Principles and Applications of Electrical Engineering
Principles and Applications of Electrical Engineering
6th Edition
ISBN: 9780073529592
Author: Giorgio Rizzoni Professor of Mechanical Engineering, James A. Kearns Dr.
Publisher: McGraw-Hill Education
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Question
Chapter 2, Problem 2.31HP
To determine

(a)

The total power dissipated by the ideal source.

Expert Solution
Check Mark

Answer to Problem 2.31HP

The power delivered for the current of 0mA is 0mW, for 10mA is 150mW, for 20mA is 300mW, for 30mA is 450mW, for 80mA is 1200mW and for 100mA is 1500mW .

Explanation of Solution

Calculation:

The formula for the total power supplied or delivered by the ideal voltage source is given by,

  pvs=vsiT ...... (1)

Substitute 0mA for iT and 15V for vs in the above expression.

  pvs=(15V)(0mA)=0W

Substitute 10mA for iT and 15V for vs in equation (1).

  pvs=(15V)(10mA)=150mW

Substitute 20mA for iT and 15V for vs in equation (1).

  pvs=(15V)(20mA)=300mW

Substitute 30mA for iT and 15V for vs in equation (1).

  pvs=(15V)(30mA)=450mW

Substitute 10mA for iT and 15V for vs in equation (1).

  pvs=(15V)(10mA)=150mW

Substitute 80mA for iT and 15V for vs in equation (1).

  pvs=(15V)(80mA)=1200mW

Substitute 100mA for iT and 15V for vs in equation (1).

  pvs=(15V)(100mA)=1500mW

Conclusion:

Therefore, the power delivered for the current of 0mA is 0mW, for 10mA is 150mW, for 20mA is 300mW, for 30mA is 450mW, for 80mA is 1200mW and for 100mA is 1500mW .

To determine

(b)

The power dissipated within the non ideal source.

Expert Solution
Check Mark

Answer to Problem 2.31HP

The power dissipated by the resistance of 100Ω for 0mA current source is 0mW, for 10mA current source is 10mW, for 20mA current source is 40mW, for current source 30mA is 90mW, for current source 80mA is 640mW and for 100mA current source is 1000mW .

Explanation of Solution

Calculation:

The formula for the power delivered by the non ideal power source is given by,

  pRS=iT2RS ...... (2)

Substitute 0mA for iT and 100Ω for Rs the above equation.

  pRS=(0mA)2(100Ω)=0W

Substitute 10mA for iT and 100Ω for Rs in equation (2).

  pRS=(10mA)2(100Ω)=10mW

Substitute 20mA for iT and 100Ω for Rs in equation (2).

  pRS=(20mA)2(100Ω)=40mW

Substitute 30mA for iT and 100Ω for Rs in equation (2).

  pRS=(30mA)2(100Ω)=90mW

Substitute 80mA for iT and 100Ω for Rs in equation (2).

  pRS=(80mA)2(100Ω)=640mW

Substitute 100mA for iT and 100Ω for Rs in equation (2).

  pRS=(100mA)2(100Ω)=1000mW

Conclusion:

Therefore, the power dissipated by the resistance of 100Ω for 0mA current source is 0mW, for 10mA current source is 10mW, for 20mA current source is 40mW, for current source 30mA is 90mW, for current source 80mA is 640mW and for 100mA current source is 1000mW .

To determine

(c)

The amount of power supplied by the load resistor.

Expert Solution
Check Mark

Answer to Problem 2.31HP

The power supplied to the load resistance of 100Ω for 0mA current source is 0mW, for 10mA current source is 140mW, for 20mA current source is 260mW, for current source 30mA is 360mW, for current source 80mA is 560mW and for 100mA current source is 500mW .

Explanation of Solution

Calculation:

The given diagram is shown in Figure 1

  Principles and Applications of Electrical Engineering, Chapter 2, Problem 2.31HP , additional homework tip  1

Apply KCL to the above circuit.

  vTvS+iTRS=0vT=vSiTRS ...... (3)

The formula for the load power is given by,

  pR0=vTiT ...... (4)

The conversion of mA into A is given by,

  1mA=103A

The conversion of 10mA into A is given by,

  10mA=10×103A

The conversion of 20mA into A is given by,

  20mA=20×103A

The conversion of 30mA into A is given by,

  30mA=30×103A

The conversion of 80mA into A is given by,

  80mA=80×103A

The conversion of 100mA into A is given by,

  100mA=100×103A

Substitute 15V for vs, 0mA for iT and 100Ω for Rs in equation (3)

  vT=15V(0× 10 3A)2(100Ω)=15V

Substitute 15V for vT and 0mA for iT in equation (4).

  pR0=(15V)(0mA)=0W

Substitute 15V for vs, 10mA for iT and 100Ω for Rs in equation (3).

  vT=15V(10× 10 3A)2(100Ω)=14V

Substitute 14V for vT and 10mA for iT in equation (4).

  pR0=(14V)(10mA)=140mW

Substitute 15V for vs, 20mA for iT and 100Ω for Rs in equation (3).

  vT=15V(20× 10 3A)2(100Ω)=13V

Substitute 13V for vT and 20mA for iT in equation (4).

  pR0=(13V)(20mA)=260mW

Substitute 15V for vs, 30mA for iT and 100Ω for Rs in equation (3).

  vT=15V(30× 10 3A)2(100Ω)=12V

Substitute 12V for vT and 30mA for iT in equation (4).

  pR0=(12V)(30mA)=360mW

Substitute 15V for vs, 80mA for iT and 100Ω for Rs in equation (3).

  vT=15V(80× 10 3A)2(100Ω)=7V

Substitute 7V for vT and 80mA for iT in equation (4).

  pR0=(9V)(80mA)=560mW

Substitute 15V for vs, 100mA for iT and 100Ω for Rs in equation (3).

  vT=15V(100× 10 3A)2(100Ω)=5V

Substitute 5V for vT and 100mA for iT in equation (4).

  pR0=(5V)(100mA)=500mW

Conclusion:

Therefore, the power supplied to the load resistance of 100Ω for 0mA current source is 0mW, for 10mA current source is 140mW, for 20mA current source is 260mW, for current source 30mA is 360mW, for current source 80mA is 560mW and for 100mA current source is 500mW .

To determine

(d)

The terminal voltage vT and the power supplied to the load resistor as a function of terminal current iT .

Expert Solution
Check Mark

Answer to Problem 2.31HP

The plot between the terminal voltage and the load current is shown in Figure 2 and plot for the load current and power delivered to the load is shown in Figure 3

Explanation of Solution

Calculation:

The plot for the terminal voltage and total current is shown in Figure 2.

  Principles and Applications of Electrical Engineering, Chapter 2, Problem 2.31HP , additional homework tip  2

The plot between for the power delivered to the load versus the total current is shown in Figure 3

  Principles and Applications of Electrical Engineering, Chapter 2, Problem 2.31HP , additional homework tip  3

Conclusion:

Therefore, the plot between the terminal voltage and the load current is shown in Figure 2 and plot for the load current and power delivered to the load is shown in Figure 3

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Chapter 2 Solutions

Principles and Applications of Electrical Engineering

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