ≈ 5.4 hp

Convert kW to W

P = 4 kW = 4,000 W

E =

I

4,000 W

27 A = 148.15 V ≈ 148.2 V

I2

4,000 W

(27 A)2

4,000 W

729 A2 = 5.49 Ω ≈ 5.5 Ωhp ≈ 5.4 hp

Course 45517001 5-37

Practice

(Cont’d)

Voltage decreased 50%

50 % = 0.50 or or

E = 0.50 × 148.15 V = 74.075 V ≈ 74.1 V P = I × E = 27 A × 74.1 V = 2000.7 W ≈ 2 kW

R1 horsepower 1 hp = 746 W P = 4 kW = 4,000 W

R1 hp = 4000 W

W hp

= 5.36 hp ≈ 5.4 hp

5-38 Course 45517001

Figure 5-9 Practice Problem 9

P = 4,476 W

R = 51.47 Ω ≈ 51.5 Ω

E = 9.325 A ≈ 9.3 A

1 hp = 746 W

P = 6 hp × 746 W hp = 4,476 W

P =

R

(25 V)2

3,600 Ω

625 V2

3,600 W = 51.47 Ω ≈ 51.5 Ω

E

4,476 W

480 V = 9.325 A ≈ 9.3 A

Course 45517001 5-39

Figure 5-10 Practice Problem 10

P = 0.1736 W = 173.6 mW I = 0.00694 A = 6.94 mA

Convert k Ω to Ω

R = 3.60 kΩ = 3,600 Ω

P =

R

(25 V)2

3,600 Ω

625 V2

3,600 W = 0.1736 W = 173.6 mW

R

25 V

3,600 Ω = 0.00694 A = 6.94 mA

Practice Problems Block 1, Lesson 6 DC Series and Parallel Circuits

Course 45517001 6-37

Figure 6-1 Practice Problem

GIVEN

RT = 2.1 kΩ == 2,100 Ω IT=22.86 mA

SOLVE

E1 = 2.29 V E2 = 18.29 V

CHECK

ET = 327.43 V

SOLVE

P1 = 0.052 W P2 = 0.418 W P3 = 0.627 W

PT = 1.097 W

6-38 Course 45517001

Practice (Cont’d)

RT = 2.1 kΩ or 2,100 Ω, given

𝐈𝐈𝐈𝐈 =

ET

RT

48 V

2,100 Ω = 0.022857 A ≈ 22.86 mA

IT = I1 = I2 = I3 = 22.86 mA

SOLVE

E1 = I1 × R1 = 22.86 mA x 100 Ω = 2286 mV ≈ 2.29 V

E2 = I2 × R2 = 22.86 mA × 800 Ω = 18,288 mV ≈ 18.29 V

E3 = I3 × R3 = 22.86 mA × 1200 Ω = 27,432 mV ≈ 27.43 V

CHECK ET

E1+ E2 + E3 = 2.29 V+18.29 V+27.43 V = 48.01 V ≈ 48 V

SOLVE

P1 = I1 × E1 = 22.86 mA × 2.29 V = 52.35 mW ≈ 0.052 W

P2 = I2 × E2 = 22.86 mA × 18.29 V = 418.1 mW ≈ 0.418 W

P3 = I3 × E3 = 22.86 mA × 27.43 V = 627.05 mW ≈ 0.627 W

PT = IT x ET = 22.86 mA x 48 V = 1097.28 mV ≈ 1.097 W

CHECK PT

P1 + P2 + P3 = 0.052 W + 0.418 W + 0.627 W = 1.097 W

Course 45517001 6-39

Figure 6-2 Practice Problem 2

RT = 47 kΩ == 47,000 Ω ET = 208v

SOLVE

IT = 4.425 mA

IT = I1 = I2 = I3

E1 = 5.31 V E2 = 159.3 V E3 = 43.37 V

CHECK

ET = E1+ E2 +E3 = 208V

P1 = 23.5 mW P2 = 705 mW P3 = 192 mW PT = 920 mW

6-40 Course 45517001

Practice (Cont’d)

CALCULATIONS

RT = 47 kΩ == 47,000 Ω, given

SOLVE

𝐈𝐈𝐈𝐈 =

ET

RT

208 V

47,000 Ω = 0.004425 A ≈ 0.004425 A

IT = I1 = I2 =I3 =4.425 mA

Convert all R values to Ω I1 = 1.2 kΩ = 1,200 Ω I2 = 36 kΩ = 36,000 Ω I3 = 9.8 kΩ = 9,800 Ω

E1 = I1 × R1 = 4.425 mA × 1200 Ω = 5310 mV = 5.31 V

E2 = I2 × R2 = 4.425 mA × 36,000 Ω = 159,300 mV = 159.3 V

E3 = I3 × R3 = 4.425 mA × 9,800 Ω = 42,365 mV = 43.37 V

CHECK ET

E1+E2+E3 = 5.31 V+159.3 V+43.37 V = 207.98 V ≈ 208 V

SOLVE

P1 = I1 x E1 = 4.425 mA × 5.31 V = 23.5 mW

P2 = I2 x E2 = 4.425 mA × 159.3 V = 704.9 mW ≈ 705 mW

P3 = I3 x E3 = 4.425 mA × 43.37 V = 191.9 mW ≈ 192 mW

PT = IT x ET = 4.425 mA × 208 V = 920.5 mW

P1 + P2 + P3 = 23.5 mW+705 mW+192 mW = 920.5 mW

Course 45517001 6-41

Figure 6-3 Practice Problem 3

RT = 1.98 MΩ == 1,980,000 Ω

ET = 480 V

SOLVE

IT = 0.242

IT = I1 = I2 = I3 0.0024242 A = 0.2424 mA E1 = 108,900 mV = 108.9 V E2 = 162,140 mV = 162.1 V E3 = 208,120 mV = 208.1 V

CHECK

ET = E1+ E2 +E3 = 480V

P1 = 26.35 mW = 0.02635 W P2 = 39.23 mW = 0.03923 W P3 = 50.36 mW = 0.05036 W PT = 115.93 mW ≈ 116 mW

6-42 Course 45517001

Practice (Cont’d)

