Power dissipated and supplied (Circuit Diagram)

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Discussion Overview

The discussion revolves around calculating power dissipation and supply in a circuit diagram involving resistors and voltage sources. Participants are addressing homework problems related to power calculations, including power dissipated in resistors and power supplied by voltage sources, as well as verifying the conservation of power in the circuit.

Discussion Character

  • Homework-related
  • Technical explanation
  • Debate/contested

Main Points Raised

  • One participant presents calculations for power dissipated in resistors and power supplied by voltage sources, questioning the accuracy of their results.
  • Another participant suggests that the currents used in the calculations may be incorrect and provides alternative values for the currents.
  • A different participant introduces a new question regarding voltage across a circuit and inquires about the validity of using source conversion as a solution technique.
  • Several participants express differing results for the voltage across a component, with one asserting their method yields a different answer than the textbook.
  • Discussion includes a query about the necessity of defining a ground in circuit analysis, with a participant noting the difference in approach when using simulation software.

Areas of Agreement / Disagreement

There is no consensus on the accuracy of the initial power calculations, as one participant suggests errors in the current values, while others provide different results for voltage calculations. The discussion remains unresolved regarding the correct approach to the homework problems.

Contextual Notes

Participants have not fully detailed their equations or methods for solving the circuit, leading to uncertainty about the correctness of their approaches. The discussion also reflects varying interpretations of circuit analysis techniques.

Who May Find This Useful

Students and individuals interested in circuit analysis, power calculations, and homework assistance in electrical engineering or physics may find this discussion relevant.

Enzo
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Homework Statement



(See attachment for diagram)
CircuitDiagram.JPG


Known Values:
R1 = 5
R2 = 15
R3 = 25
R4 = 5
V1 = 40
V2 = 20
I1 = 71/7
I2 = 25/7
I3 = 4

1) Find the power dissipated in each resistor

2) Find the power associated with each voltage source including the independent
source and specify whether it is delivered or dissipated.

3) Verify that the total delivered power equal the total dissipated power

Homework Equations


P=RI2
P=VI

The Attempt at a Solution



Power Dissipated:

(In the resistors)
P=RI2
R1*(I1- I3)2= 5(71/7-4)2 = 188.7
R2*(I1- I2)2=15(71/7-25/7)2 = 647.8
R3*(I3- I2)2=25(4-25/7)2 = 4.59
R4*(I3)2=5(4)2= 80

(By the voltage source in the middle of the branch)
P=VI
(25I2)(I2)=(25*25/7)(25/7)=318.8

Total Power Dissipated:1239.9

Power Supplied:
V1= 40(I1) = 405.7
V2= 20(I2) = 71.4
25I2* I1=905.6

Total Power Supplied: 1382.7

What am I doing wrong here in calculating the power supplied/absorbed?
 
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Diagram:

f_CircuitDiagm_ac5c192.jpg
 
Enzo said:

Homework Statement



(See attachment for diagram)
View attachment 15233

Known Values:
R1 = 5
R2 = 15
R3 = 25
R4 = 5
V1 = 40
V2 = 20
I1 = 71/7
I2 = 25/7
I3 = 4

1) Find the power dissipated in each resistor

2) Find the power associated with each voltage source including the independent
source and specify whether it is delivered or dissipated.

3) Verify that the total delivered power equal the total dissipated power

Homework Equations


P=RI2
P=VI

The Attempt at a Solution



Power Dissipated:

(In the resistors)
P=RI2
R1*(I1- I3)2= 5(71/7-4)2 = 188.7
R2*(I1- I2)2=15(71/7-25/7)2 = 647.8
R3*(I3- I2)2=25(4-25/7)2 = 4.59
R4*(I3)2=5(4)2= 80

(By the voltage source in the middle of the branch)
P=VI
(25I2)(I2)=(25*25/7)(25/7)=318.8

Total Power Dissipated:1239.9

Power Supplied:
V1= 40(I1) = 405.7
V2= 20(I2) = 71.4
25I2* I1=905.6

Total Power Supplied: 1382.7

What am I doing wrong here in calculating the power supplied/absorbed?

HINT: What's the current through R2 / the dependent power supply?
 
You haven't solved the network correctly. The currents are:

I1 = 193/7
I2 = 79/7
I3 = 12

You probably made an error in setting up the equations to solve the network, but since you didn't show them, I can't tell you exactly where your error is.
 
Thanks for the above responses - It was a tricky one for me...

Hope nobody minds, but I've got another question, but I don't want to congest the forum with my homework threads, so I'll just post it in here.

1. Homework Statement
Find the voltage across Vab (This is for a thevenins circuit).

Can I solve this using source conversion? (Its the technique that I'm most unfamiliar with, so I decided to try and solve this problem using it...I could have easily used nodal or mesh, I think)

f_Questionm_c636227.jpg


2. Homework Equations

3. The Attempt at a Solution
My solution:

f_MySolutionm_30cb92c.jpg


Answer in the book is 120V... I also want to check if I can actually do the first step of my solution..combing the resistors and the currents?
 

Attachments

  • f_Questionm_c636227.jpg
    f_Questionm_c636227.jpg
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I think the book is wrong. I get 150 volts also, by a different method.
 
I got 150V as well. I'm assuming the nodal voltage at Va is the same as that above the 60k resistor since no current passes through the 15k resistor.
 
Awesome guys..thanks for the responses...I was looking at this problem in complete bewilderment after the book gave me 120V.

I'm actually really curious as whether I can do the first step of the solution (ie, the add resistors and current sources that are in parallel like that..)?
 
Well you managed to get the same answer, so it is valid. This is known as source transformation.

I didn't use any source transformation in my working, by the way.
 
  • #10
I was wondering, why don't we have to define a ground? I am using a program Pspice to check my answers to similar problems and it always makes me define a ground. Yet, circuits like a flashlight obviously don't need a ground node. Perhaps I'm doing something wrong on the program.
 
  • #11
A ground node is just a reference node from which all other node voltages are evaluated. In circuits, only potential differences and not absolute potentials matter.
 

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