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Textbook Question
Chapter 2, Problem 2.1P

Reduce each of the networks shown in Figure P2.1 to a single equivalent resistance by combining resistances in series and parallel.
* Denotes that answers are contained in the Student Solutions flies. See Appendix E for more information about accessing the Student Solutions

Chapter 2, Problem 2.1P, Reduce each of the networks shown in Figure P2.1 to a single equivalent resistance by combining

FIgure P2.1

Expert Solution
Check Mark
To determine

(a)

The equivalent resistance by combining resistance in series.

Answer to Problem 2.1P

The value of equivalent resistance is 20Ω.

Explanation of Solution

Calculation:

The required diagram is shown in Figure 1.

Modified Masteringengineering With Pearson Etext -- Standalone Access Card -- For Electrical Engineering: Principles & Applications, Chapter 2, Problem 2.1P , additional homework tip  1

Mark the resistance R1, R2, R3, R4, R5, R6 and R7 in the above figure.

The required diagram is shown in Figure 2.

Modified Masteringengineering With Pearson Etext -- Standalone Access Card -- For Electrical Engineering: Principles & Applications, Chapter 2, Problem 2.1P , additional homework tip  2

The value of resistance Req1 is calculated as,

  Req1=R7R6

Substitute 12Ω for R6 and 24Ω for R7 in the above equation.

  Req1=12Ω24Ω=( 12Ω)( 24Ω)12Ω+24Ω=8Ω

The required diagram is shown in Figure 3.

Modified Masteringengineering With Pearson Etext -- Standalone Access Card -- For Electrical Engineering: Principles & Applications, Chapter 2, Problem 2.1P , additional homework tip  3

The value of resistance Req2 is calculated as,

  Req2=R4+R5+Req1

Substitute 3Ω for R4, 4Ω for R5 and 8Ω for Req1 in the above equation.

  Req2=3Ω+4Ω+8Ω=15Ω

The required diagram is shown in Figure 4.

Modified Masteringengineering With Pearson Etext -- Standalone Access Card -- For Electrical Engineering: Principles & Applications, Chapter 2, Problem 2.1P , additional homework tip  4

The value of resistance Req3 is calculated as,

  Req3=R3Req2

Substitute 15Ω for Req2 and 30Ω for R3 in the above equation.

  Req3=15Ω30Ω=( 15Ω)( 30Ω)15Ω+30Ω=10Ω

The required diagram is shown in Figure 5.

Modified Masteringengineering With Pearson Etext -- Standalone Access Card -- For Electrical Engineering: Principles & Applications, Chapter 2, Problem 2.1P , additional homework tip  5

The value of resistance Req is calculated as,

  Req=R1+R2+Req3

Substitute 3Ω for R1, 7Ω for R2 and 10Ω for Req3 in the above equation.

  Req=3Ω+7Ω+10Ω=20Ω

The required diagram is shown in Figure 6.

Modified Masteringengineering With Pearson Etext -- Standalone Access Card -- For Electrical Engineering: Principles & Applications, Chapter 2, Problem 2.1P , additional homework tip  6

Conclusion:

Therefore, the value of equivalent resistance is 20Ω .

Expert Solution
Check Mark
To determine

(b)

The equivalent resistance by combining resistance in series.

Answer to Problem 2.1P

The value of equivalent resistance is 23Ω.

Explanation of Solution

Calculation:

The required diagram is shown in Figure 7.

Modified Masteringengineering With Pearson Etext -- Standalone Access Card -- For Electrical Engineering: Principles & Applications, Chapter 2, Problem 2.1P , additional homework tip  7

The value of resistance Req1 is calculated as,

  Req1=R7R8

Substitute 15Ω for R7 and 60Ω for R8 in the above equation.

  Req1=15Ω60Ω=( 15Ω)( 60Ω)15Ω+60Ω=12Ω

The required diagram is shown in Figure 8.

Modified Masteringengineering With Pearson Etext -- Standalone Access Card -- For Electrical Engineering: Principles & Applications, Chapter 2, Problem 2.1P , additional homework tip  8

The value of resistance Req2 is calculated as,

  Req2=R6+Req1

Substitute 6Ω for R6 and 12Ω for Req1 in the above equation.

  Req2=6Ω+12Ω=18Ω

The equivalent resistance is shown in Figure 9.

Modified Masteringengineering With Pearson Etext -- Standalone Access Card -- For Electrical Engineering: Principles & Applications, Chapter 2, Problem 2.1P , additional homework tip  9

The value of resistance Req3 is calculated as,

  Req3=R5Req2

Substitute 18Ω for Req2 and 9Ω for R5 in the above equation.

  Req3=9Ω18Ω=( 9Ω)( 18Ω)9Ω+18Ω=6Ω

The required diagram is shown in Figure 10.

Modified Masteringengineering With Pearson Etext -- Standalone Access Card -- For Electrical Engineering: Principles & Applications, Chapter 2, Problem 2.1P , additional homework tip  10

The value of resistance Req4 is calculated as,

  Req4=R4+Req3

Substitute 6Ω for R4 and 6Ω for Req3 in the above equation.

  Req4=6Ω+6Ω=12Ω

The equivalent resistance is shown in Figure 11.

Modified Masteringengineering With Pearson Etext -- Standalone Access Card -- For Electrical Engineering: Principles & Applications, Chapter 2, Problem 2.1P , additional homework tip  11

The value of resistance Req5 is calculated as,

  Req5=R3Req4

Substitute 12Ω for Req4 and 24Ω for R5 in the above equation.

  Req5=12Ω24Ω=( 12Ω)( 24Ω)12Ω+24Ω=8Ω

The required diagram is shown in Figure 12.

Modified Masteringengineering With Pearson Etext -- Standalone Access Card -- For Electrical Engineering: Principles & Applications, Chapter 2, Problem 2.1P , additional homework tip  12

The value of resistance Req is calculated as,

  Req=R1+R2+Req5

Substitute 10Ω for R1, 5Ω for R2 and 8Ω for Req5 in the above equation.

  Req=10Ω+5Ω+8Ω=23Ω

The required diagram is shown in Figure 13.

Modified Masteringengineering With Pearson Etext -- Standalone Access Card -- For Electrical Engineering: Principles & Applications, Chapter 2, Problem 2.1P , additional homework tip  13

Conclusion:

Therefore, the value of equivalent resistance is 23Ω .

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Chapter 2 Solutions

Modified Masteringengineering With Pearson Etext -- Standalone Access Card -- For Electrical Engineering: Principles & Applications

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