CALCULATIONS

RT = 1.98 MΩ == 1,980,000 Ω, (Given)

SOLVE

𝐈𝐈𝐈𝐈 =

ET

RT

480 V

1,980,000 Ω = 0.0024242 A = 0.2424 mA

IT = I1 = I2 = I3 = 0.0024242 A = 0.2424 mA

R1 = 450 kΩ = 450,000 Ω R2 = 670 kΩ = 670,000 Ω R3 = 860 kΩ = 860,000 Ω

E1 = I1 × R1 = 0.242 mA × 450,000 Ω = 108,900 mV = 108.9 V E2 = I2 × R2 = 0.242 mA × 670,000 Ω = 162,140 mV = 162.1 V E3 = I3 × R3 = 0.242 mA × 860,000 Ω = 208,120 mV = 208.1 V

CHECK ET

ET = E1+E2+E3 = 108.9 V + 162.1 V + 208.1 V = 479.1 V ≈ 480 V

SOLVE

P1 = I1 × E1 = 0.242 mA × 108.9 V = 26.35 mW = 0.02635 W P2 = I2 × E2 = 0.242 mA × 162.1 V = 39.23 mW = 0.03923 W P3 = I3 × E3 = 0.242 mA × 208.1 V = 50.36 mW = 0.05036 W PT = P1 + P2 + P3 = 26.35 mW + 39.2 mW + 50.36 mW PT = 115.94 mW ≈ 116 mW

CHECK PT

𝐏𝐏𝐏𝐏 =

ET2

RT

(480 V)2

1,980,000 Ω

230,400 V2

1,980,000 Ω

PT = 0.11636 W = 116.36 mW

Course 45517001 6-43

Figure 6-4 Practice Problem 4

RT = 9.42 MΩ == 9,420,000 Ω

E2 = 470 V

SOLVE

I2 = 0.1 mA

IT= I1 = I2 = I3

E1 = 390 V E2 = 470 V E3 = 82 V

ET = 942 V

CHECK

ET = 942V

P1 = 39 mW P2 = 47 mW P3 = 8.2 mW PT = 94.2 mW

6-44 Course 45517001

Practice (Cont’d)

CALCULATIONS

Convert all R values to Ω

R1 = 3.9 MΩ = 3,900,000 Ω

R2 = 4.7 M Ω = 4,700,000 Ω R3 = 820 kΩ = 820,000 Ω

R2

480 V

4,700,000 Ω = 0.0001A = 0.1 mA

IT = I1 = I2 = I3 = 0.1 mA

E1 = I1 × R1 = 0.1 mA x 3,900,000 Ω = 390,000 mV = 390 V

E2 = I2 × R2 = 0.1 mA x 4,700,000 Ω = 470,000 mV = 470 V

E3 = I3 × R3 = 0.1 mA x 820,000 Ω = 82,000 mV = 82 V

ET = IT × RT = 0.1 mA x 9.42 MΩ = 0.942 mV = 942 V

CHECK ET

E1+E2+E3 = 390 V + 470 V + 82 V = 942 V

SOLVE

P1 = I1 × E1 = 0.1 mA × 390 V = 39 mW P2 = I2 × E2 = 0.1 mA × 470 V = 47 mW

P3 = I3 × E3 = 0.1 mA × 82 V = 8.2 mW

PT = IT × ET = 0.1 mA × 942 V = 94.2 mW

P1 + P2 + P3 = 39 mW + 47 mW + 8.2 mW = 94.2 mW

Course 45517001 6-45

Figure 6-5 Practice Problem 5

RT = 9,060 Ω == 9.06 kΩ I2 = 30mA

SOLVE

I2 = 0.1 mA

IT = I1 = I2 = I3

E1 = 69 V E2 = 174V E3 = 28.8V

ET = 271.8V

CHECK

ET = E1+ E2 +E3 = 271.8V

P1 = 2.07 W P2 = 5.22 W P3 = 864 mW

PT = 8.154 W

6-46 Course 45517001

Practice (Cont’d)

CALCULATIONS

RT = 9,060 Ω = 9.06 kΩ

IT = I1 = I2 = I3 = 30 mA

Convert all R values to Ω R1 = 2.3 kΩ = 2,300 Ω R2 = 5.8 kΩ = 5,800,000 Ω R3 = 960 Ω

SOLVE

E1 = I1 × R1 = 30 mA × 2,300 Ω = 69,000 mV = 69 V

E2 = I2 × R2 = 30 mA × 5,800 Ω = 174,000 mV = 174 V

E3 = I3 × R3 = 30 mA × 960 Ω = 28,800 mV = 28.8 V

CHECK ET

ET = E1+E2+E3 = 69 V + 174 V + 28.8 V = 271.8 V

SOLVE

P1 = I1 × E1 = 30 mA × 69 V = 2,070 mW = 2.07 W

P2 = I2 × E2 = 30 mA × 174 V = 5,220 mW = 5.22 W

P3 = I3 × E3 = 30 mA × 28.8 V = 864 mW = 0.864 W

PT = IT × ET = 30 mA × 271.8 V = 8,154 mW = 8.154 W

PT = P1 + P2 + P3 = 2.07 W + 5.22 W + 0.864 W = 8.154 W

Course 45517001 6-47

Figure 6-6 Practice Problem 6

RT = 2.36 MΩ == 2,360,000 Ω

P2 = 85.2 mW = 0.0852 W I2 = IT= 0.000426 = A 0.426mA=426uA

IT = I1 = I2= I3

E1 = 166,140 mV = 166.14 V E2 = 200,220 mV = 200.22 V E3 = 639,000 mV = 639.00 V ET = 1,005,360 mV = 1,005.36 V

CHECK

ET = E1 + E2 + E3 = 1005.36 V

P1 = 70.78 mW = 0.07078 W P2 = 85.29 mW = 0.08529 W P3 = 272.21 mW = 0.27221 W PT = 428.28 mW ≈ 428 mW

6-48 Course 45517001

Practice (Cont’d)

CALCULATIONS

Convert all R values to MΩ R1 = 390 kΩ = 0.390 MΩ = 390,000 Ω R2 = 0.47 MΩ = 470 k Ω = 470,000 Ω R3 = 1.5 MΩ = 1,500 k Ω = 1,500,000 Ω

