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 3.1HP

Use node voltage analysis to find the voltages V 1 and V 2 for the circuit of Figure P3.1.
Chapter 3, Problem 3.1HP, Use node voltage analysis to find the voltages V1 and V2 for the circuit of Figure P3.1.

Expert Solution & Answer
Check Mark
To determine

The voltages V1 and V2 of the circuit using node voltage analysis.

Answer to Problem 3.1HP

The value of voltage V1 is 4.8V and the value of voltage V2 is 2.4V .

Explanation of Solution

Calculation:

The given diagram is shown in Figure 1

  Principles and Applications of Electrical Engineering, Chapter 3, Problem 3.1HP

Apply KCL at the node V1 in the above circuit.

  4V13V1V21=04V1+3V23=44V1+3V2=12   .......... (1)

Apply KCL at the node V2 .

  V21+1+V2V11+V22=0V2+V2V1=02V2V1=0V1=2V2   .......... (2)

Substitute 2V2 for V1 in equation (1)

  4(2V2)+3V2=128V2+3V2=12V2=125VV2=2.4V

Substitute 2.4V for V2 in equation (2).

  V1=2(2.4V)=4.8V

Conclusion:

Therefore, the value of voltage V1 is 4.8V and the value of voltage V2 is 2.4V .

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Chapter 3 Solutions

Principles and Applications of Electrical Engineering

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