How can I represent these resistances on a diagram?

In summary, the conversation is about a problem involving designing an electric grill and choosing a resistance circuit. The circuits shown have individual resistors with 2 Ohms of resistance and the terminals a-b are connected to a potential of 120V. The problem asks to find the power dissipated by each model of circuit and how long it takes to prepare delicious ramen using each model. The conversation then shifts to discussing the formulas used to find the power and equivalent resistance, and the correctness of a diagram provided. The conversation ends with a request for the person to identify two paths between points on the diagram for further discussion.
  • #1
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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  • #2
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.
 
  • #3
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?
 
  • #4
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
 
  • #5
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.
 
  • #6
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
 
  • #7
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?
 
  • #8
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?
 
  • #9
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 .
 

1. How do I represent resistances in a circuit diagram?

In a circuit diagram, resistances are represented by zigzag lines. The number next to the line indicates the value of the resistance in ohms.

2. Can I use symbols to represent resistances in a circuit diagram?

Yes, there are common symbols used to represent resistances in a circuit diagram. The most commonly used symbol is a zigzag line, but there are also symbols for specific types of resistances such as variable resistors and thermistors.

3. How do I indicate the direction of current flow in a circuit diagram?

In a circuit diagram, the direction of current flow is indicated by an arrow pointing in the direction of the flow. The arrow should be placed on the line representing the component through which the current is flowing.

4. Is there a standard for the placement of resistances in a circuit diagram?

Yes, there is a standard for the placement of resistances in a circuit diagram. Generally, resistances are placed in a vertical position, with the positive terminal at the top and the negative terminal at the bottom. However, in more complex diagrams, the placement may vary depending on the specific circuit design.

5. How do I represent a variable resistance in a circuit diagram?

A variable resistance can be represented in a circuit diagram by using a specific symbol, which looks like a zigzag line with an arrow pointing towards it. The arrow indicates that the resistance can be adjusted or varied.

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