RT = R1 + R2 + R3 = 0.39 MΩ + 0.47 MΩ + 1.5 MΩ = 2.36 MΩ

I2 =�P2 R2

= �0.0852 W 470,000 Ω

= √0.000000181 = 0.000426 A = 0.426 mA

IT = I1 = I2 = I3 = 0.000426 A = 0.426 mA

E1 = I1 × R1 = 0.426 mA × 390,000 Ω = 166,140 mV = 166.14 V

E2 = I2 × R2 = 0.426 mA × 470,000 Ω = 200,220 mV = 200.22 V

E3 = I3 × R3 = 0.426 mA × 1,500,000 Ω = 639,000 mV = 639.00 V

ET = IT × RT = 0.426 mA × 2,360,000 Ω =1,005,360 mV = 1,005.36 V

CHECK ET

ET =E1+E2+E3 = 166.14 + 200.22+ 639.00 V = 1005.36 V

SOLVE

P1 = I1 × E1 = 0.426 mA × 166.14 V = 70.78 mW = 0.07078 W

P2 = I2 × E2 = 0.426 mA × 200.22 V = 85.29 mW = 0.08529 W

P3 = I3 × E3 = 0.426 mA × 639.00 V = 272.21 mW = 0.27221 W

PT = IT × ET = 0.426 mA × 1005.36 V = 428.28 mW ≈ 428 mW

CHECK PT

PT = P1 + P2 + P3 = 70.78 mW + 85.29 mW + 272.21 mW PT = 428.28 mW ≈ 428 mW

Course 45517001 6-49

Figure 6-7 Practice Problem 7

SOLVE

E2 = 480 V

E1 = E2 = ET

I1 = 0.48 A = 480 mA I2 = 0.48 A = 480 mA IT = 0.96 A = 960 mA P1 = 230.4 W P2 = 230.4 W

PT = 460.8 W

6-50 Course 45517001

Practice (Cont’d)

CALCULATIONS

Convert kΩ 𝐭𝐭𝐭𝐭 Ω R1 = R2 = 1 kΩ = 1,000 Ω

SOLVE

For parallel circuits where resistors all parallel and of the same value:

𝐑𝐑𝐑𝐑 =

V N

1,000 Ω

= 500 Ω or For all parallel circuits:

𝐑𝐑𝐑𝐑 =

R1 × R2 R1 + R2

1,000 Ω × 1,000 Ω 1,000 Ω + 1,000 Ω

1,000,000 Ω2

2,000 Ω

E2 = E1 = ET = 480 V

E1 R1

480 V

1,000 Ω = 0.48 A = 480 mA

R2

480 V

= 0.48 A = 480 mA

RT

480 V 500 Ω

= 0.96 A = 960 mA

P1 = I1 × E1 = 0.48 A × 480 V = 230.4 W

P2 = I2 × E2 = 0.48 A × 480 V = 230.4 W

PT = IT × ET = 0.96 A × 480 V = 460.8 W

PT = P1 + P2 + P3 = 230.4 W + 230.4 W = 460.8 W

Course 45517001 6-51

Figure 6-8 Practice Problem 8

RT = 276.9 × ≈ 277 Ω

SOLVE

I2 = 0.025 A = 25 mA E2 = 10 V

E1 = E2 = ET

I1 = 0.0111 A = 11.1 mA IT = 36.1 mA = 0.0361 A

CHECK

IT = 36.1 mA P1 = 0.111 W = 111 mW P2 = 0.250 W = 250 mW PT = 0.361 W = 361 mW

6-52 Course 45517001

Practice (Cont’d)

CALCULATIONS

Convert kΩ 𝐭𝐭𝐭𝐭 Ω R1 = R2 = 1 kΩ = 1,000 Ω

SOLVE

𝐑𝐑𝐑𝐑 =

R1 × R2 R1 + R2

400 kΩ × 900 kΩ 400 kΩ + 900 kΩ

360,000 (kΩ)2

1,300 kΩ

𝐑𝐑𝐑𝐑 = 276.9 Ω ≈ 277 Ω

Convert mA to A I2 = 25 mA = 0.025 A

E2 = I2 × R2 = 0.025 A × 400 Ω = 10 V

ET = E1 = E2 = 10 V

R1

10 V

900 Ω = 0.0111 A = 11.1 mA

R2

10 V

400 Ω = 0.025 A = 25 mA

𝐈𝐈𝐈𝐈 = I1 + I2 = 11.1 mA + 25 mA = 36.1 mA = 0.0361 A

P1 = I1 × E1 = 0.0111 A × 10 V = 0.111 W = 111 mW

P2 = I2 × E2 = 0.025 A × 10 V = 0.250 W= 250 mW

PT = IT × ET = 0.0361 A × 10 V = 0.361 W = 361 mW

PT = P1 + P2 + P3 = 111 mW + 250 mW = 361 mW

Course 45517001 6-53

Figure 6-9 Practice Problem 9

RT = 4.42 kΩ = 4,420 Ω

ET = E1 = E2 = 48 V

I1= 0.00527 A = 5.27 mA I2= 0.00558 A = 5.58 mA IT = 10.85 mA = 0.01085 A

CHECK

IT = 36.1 mA P1 = 0.253 W = 253 mW P2 = 0.268 W = 268 mW PT = 0.361 W = 361 mW

6-54 Course 45517001

Practice (Cont’d)

