SOLVE THIS!!!!!!!

[Les Jones]

Hi Ian and John,
I thought I might learn a technique that I did not know.

Les.

 

[Inquisitive]

Les Jones My solution was very similar to yours. I just wrote the equations for each case, equated them, and solved for one variable. Then substituted to get the other variable.
I would have shown my work, but I am terrible with LaTex.


John

Then just do it in Spandex if'n it suites ya.

 

[Ratchit]

Two equations and two unknowns. jpanhalt and Les Jones are both correct.

Ratch

 

[jpanhalt]

Then just do it in Spandex if'n it suites ya.

I assume you are not aware of LaTeX : https://en.wikipedia.org/wiki/LaTeX

John

 

[Inquisitive]

Yes I'am. The software uses an upper case T and X in the name. However the polymer does not. It was an attempt at humor.

https://en.wikipedia.org/wiki/Latex

Les

 

[jpanhalt]

Yes I'am. The software uses an upper case T and X in the name. However the polymer does not. It was an attempt at humor.

https://en.wikipedia.org/wiki/Latex

Les

Capital "X" ?

I thought it was a very poor joke. Let's drop it.

Regards, John

 

[Les Jones]

Hi Spec,
You have the correct interpretation of the question. And there is a solution. Also it is NOT a trick question.

Les.

 

[spec]

Hi Spec,
You have the correct interpretation of the question. And there is a solution. Also it is NOT a trick question.

Les.

Yes Les thanks,

I realized that I had made a wrong calc and had deleted my post already.

John has the right answer- he is too fast for me

spec

 

[jpanhalt]

Les Jones

Here is a slightly different way to solve, as you wanted to see different ways. As I mentioned, I originally used the "equate" method to combine the equations. Another method is to solve for one variable in terms of the other(s) and then use substitution. In this instance, both methods are about the same amount of work. I have found the substitution method to be more useful in solving less obvious equations. Just personal preference. As I am really poor at LaTeX, here is an image. It refers to the schematic I posted above.

This site used to have a great and simple to use LaTeX editor, but it disappeared with the update of the software.

John

Les Jones It just occurred to me that my comment could be misinterpreted. It was not meant to be instructional to you! (embarrassed) John

 

[gophert]

Les Jones My solution was very similar to yours. I just wrote the equations for each case, equated them, and solved for one variable. Then substituted to get the other variable.
I would have shown my work, but I am terrible with LaTex.


John

Just snap a photo of your hand-work.

EDIT: Never mind, I didn't scroll down far enough.

 

[spec]

in a circuit two resistors each of 2 Ω is connected in parallel to a cell such that a current of 1.2 Amps flows through it now when those same two resistors are connected in series a current of 0.4 Amps is detected in the ammeter.(taking resistance of wire as zero)calculate
1)internal resistance of cell
2)EMF of cell
GOOD LUCK!!!!!!

As a general point of interest, assuming that the two schematics above correctly represent the two configurations that you describe, am I right is saying that practical solutions for VB and ri (battery internal resistance) are only possible if the current in CIRCUIT #1 is greater than the current in CIRCUIT #2 (RT represents the resulting resistance of the parallel and serial configurations of the two 2 Ohm resistors respectively)?

spec

 

[Les Jones]

Hi John,
As you say, similar to my method but not the same. We both set out to get an equation with only one unknown to to solve. I find sometimes you come across a way of solving a problem that you had not thought of. When I do I try to make a note of it for future use. (Although I very often forget where I have made a note.)

Les.

 

[misterT]

This site used to have a great and simple to use LaTeX editor, but it disappeared with the update of the software.

World Wide Web was invented so that scientists could share scientific information easily.. still can't do that properly.

 

[jpanhalt]

Some background: I used to have a link on my PC to an online editor. I discovered yesterday that link had become non-functional. A quick search yesterday didn't fix that, so I posted the image.
This editor is not too bad: https://www.codecogs.com/latex/eqneditor.php

John

 

[Ratchit]

View attachment 97416​

As a general point of interest, assuming that the two schematics above correctly represent the two configurations that you describe, am I right is saying that practical solutions for VB and ri (battery internal resistance) are only possible if the current in CIRCUIT #1 is greater than the current in CIRCUIT #2 (RT represents the resulting resistance of the parallel and serial configurations of the two 2 Ohm resistors respectively)?

spec

I must be missing something. I cannot understand why all the angst about finding the voltage and internal resistance of a circuit with two different operating points. The solution boils down to two linear equations solvable by a variety of elementary algebra methods. I can't understand what goes on here.

Ratch

 

[spec]

I must be missing something. I cannot understand why all the angst about finding the voltage and internal resistance of a circuit with two different operating points. The solution boils down to two linear equations solvable by a variety of elementary algebra methods. I can't understand what goes on here.

Ratch

No angst Ratch. I was just interested to establish what values for I1 and I2 you could specify and still solve for practical values of ri. In my schematic, of post 31, call the current of CIRCUIT #1, I1 and the current of CIRCUIT #2, I2.

Using simultaneous equations and all that you end up with ri = ( [I1*1]-[I2*4])/ (I2-I1).

From this equation you can see that:
(1) if I1 = I2 the denominator will be zero and the numerator will be -3, so ri will be minus infinity.
(2) if I1 is less than I2 the denominator will be positive and the numerator will be negative so ri will be negative.
Thus I1 must be more than I2 for the solution to produce a real value for re.
That is all.

spec

 

[Ratchit]

No angst Ratch. I was just interested to establish what values for I1 and I2 you could specify and still solve for practical values of ri. In my schematic, of post 31, call the current of CIRCUIT #1, I1 and the current of CIRCUIT #2, I2.

Using simultaneous equations and all that you end up with ri = ( [I1*1]-[I2*4])/ (I2-I1).

From this equation you can see that:
(1) if I1 = I2 the denominator will be zero and the numerator will be -3, so ri will be minus infinity.
(2) if I1 is less than I2 the denominator will be positive and the numerator will be negative so ri will be negative.
Thus I1 must be more than I2 for the solution to produce a (real) positive value for re.
That is all.

spec

If you are willing to accept negative resistance, then your currents can be anything you want. Otherwise, you cannot expect to have more current after increasing the resistance.

Ratch

 

[Les Jones]

Hi Spec,
Your equation gives the correct value for the internal resistance of the cell .
ri = ( [I1*1]-[I2*4])/ (I2-I1). gives the correct answer.
= (1.2 * 1)- (0.4 *4)/(0.4-1.2)
= 1.2-1.6/-0.8
= -0.4/-0.8
= 0.5

Les.

 

[JoeJester]

That must have been an old lead-acid battery question.


I like that question.

 

[spec]

That must have been an old lead-acid battery question.


I like that question.

Hi Joe,

No it wasn't a question that I knew of before. As I said, I was just interested to know what two currents you could specify with out getting into a pickle. Right away, I suspected that tech had tried to catch us out, but no he was too honest for that