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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Chapter 3, Problem 3.46HP
To determine

The value of the node voltages Va and Vb .

Expert Solution & Answer
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Answer to Problem 3.46HP

The value of node voltage Va is 2.708V and Vb is 3.415V .

Explanation of Solution

Calculation:

The given diagram is shown in Figure 1

  Principles and Applications of Electrical Engineering, Chapter 3, Problem 3.46HP , additional homework tip  1

Replace all the voltage sources except source V1 with their internal resistance and redraw the circuit.

The required diagram is shown in Figure 2

  Principles and Applications of Electrical Engineering, Chapter 3, Problem 3.46HP , additional homework tip  2

Apply nodal at the node Vb1 .

  V b1210+4+V b16+V b16=0Vb1[114+16+16]=214Vb1[1742]=214Vb1=617V

The expression to calculate the current through the resistance R1 is given by,

  IR1=2VVb110Ω+4Ω

Substitute 617V for Vb1 in the above equation.

  IR1=2V6 17V10Ω+4Ω=0.1176A

The expression to calculate the value of the voltage Va1 is given by,

  Va1=IR1(10Ω)+2V

Substitute IR1 for IR1 in the above equation.

  Va1=(0.1176A)(10Ω)+2V=0.824V

Short circuit the source V1 and consider only voltage source V2 .

The required diagram is shown in Figure 3

  Principles and Applications of Electrical Engineering, Chapter 3, Problem 3.46HP , additional homework tip  3

Apply nodal at the node Vb2 .

  V b210+4+V b26+V b246=0Vb2[114+16+16]=46Vb2[1742]=46Vb2=1.65V

The value of the voltage Va2 is given by,

  Va2=(Vb2)(10Ω10Ω+4Ω)

Substitute 1.65V for Vb2 in the above equation.

  Va2=(1.65V)( 10Ω 10Ω+4Ω)=1.178V

Replace all the sources shown in Figure 1 and consider only the current source, then redraw the circuit.

The required diagram is shown in Figure 4

  Principles and Applications of Electrical Engineering, Chapter 3, Problem 3.46HP , additional homework tip  4

Apply nodal at the node Va3 .

  V a310+2+V a3V b34=0Va3[110+14]Vb3[14]=2Va3[720]Vb3[14]=2   ....... (1)

Apply nodal at the node Vb3 .

  V b36I1+V b34+V b3V a34=0Va3[14]+Vb3[16+16+14]=2Va3[14]+Vb3[712]=2   ....... (2)

From equation (1) and equation (2) the node voltages are,

  Va3=4.71VVb3=1.412V

The expression to calculate the node voltage Va is given by,

  Va=Va1+Va2+Va3

Substitute 0.824V for Va1, 1.178V for Va2 and 4.71 for Va3 in the above equation.

  Va=0.824V+1.178V4.71=2.708V

The expression to calculate the node voltage Vb is given by,

  Vb=Vb1+Vb2+Vb3

Substitute 617V for Vb1, 1.65V for Vb2 and 1.412 for Vb3 in the above equation.

  Vb=617V+1.65V+1.412=3.415V

Conclusion:

Therefore, the value of node voltage Va is 2.708V and Vb is 3.415V .

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Principles and Applications of Electrical Engineering

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