CALCULATIONS

Convert kΩ 𝐭𝐭𝐭𝐭 Ω R1 = R2 = 1 kΩ = 1,000 Ω

SOLVE

𝐑𝐑𝐑𝐑 =

R1 × R2 R1 + R2

9.1 kΩ × 8.6 kΩ

9.1 kΩ + 8.6 kΩ

78.26 (kΩ)2

17.7 kΩ = 4.42 kΩ = 4,420 Ω

ET = E1 = E2 = 48 V

R1 = 9.1 kΩ = 9,100 Ω R2 = 8.6 kΩ = 8,600 Ω

R1

48 V

9,100 Ω = 0.00527 A = 5.27 mA

R2

48 V

8,600 Ω = 0.00558 A = 5.58 mA

𝐈𝐈𝐈𝐈 = I1 + I2 = 5.27 mA + 5.58 mA = 10.85 mA = 0.01085 A

𝐏𝐏𝐏𝐏 =

E12

R1

(48 V)2

9,100 Ω

2,304 V2

9,100 Ω = 0.253 W = 253 mW

E22

R2

(48 V)2

8,600 Ω

2,304 V2

8,600 Ω = 0.268 W = 268 mW

𝐏𝐏𝐏𝐏 =P1 + P2 + P3 = 253 mW + 268 mW = 521 mW

PT = IT × ET = 0.01085 A × 48 V = 0.5208 W = 520.8 mW ≈ 521 mW

Course 45517001 6-55

Figure 6-10 Practice Problem 10

RT = 24.785 kΩ = 24,785 Ω

ET = E1 = E2 = 120 V

I1= 0.00308 A = 3.08 mA I2= 0.00176 A = 1.76 mA IT = 4.84 mA = 0.00484 A P1 = 0.369 W = 369 mW P2 = 0.212 W = 212 mW PT = 581 mW = 5.81 W

CHECK

PT = 5.81 W

6-56 Course 45517001

Practice (Cont’d)

CALCULATIONS

SOLVE

𝐑𝐑𝐑𝐑 =

R1 × R2 R1 + R2

39 kΩ × 68 kΩ 39 kΩ + 68 kΩ

2,652(kΩ)2

107 kΩ

RT = 24.785 kΩ = 24,785Ω

ET = E1 = E2 = 120 V

Convert all R values to Ω R1 = 39 kΩ = 39,000 kΩ R2 = 68 kΩ = 68,000 kΩ

R1

120 V

39,000 Ω = 0.00308 A = 3.08 mA

R2

120 V

68,000 Ω = 0.00176 A = 1.76 mA

𝐈𝐈𝐈𝐈 = I1 + I2 = 3.08 mA + 1.76 mA = 4.84 mA = 0.00484 A

R1

(120 V)2

39,000 Ω

14,400 V2

39,000 Ω = 0.369 W = 369 mW

R2

(120 V)2

68,000 Ω

14,400 V2

68,000 Ω = 0.212 W = 212 mW

𝐏𝐏𝐏𝐏 =P1 + P2 + P3 = 369 mW + 212 mW = 581 mW = 5.81 W

PT = IT × ET = 0.00484 A × 120 V = 0.581 W = 581 mW

Course 45517001 6-57

Figure 6-11 Practice Problem 11

Practice SOLVE RT = 360.1 kΩ

ET = E1 = E2 = 305.5 V

I1 = 0.000065 A = 65 uA I2 = 0.000783 A = 783 uA IT = 0.000848 A = 848 uA P1 = 0.01985 W = 19.85 mW P2 = 0.239 W = 239 mW PT = 259 mW = 0.259 W

6-58 Course 45517001

Practice (Cont’d)

CALCULATIONS

Convert all R values to 𝐤𝐤Ω R1= 4.7 MΩ = 4,700 kΩ R2= 390 kΩ

SOLVE

𝐑𝐑𝐑𝐑 =

R1 × R2 R1 + R2

4,700 kΩ × 390 kΩ 4,700 kΩ + 390 kΩ

1,833,000 (kΩ)2

5,090 kΩ = 360.1 kΩ

Convert I and R to standard values I1 = 65 uA = 0.000065 A

R1 = 4.7 MΩ = 4,700,000 Ω

R2 = 390 kΩ = 390,000 Ω

E1 = I1 × R1 = 0.000065 A × 4,700,000 Ω = 305.5 V

ET = E1 = E2 = 305.5 V

Given I1 = 65 uA = 0.000065 A

R2

305.5 V

390,000 Ω = 0.000783 A = 783 uA

IT = I1+I2 = 0.000065 A + 0.000783 A = 0.000848 A = 848 uA

R1

(305.5 V)2

4,700,000 Ω

93,330.250 V2

4,700,000 Ω = 0.01985 W = 19.85 mW

R2

(305.5 V)2

390,000 Ω

93,330.250 V2

390,000 Ω = 0.2393 W = 239.3 mW

𝐏𝐏𝐏𝐏 =P1 + P2 + P3 = 19.85 mW + 239.3 mW = 259.15 mW

PT = IT × ET = 0.000848 A × 305.5 V = 0.259 W = 259 mW

Course 45517001 6-59

Figure 6-12 Practice Problem 12

RT = 1.79 kΩ = 179,000 Ω E2 = 211.5 V

ET = E1 = E2 = 211.5 V

I1 = 0.07293 A = 72.93 m I2= 0.045 A = 45 mA IT = 117.93 mA = 0.11793 A P1 = 15.42 W P2 = 9.52 W

PT = 24.94 W

PT = 24.94 W

6-60 Course 45517001

Practice (Cont’d)

CALCULATIONS

SOLVE

𝐑𝐑𝐑𝐑 =

R1 × R2 R1 + R2

2.9 kΩ × 4.7 kΩ

2.9 kΩ + 4.7 kΩ

13.63(kΩ)2

7.6 kΩ = 1.79 kΩ

Convert R1 and R2 to Ω R1 = 2.9 kΩ = 2,900 Ω R2 = 4.7 kΩ = 4,700 Ω

E2 = √P2 × R2 = �9.5175 W × 4,700 Ω = 211.5 V

ET = E1 = E2 = 211.5 V

R1

211.5 V

2,900 Ω

= 0.07293 A = 72.93 mA

R2

211.5 V

4,700 Ω

= 0.045 A = 45 mA

𝐈𝐈𝐈𝐈 = I1 + I2 = 72.93 mA + 45 mA = 117.93 mA = 0.11793 A

R1

(211.5 V)2

2,900 Ω

44,732.25V2

2,900 Ω = 15.42 W

R2

(120 V)2

4,700 Ω

44,732.25 V2

= 9.52 W

𝐏𝐏𝐏𝐏 =P1 + P2 + P3 = 15.42 W + 9.52 W = 24.94 W

PT = IT × ET = 0.11793 A × 211.5 V = 24.94 W

Course 45517001 6-61

Figure 6-13 Practice Problem 13

RT = 33.33 Ω

ET = E1 = E2 = E3 = E4 = 75 V

I1 = 0.75 A = 750 mA I2 = 0.75 A = 750 mA I3 = 0.375 A = 375 mA I4 = 0.375 A = 375 mA IT = 2.25 A = 2,250 mA P1 = 56.25 W P2 = 56.25 W P3 = 28.125 W P4 = 28.125 W

