Power dissipated and supplied (Circuit Diagram)

[Enzo]

Homework Statement

(See attachment for diagram)

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?

 

[Enzo]

Diagram:

 

[MATLABdude]

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?

 

[The Electrician]

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.

 

[Enzo]

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)

2. Homework Equations

3. The Attempt at a Solution
My solution:

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?

 

[The Electrician]

I think the book is wrong. I get 150 volts also, by a different method.

 

[Defennder]

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.

 

[Enzo]

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..)?

 

[Defennder]

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.

 

[enian]

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.

 

[Defennder]

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.