How can I represent these resistances on a diagram?

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SUMMARY

The discussion centers on calculating the equivalent resistance and power dissipation in an electric grill circuit using resistors of 2 Ohms each connected to a 120V potential. Participants applied Ohm's Law and power formulas, specifically P = V^2/R and P = I^2R, to derive the power dissipated, which was calculated to be 7200 Watts for one model. The conversation emphasized the importance of accurately representing circuit diagrams and understanding series and parallel resistor configurations to solve for equivalent resistance.

PREREQUISITES
  • Understanding of Ohm's Law and power formulas (P = V^2/R, P = I^2R)
  • Familiarity with series and parallel resistor configurations
  • Ability to interpret and create electrical circuit diagrams
  • Knowledge of equivalent resistance calculations
NEXT STEPS
  • Learn how to derive equivalent resistance in complex circuits using Kirchhoff's laws
  • Study the impact of varying resistor values on total circuit resistance
  • Explore advanced circuit simulation tools like LTspice or Multisim
  • Investigate power loss in resistive circuits and methods to minimize it
USEFUL FOR

Electrical engineering students, hobbyists designing circuits, and professionals working on circuit analysis and optimization will benefit from this discussion.

GaussianSurface
1. The problem statement, all variables and given/known d
Hi have this problem but I don't know how to represent, however I've tried.
The problem says: You are designing an electric grill and you have to choose a resistance circuit between the circuits shown the figure. Consider that every individual resistor has 2 Ohms of resistance and the terminals a-b are connected to a potential of 120V.
And then:1) Find the power dissipated by each model of circuit.
2) If using the model 1 we can prepare a delicious ramen in 5 min. How much time it takes to prepare with the other models.

2. Homework Equations
Ohm's Law: 1/Re: 1/R1+ 1/R2 + 1/Rn...
Power: P= V^2/R
P= W/t then t= W/p

The Attempt at a Solution


It's the second photo
What I want to know is how the othee models are represented on a electric diagram and know if it's correct my first attemp with the model 1
 

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GaussianSurface said:
It's the second photo
Too faint for me to read. It needs more contrast.
Please only use images for diagrams and textbook extracts. For your workng, please take the trouble to type it in.
 
haruspex said:
Too faint for me to read. It needs more contrast.
Please only use images for diagrams and textbook extracts. For your workng, please take the trouble to type it in.
haruspex said:
Too faint for me to read. It needs more contrast.
Please only use images for diagrams and textbook extracts. For your workng, please take the trouble to type it in.
Then, what I've done is placed the resitances in such way they are like in the diagram I did and the equivalent resistance is 2 Ohms.
Then I applyed the formula of power in order to find the first statement.
Like this: Re= 1/1/4 + 1/ which is equal to 2 ohms.
Now I put it on the formula:
P= 60^2Ampers * (2 ohms) then i got 7200 whatts.
Is this correct?. Am using well the formulas?
And finally Is the diagram i did ok?
 
GaussianSurface said:
Then, what I've done is placed the resitances in such way they are like in the diagram I did and the equivalent resistance is 2 Ohms.
Then I applyed the formula of power in order to find the first statement.
Like this: Re= 1/1/4 + 1/ which is equal to 2 ohms.
Now I put it on the formula:
P= 60^2Ampers * (2 ohms) then i got 7200 whatts.
Is this correct?. Am using well the formulas?
And finally Is the diagram i did ok?
I forgot to say the Voltage is equal to 120 so, by applying the formula I= V/R it's 60 Ampers
 
GaussianSurface said:
1/1/4 + 1/
I have no idea what that means, but 2 Ohms and 7200 Watts are correct.
As I wrote, the diagram is too faint for me.
 
haruspex said:
I have no idea what that means, but 2 Ohms and 7200 Watts are correct.
As I wrote, the diagram is too faint for me.
Im sorry it was 1/4 I'm typing from my cellphone and I have some mistakes so now that I'm going I am good way, how can I find the power of the next models that are in the figure? they've got another nodes and so... Please help me
 
GaussianSurface said:
Im sorry it was 1/4
Then how did you get 2 Ohms?
For model 2, can you find two points connected by two paths each with a known resistance?
 
haruspex said:
Then how did you get 2 Ohms?
For model 2, can you find two points connected by two paths each with a known resistance?
By applying the formula of equivalent resistance.
And yes there's node in the middle of the model two... But how can it affect to the result?
 
GaussianSurface said:
By applying the formula of equivalent resistance.
And yes there's node in the middle of the model two... But how can it affect to the result?
To be able to discuss this we will need some labels on the diagram.
We already have a and b. Label the top left point C, top right D, lower left E, lower right F, central point O.
In model 2 there is also a vertex at the bottom. Call this P.
Identify two paths connecting two of these points with a known resistance on each path.
 