PT = 168.75 W

PT = 168.75 W

6-62 Course 45517001

Practice (Cont’d)

CALCULATIONS

SOLVE

𝐑𝐑𝐑𝐑 =

R1 + 1

R2 + 1 R3 + 1

R4

100 Ω + 1 100 Ω + 1

200 Ω + 1 200 Ω

RT =

0.01 Ω + 0.01 Ω + 0.005 Ω + 0.005 Ω

0.03 Ω

= 33.33 Ω

ET = E1 = E2 = E3 = E4 = 75 V

R1

75 V

100 Ω = 0.75 A = 750 mA

R2

75 V

= 0.75 A = 750 mA

E3 R3

75 V

= 0.375 A = 375 mA

E4 R4

75 V

= 0.375 A = 375 mA

𝐈𝐈𝐈𝐈 = I1 + I2 + I3 + I4 = 0.75 A + 0.75 A + 0.375 A + 0.375 A

IT = 2.25 A = 2,250 mA

R1

(75 V)2

100 Ω

5,625 V2

= 56.25 W

R2

(75 V)2

100 Ω

5,625 V2

E32

R3

(75 V)2

200 Ω

5,625 V2

= 28.125 W

E42

R4

(75 V)2

200 Ω

5,625 V2

= 28.125 W

𝐏𝐏𝐏𝐏 =P1 + P2 + P3 + P4 = 56.25 W + 56.25 W + 28.125 W + 28.125 W

PT = 168.75 W

CHECK PT

PT = IT × ET = 2.25 A × 75 V = 168.75 W

Course 45517001 6-63

Figure 6-14 Practice Problem 14

RT = 561.8 kΩ E2 = 175 V

ET = E1 = E2 = E3 = E4= 175 V

I1 = 0.00308 A = 3.08 mA I2 = 0.00176 A = 1.76 mA IT = 0.31205 A ≈ 312 mA P1 = 30.625 W P2 = 6.125 W P3 = 10.208 W P4 = 7.656 W

PT = 54.61 W

PT = 54.61 W

6-64 Course 45517001

Practice (Cont’d)

CALCULATIONS

SOLVE

𝐑𝐑𝐑𝐑 =

R1 + 1

R2 + 1 R3 + 1

R4

1 kΩ + 1 5 kΩ + 1

3 kΩ + 1 4 kΩ

1 kΩ + 0.2 kΩ + 0.33 kΩ + 0.25 kΩ

1.78 kΩ

RT = 561.8 kΩ

R1 = 1 kΩ = 1,000 Ω R2 = 5 kΩ = 5,000 Ω R3 = 3 kΩ = 3,000 kΩ R4 = 4 kΩ = 4,000 kΩ

Convert mA to A I2 = 35 mA = 0.035 A

E2 = I2 × R2 = 0.035 A × 5,000 Ω = 175 V

ET = E1 = E2 = E3 = E4 = 175 V

R1

175 V

= 0.175 A = 175 mA

R2

175 V

5,000 Ω = 0.035 A = 35 mA (Given)

R1

175 V

3,000 Ω = 0.0583 A = 53.8 mA

R2

175 V

4,000 Ω = 0.04375 A = 43.75 mA

𝐈𝐈𝐈𝐈 = I1 + I2 + I3 + I4 = 0.175 A + 0.035 A + 0.0583 A + 0.04375 A

IT = 0.31205 A ≈ 312 mA

Course 45517001 6-65

Practice (Cont’d) 𝐏𝐏𝐏𝐏 =

E12

R1

(175 V)2

1,000 Ω

30,625 V2

= 30.625 W

R2

(175 V)2

5,000 Ω

30,625 V2

5,000 Ω = 6.125 W

R3

(175 V)2

3,000 Ω

30,625 V2

= 10.208 W

R4

(175 V)2

4,000 Ω

30,625 V2

4,000 Ω = 7.656 W

𝐏𝐏𝐏𝐏 =P1 + P2 + P3 + P4 = 30.625 W + 6.125 W + 10.208 W + 7.656

PT = 54.61 W

PT = IT × ET = 0.31205 A × 175 V = 54.61 W

6-66 Course 45517001

Figure 6-15 Practice Problem 15

RT = 10 kΩ

ET = E1 = E2 = E3 = E4= 2,250 V

IT = 225 mA = 0.225 A I1 = 0.1125 A = 112.5 mA I2 = 0.0375 A = 37.5 mA I3 = 0.05 A = 50 mA I4 = 0.025 A = 25 mA IT = 0.225 A = 225 mA (Given) P1 = 253.125 W P2 = 84.375 W P3 = 112.5 W P4 = 56.25 W

PT = 506.3125 W

PT = 506.25 W ≈ 506.3 W

Course 45517001 6-67

Practice (Cont’d)

CALCULATIONS

SOLVE

𝐑𝐑𝐑𝐑 =

R1 + 1

R2 + 1 R3 + 1

R4

20 kΩ + 1 60 kΩ + 1

45 kΩ + 1 90 kΩ

0.05 kΩ + 0.0167 kΩ + 0.0222 kΩ + 0.0111 kΩ

0.997 kΩ

R1 = 20 kΩ = 20,000 Ω R2 = 60 kΩ = 60,000 Ω R3 = 45 kΩ = 45,000 kΩ R4 = 90 kΩ = 90,000 kΩ RT = 10 kΩ = 10,000 kΩ

Convert mA to A IT = 225 mA = 0.225 A

ET = IT × R2 = 0.225 A × 10,000 Ω = 2,250 V

ET = E1 = E2 = E3 = E4 = 2,250 V

R1

2,250 V

20,000 Ω = 0.1125 A = 112.5 mA

R2

2,250 V

60,000 Ω = 0.0375 A = 37.5 mA

R3

2,250 V

45,000 Ω = 0.05 A = 50 mA

R2

2,250 V

90,000 Ω = 0.025 A = 25 mA

𝐈𝐈𝐈𝐈 = I1 + I2 + I3 + I4 = 0.1125 A + 0.0375 A + 0.05 A + 0.025 A

IT = 0.225 A = 225 mA (Given)

Continued on next page

6-68 Course 45517001

Practice (Cont’d) 𝐏𝐏𝐏𝐏 =

E12

R1

(2,250V)2

20,000 Ω

5,062,500 V2

20,000 Ω = 253.125 W

R2

(2,250 V)2

60,000 Ω

5,062,500 V2

60,000 Ω = 84.375 W

R3

(2,250 V)2

45,000 Ω

5,062,500 V2

45,000 Ω = 112.5 W

R4

(2,250V)2

90,000 Ω

5,062,500 V2

90,000 Ω

𝐏𝐏𝐏𝐏 =P1 + P2 + P3+ P4 = 253.125 W + 84.375 W + 112.5 W + 56.25

PT = 506.3125 W

PT = IT × ET = 0.225 A × 2,250 V = 506.25 W ≈ 506.3 W

Course 45517001 6-69

Figure 6-16 Practice Problem 16

RT = 0.2 MΩ = 200 kΩ E4 = 42 V

ET = E1 = E2 = E3 = E4= 42 V

I1 = 0.00007 A = 70 µA I2 = 0.00007 A = 70 µA I3 = 0.000035 A = 35 µA I4 = 0.000035 A = 35 µA IT = 210 µA P1 = 0.00294 W = 2.94 mW P2 = 0.00294 W = 2.94 mW P3 = 0.00147 W = 1.47 mW P4 = 0.00147 W = 1.47 mW PT = 8.82 mW

PT = 0.00882 W = 8.82 mW

6-70 Course 45517001

Practice (Cont’d)

CALCULATIONS

Convert all R values to MΩ R1 = 600 kΩ = 0.6 MΩ R2 = 600kΩ = 0.6 MΩ

R3 = 1.2 MΩ

R4 = 1.2 MΩ

SOLVE

𝐑𝐑𝐑𝐑 =

R1 + 1

R2 + 1 R3 + 1

R4

0.6 MΩ + 1

0.6 MΩ + 1

1.2 MΩ + 1

1.2 MΩ

1.67 MΩ + 1.67 MΩ + 0.83 MΩ + 0.83 MΩ

5 MΩ

RT = 0.2 MΩ = 200 kΩ

R1 = 600 kΩ = 600,000 Ω R2 = 600 kΩ = 600,000 Ω R3 = 1.2 MΩ = 1,200,000 kΩ R4 = 1.2 MΩ = 1,200,000 kΩ

Convert µA to A IT = 35 µA = 0.000035 A

E4 = I4 × R4 = 0.000035 A × 1,200,000 Ω = 42 V

ET = E1 = E2 = E3 = E4 = 42 V

R1

42 V

600,000 Ω = 0.00007 A = 70 µA

R2

42 V

= 0.00007 A = 70 µA

R3

42 V

1,200,000 Ω = 0.000035 A = 35 µA

R4

42 V

1,200,000 Ω = 0.000035 A = 35 µA

Continued on next page.

Course 45517001 6-71

Practice (Cont’d)

𝐈𝐈𝐈𝐈 = I1 + I2 + I3 + I4 = 70 µA + 70 µA + 35 µA + 35 µA IT = 210 µA

R1

(42 V)2

600,000 Ω

1,764 V2

= 0.00294 W = 2.94 mW

R2

(42 V)2

600,000 Ω

1,764 V2

= 0.00294 W = 2.94 mW

R3

(42 V)2

1,200,000 Ω

1,764 V2

= 0.00147 = 1.47 mW

R4

(42 V)2

1,200,000 Ω

1,764 V2

= 0.00147 W = 1.47 mW

𝐏𝐏𝐏𝐏 =P1 + P2 + P3+ P4 = 2.94 mW + 2.94 mW + 1.47 W +1.47 W PT = 8.82 mW

CHECK PT

IT = 210 µA = 0.00210

PT = IT × ET = 0.000210 A × 42 V = 0.00882 W = 8.82 mW

6-72 Course 45517001

Figure 6-17 Practice Problem 17

RT = 18.59 Ω

E2 = 332.5 V

ET = E1 = E2 = E3 = E4= 332.5 V

I1 = 3.325 A ≈ 3.33 A I2 = 3.5 A I3 = 4.43 A I4 = 6.65 A

IT = 17.905 A ≈ 17.91 A

P1 = 1105.6 W ≈ 1.106 kW P2 = 1,163.75 W = 1.16 kW P3 = 1,474.1 W = 1.47 kW P4 = 2,211.125 W = 2.21 kW PT = 5.95 kW

PT = 5.96 kW

Course 45517001 6-73

Practice (Cont’d)