  • #10
haruspex said:
To be able to discuss this we will need some labels on the diagram.
We already have a and b. Label the top left point C, top right D, lower left E, lower right F, central point O and the
In model 2 there is also a vertex at the bottom. Call this P.
Identify two paths connecting two of these points with a known resistance on each path.
Mmm... Well, by considering what you wrote the two points woud be the vertex P and the lower point F
 
  • #11
GaussianSurface said:
Mmm... Well, by considering what you wrote the two points woud be the vertex P and the lower point F
I asked you to identify two paths, not two points. And I should have specified that to be useful they should paths in which the current flows one way, not different ways on different parts of the path.
 
  • #12
haruspex said:
I asked you to identify two paths, not two points. And I should have specified that to be useful they should paths in which the current flows one way, not different ways on different parts of the path.
I tried to solve it and the result was 2ohms as well as the first :/
 
  • #13
GaussianSurface said:
I tried to solve it and the result was 2ohms as well as the first :/
And I would still like you to identify two paths.
I'll make it easier. Find two paths between O and F for which the resistances are known.
 
  • #14
GaussianSurface said:
By applying the formula of equivalent resistance
Yes, I understand that you applied that, but I would like to see all the steps. From what you did post, I have reason to suspect you got the right answer by accident.
 
  • #15
haruspex said:
Yes, I understand that you applied that, but I would like to see all the steps. From what you did post, I have reason to suspect you got the right answer by accident.
I wrote the steps
Here they go once:
First I tried to transform the circle diagram into a "square diagram" (it's easier to me to see from there the resistances)
Second y saw that there were 2 ohms in serie so I added the exponents from the two sides.
Third i had 4 ohms in parallel I applyed the Ohm's once again but this time for paralles resistances and I found the R
Resistance equivalent.
Fourth by Ohms law i solved for I= VR so i found the ampers
Fifth i applyed the formula of power it g
 
  • #16
GaussianSurface said:
I wrote the steps
Here they go once:
First I tried to transform the circle diagram into a "square diagram" (it's easier to me to see from there the resistances)
Second y saw that there were 2 ohms in serie so I added the exponents from the two sides.
Third i had 4 ohms in parallel I applyed the Ohm's once again but this time for paralles resistances and I found the R
Resistance equivalent.
Fourth by Ohms law i solved for I= VR so i found the ampers
Fifth i applyed the formula of power it g
P= I^2R so i found finally the power dissipated
 
  • #17
GaussianSurface said:
Second y saw that there were 2 ohms in serie so I added the exponents from the two sides.
Third i had 4 ohms in parallel I applyed the Ohm's once again but this time for paralles resistances and I found the R
Ok, that is not really valid, as I suspected. You got away with it because all the resistances are the same. If I were to change one of them to 1 Ohm your method would give 12/7 instead of the correct answer, 5/3.

You cannot start with, say, the uppermost two in series (aCODb) because other parts of the circuit join at O.
You need to start with the two paths from a to O and the two paths from O to b. E.g. the path aCO and the path aEO can be combined using the parallel resistors rule.
 
  • #18
haruspex said:
Ok, that is not really valid, as I suspected. You got away with it because all the resistances are the same. If I were to change one of them to 1 Ohm your method would give 12/7 instead of the correct answer, 5/3.

You cannot start with, say, the uppermost two in series (aCODb) because other parts of the circuit join at O.
You need to start with the two paths from a to O and the two paths from O to b. E.g. the path aCO and the path aEO can be combined using the parallel resistors rule.
I know you want to explain by that way but I'm not used to see the diagramas with circles so I what I asked you was to represent them into a normal diagram in order to compute the resistances. I still do not understand your methodology, sorry.
 
  • #19
GaussianSurface said:
I know you want to explain by that way but I'm not used to see the diagramas with circles so I what I asked you was to represent them into a normal diagram in order to compute the resistances. I still do not understand your methodology, sorry.
Then please post a readable diagram that suits you better. Don't forget to label the nodes.
 
  • #20
haruspex said:
Then please post a readable diagram that suits you better. Don't forget to label the nodes.
That's what I came here. To know how to solve it :S
 
  • #21
GaussianSurface said:
That's what I came here. To know how to solve it :S
This is a homework forum. We do not just post answers. You post a diagram and I'll tell you if and where it is wrong.
 
  • #22
Part of the problem is that the original circuit diagrams are not very clear and mostly without any labelling of the nodes .

So I've re-drawn them to a more professional standard for you :

round resistance circuits.jpg


Can you see any way to proceed now ?

Hint 1 : Look for resistances that have no current flowing in them .
Hint 2 : Look for series and parallel combinations of resistances .
 

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