CALCULATIONS

SOLVE

RT =

R1 + 1

R2 + 1 R3 + 1

R4

100 Ω + 1 95 Ω + 1

75 Ω + 1

0.01 Ω + 0.0105 Ω + 0.0133 Ω + 0.02 Ω

0.0538 Ω

E2 = I2 × R2 = 3.5 A × 95= 332.5 V

ET = E1 = E2 = E3 = E4 = 332.5 V

R1

332.5 V

= 3.325 A ≈ 3.33 A

𝐈𝐈𝟐𝟐 =

R2

332.5 V

95 Ω = 3.5 A

𝐈𝐈𝐈𝐈 =

E3 R3

332.5 V

75 Ω = 4.43 A

R4

332.5 V

= 6.65 A

𝐈𝐈𝐈𝐈 = I1 + I2 + I3 + I4 = 3.325 A + 3.5 A + 4.43 A + 6.65 A

IT = 17.905 A ≈ 17.91 A

6-74 Course 45517001

Practice (Cont’d) 𝐏𝐏𝐏𝐏 =

E12

R1

(332.5 V)2

100 Ω

110,556.25 V2

= 1105.6 W ≈ 1.11 kW

R2

(332.5 V)2

95 Ω

110,556.25 V2

95 Ω = 1,163.75 W = 1.16 kW

R3

(332.5 V)2

75 Ω

110,556.25 V2

75 Ω = 1,474.1 W = 1.47 kW

R4

(332.5 V)2

50 Ω

110,556.25 V2

= 2,211.125 W = 2.21 kW

𝐏𝐏𝐏𝐏 =P1 + P2 + P3+ P4 = 1.11 kW + 1.16 kW + 1.47 W +2.21 W PT = 5.95 kW

CHECK PT

PT = IT × ET = 17.91 A × 332.5 V = 5,955.1 W ≈ 5.96 kW

Course 45517001 6-75

Figure 6-18 Practice Problem 18

RT = 2.095 kΩ ≈ 2.1 kΩ E4 = 305.4 V ≈ 305 V

ET = E1 = E2 = E3 = E4= 305.4 V ≈ 305 V

I1 = 0.0191 A = 19.1 mA I2 = 0.0649 A = 64.9 mA I3 = 0.0311 A = 31.1 mA I4 = 0.0305 A = 30.5 mA IT = 145.6 mA P1 = 0.00294 W = 2.94 mW P2 = 0.00294 W = 2.94 mW P3 = 0.00147 W = 1.47 mW P4 = 0.00147 W = 1.47 mW PT = 8.82 mW

PT = 0.00882 W = 8.82 mW

6-76 Course 45517001

Practice (Cont’d)

CALCULATIONS

Convert all R values to MΩ R1 = 600 kΩ = 0.6 MΩ R2 = 600kΩ = 0.6 MΩ

R3 = 1.2 MΩ

R4 = 1.2 MΩ

SOLVE

RT =

R1 + 1

R2 + 1 R3 + 1

R4

16 kΩ + 1

4.7 kΩ + 1

9.8 kΩ + 1 10 kΩ

0.0625 kΩ + 0.2128 kΩ + 0.1020 kΩ + 0.1 kΩ

0.4773 kΩ

R1 = 1.6 kΩ = 1,600 Ω R2 = 4.7kΩ = 4,7000 Ω R3 = 9.8 kΩ = 9,800 Ω R4 = 10 kΩ = 10,000 Ω

𝐄𝐄𝐄𝐄 = √P3 × R3 = �9.5175 W × 9,800 Ω = 305.4 V

ET = E1 = E2 = E3 = E4 = 305.4 V ≈ 305 V

Course 45517001 6-77

Practice (Cont’d) 𝐈𝐈𝐈𝐈 =

E1 R1

305 V

1,600 Ω = 0.0191 A = 19.1 mA

R2

42 V

= 0.0649 A = 64.9 mA

R3

42 V

9,800 Ω = 0.0311 A = 31.1 mA

R4

42 V

10,000 Ω = 0.0305 A = 30.5 mA

𝐈𝐈𝐈𝐈 = I1 + I2 + I3 + I4 = 19.1 mA + 64.9 mA + 31.1 mA + 30.5 mA IT = 145.6 mA

R1

(305 V)2

1,600 Ω

93,025 V2

1,600 Ω = 5.814 W

R2

(305 V)2

4,700 Ω

93,025 V2

= 19.79 W

R3

(305 V)2

9,800 Ω

93,025 V2

9,800 Ω = 9.49 W

R4

(305 V)2

10,000 Ω

93,025 V2

10,000 Ω = 9.3 W

𝐏𝐏𝐏𝐏 =P1 + P2 + P3+ P4 = 5.814 W + 19.79 W + 9.49 W + 9.3 W

PT = 44.394 W

PT = IT × ET = 0.1456 A × 305 V = 44.4 W

6-78 Course 45517001

Figure 6-19 Practice Problem 19

RT = 250 Ω

ET = 120 V

E1 = 48 V E2 = 24 V E3 = 24 V E4= 48 V

IT = 0.48 A

I1 = 0.48 A I2 = 0.24 A I3 = 0.24 A I4 = 0.48 A

PT = 57.6 W

P1 = 23.04 W P2 = 5.76 W P3 = 5.76 W P4 = 23.04 W

Continued on next page.

Course 45517001 6-79

Practice (Cont’d)

CALCULATIONS

SOLVE

REQ =

R1 × R2 R1 + R2 or R2 × R3 R2 + R3

100 Ω × 100 Ω 100 Ω + 100 Ω

10,000 Ω2

RT = 100 Ω + 50 Ω + 100 Ω = 250 Ω

RT

120 V 250 Ω

= 0.48 A

IT = I1 = IREQ = I4 = 0.48 A

IT = I2 + I3

I2 = I3 = 0.24 A

(ET)2

RT

(120 V)2

250 Ω = 57.6 W

𝐏𝐏𝐏𝐏 = (I1)2 × R1 = (0.48 A)2 × 100 Ω = 23.04 W

𝐏𝐏𝐏𝐏 = (I2)2 × R2 = (0.24 A)2 × 100 Ω = 5.76 W

𝐏𝐏𝐏𝐏 = (I3)2 × R3 = (0.24 A)2 × 100 Ω = 5.76 W

𝐏𝐏𝐏𝐏 = (I4)2 × R4 = (0.48 A)2 × 100 Ω = 23.04 W

𝐄𝐄𝐄𝐄 = 120 V (Given)

𝐄𝐄𝐄𝐄 = I1 × R1 = 0.48 A × 100 Ω = 48 V

𝐄𝐄𝐄𝐄 = I2 × R2 = 0.24 A × 100 Ω = 24 V

𝐄𝐄𝐄𝐄 = I3 × R3 = 0.24 A × 100 Ω = 24 V

𝐄𝐄𝐄𝐄 = I4 × R4 = 0.48 A × 100 Ω = 48 V

6-80 Course 45517001

Figure 6-20 Practice Problem 20

RT = 200 kΩ = 2,000 Ω REQ = 1 kΩ = 1,000 Ω I1 = 0.15 A = 15 mA

EREQ = E2 = E3= 150 V

E1 = E4 = 75 V

ET = 300 V

I2 = 0.5 A = 50 mA I3 = 0.1 A = 100 mA I4 = 0.15 A = 150 mA

PT = 45 W

P1 = 11.25 W P2 = 7.5 W P3 = 15 W P4 = 11.25 W

Course 45517001 6-81

Practice (Cont’d)

CALCULATIONS

𝐑𝐑𝐑𝐑𝐑𝐑 =

R1 × R2 R1 + R2 or R2 × R3 R2 + R3

3 kΩ × 1.5 kΩ 3 kΩ + 1.5 kΩ

4.5 Ω2

4.5 Ω

REQ = 1 kΩ = 1,000 Ω

R1 = 500 Ω R2 = 1.5 kΩ = 1,500 Ω R3 = 3 kΩ = 3,000 Ω R4 = 500 Ω

𝐄𝐄𝐄𝐄 = 75 V (Given)

R1

75 V

500 Ω = 0.15 A = 150 mA

𝐄𝐄𝐄𝐄𝐄𝐄𝐄𝐄 = I1 × REQ = 0.15 A × 1,000 Ω = 150 V

𝐄𝐄𝐄𝐄𝐄𝐄𝐄𝐄 = E2 = E3 = 150 V

E4 = I4 × R4 = 0.15 A × 5000 Ω = 75 V

RT = 500 Ω + 1,000 Ω + 5,000 Ω = 2,000 Ω

R2

150 V

= 0.05 A = 50 mA

R3

150 V

1,500 Ω = 0.1 A = 100 mA

𝐈𝐈𝟒𝟒 =

R4

75 V 500 Ω

= 0.15 A = 150 mA

ET = E1 + REQ + E4 = 75 V+150 V+75 V = 300 V

PT = IT × ET = 0.15 A × 300 V = 45 W

P1 = I1 × E1 = 0.15 A × 75 V = 11.25 W P2 = I2 × E2 = 0.05 A × 150 V = 7.5 W P3 = I3 × E3 = 0.1 A × 150 V = 15 W P4 = I4 × E4 = 0.15 A × 75 V = 11.25 W

6-82 Course 45517001

Figure 6-21 Practice Problem 21

REQ = 4.15 kΩ = 4,150 Ω I3 = 2.50 mA = 0.0025 A

RT = 15,750 Ω

I1= 5.6 mA I2 = 3.1 mA I3 = 2.5 mA I4 = 5.6 mA IT = 5.6 mA

ET = 88.21 V

E1 = 38.08 V E2 = 23.25 V E3 = 23.25 V E4 = 26.9 V

PT = 0.49 W

P1 = 0.213 W P2 = 0.072 W P3 = 0.058 W P4 = 0.151 W

Continued on next page.

Course 45517001 6-83

Practice (Cont’d)

CALCULATIONS

SOLVE

𝐑𝐑𝐑𝐑𝐑𝐑 =

R1 × R2 R1 + R2 or R2 × R3 R2 + R3

7.5 kΩ × 9.3 kΩ

7.5 kΩ + 9.3 kΩ

69.75 Ω2

16.8 Ω

REQ = 4.15 kΩ = 4,150 Ω

Convert all R values to MΩ R1 = 6.8 kΩ = 6,800 Ω R2 = 7.5 kΩ = 7,500 Ω R3 = 9.3 kΩ= 9,300 Ω R4 = 4.8 kΩ= 4,800 Ω

RT = R1 + REQ + R4

RT = 6,800 Ω + 4,150 Ω + 4,800 Ω = 15,750 Ω

Convert mA to A I3 = 2.50 mA = 0.0025 A (Given)

E3 = I3 × R3 = 0.0025 A × 9,300 Ω = 23.25 V

𝐄𝐄𝐄𝐄𝐄𝐄𝐄𝐄 = E2 = E3 = 23.25 V

E1 = I1 × R1 = 0.0056 A × 6,800 Ω = 38.08 V E4 = I4 × R4 = 0.0056 A × 4,800 Ω = 26.88 V

R2

23.25 V

7,500 Ω

= 0.0031 A = 3.1 mA

𝐈𝐈𝐈𝐈𝐈𝐈𝐈𝐈 = I2+I3 = 0.0031 A + 0.0025 A = 5.6 mA

IT = I1 = IREQ = I4 = 5.6 mA = 0.0056 A

E3 = I4 × R4 = 0.0025 A × 9,300 Ω = 23.25 V

EREQ = E2 = E3 = 23.25 V

E1 = I1 × R1 = 0.0056 A × 6,800 Ω = 38.08 V E4 = I4 × R4 = 0.0056 A × 4,800 Ω = 26.88 V

𝐄𝐄𝐄𝐄 = I1+ I2 + I3 + I4 = 38.08 V + 23.25 V + 23.25 V +6.88 V

ET = 88.21 V

6-84 Course 45517001

Practice (Cont’d) 𝐏𝐏𝐏𝐏 =

E12

R1

(38.08 V)2

6,800 Ω = 0.213 W = 213 mW

R2

(23.25 V)2

7,500 Ω = 0.072 W = 72 mW

R3

(23.25 V)2

9,300 Ω = 0.058 W = 58 mW

R4

(26.88 V)2

4,800 Ω = 0.151 W = 151 mW

𝐏𝐏𝐏𝐏 =P1 + P2 + P3+ P4 = 213 mW + 72 mW + 58 mW + 151 mW PT = 494 mW

Course 45517001 6-85

Figure 6-22 Practice Problem 22

REQ = 290 kΩ = 2,900,000 Ω RT = 2,660 kΩ = 2,660,000 Ω

EREQ = E2 = E3 = 24.5 V

I2 = 0.01633 A = 16.33 mA I3 = 0.06806 A = 68.06 mA IREQ = 0.08439 A = 84.39 mA IREQ = IT= I1= I4= 0.08439 A = 84.39 mA E1 = 160.3 V E2 = 24.5 V E3 = 24.5 V E4 = 39.7 V

ET = 224.5 V

PT = 19.92 mW = 0.01992 W P1 = 0.0135 W = 13.5 mW P2 = 0.004 W = 0.45 mW P3 = 0.00167 W = 1.67 mW P4 = 0.00335 W = 3.35 mW PT = 0.01892 W = 18.92 mW

6-86 Course 45517001