Equivalent formula adjustments

Corrosio

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Jun 23, 2022
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Have a nice day to everyone present.
I need to remember the basic skill - expressing a formula from a basic formula.

I cite Ohm's law as an example.:

To calculate the electric current I : I = U / R [A; V; Ω]
From this basic formula I can determine the relationship for the rest of the quantities.

Thus, the expression of electrical resistance "R" using an equivalent treatment: I = U/R // *U ~ I * U = R
I remember this very well.
Now I'm going to give a more complicated example that I can't handle.

The basic formula for calculating the output voltage "U_OUT" is:

U_OUT = (U_IN × R1) / (R1 + R2)

How to express the relation for resistor R1 from this formula? When I move the first expression with the occurrence of the unknown, after that the search will be unknown on both sides of the equation.

I try it like this:
U_OUT = (U_IN × R1) / (R1 + R2) // ×(R1 + R2)
U_OUT × (R1 + R2) = U_IN × R1 // ÷U_IN

U_OUT × (R1 + R2) / U_IN = R1


Now I have a value for R1 on both sides of the equation. How to combine the occurrence of the value R1 so that it is repeated only once in the equation and its occurrence is only on one side of the equation? I know there is a "blame before the parenthesis" operation, but I don't remember the exact procedure.
Can you please advise how to express an R1 unknown from a fraction when this unknown value is in the fraction both on the numerator side and on the denominator side?

Thank you very much.
 
Thus, the expression of electrical resistance "R" using an equivalent treatment: I = U/R // *U ~ I * U = R
I remember this very well.
Either I don't understand your notation here, or you are just wrong. Can you explain what this means, and why?

I try it like this:
U_OUT = (U_IN × R1) / (R1 + R2) // ×(R1 + R2)
U_OUT × (R1 + R2) = U_IN × R1 // ÷U_IN

U_OUT × (R1 + R2) / U_IN = R1
I'm beginning to understand your notation; after "//" you show an operation to be done on both sides, and "~" is followed by the result. This means that what I showed above is indeed wrong!

The work here is correct, but not useful, as you note.

What you need to do (after the first multiplication) is to expand (distribute) to eliminate parentheses (brackets), so that R1 will appear in two terms. Gather those terms together on one side, with all other terms on the other side, then factor and divide. This is certainly one of the harder problems of its type.

You should really find a good site to review basic algebra and work through it.
 
Have a nice day to everyone present.
I need to remember the basic skill - expressing a formula from a basic formula.

I cite Ohm's law as an example.:

To calculate the electric current I : I = U / R [A; V; Ω]
From this basic formula I can determine the relationship for the rest of the quantities.

Thus, the expression of electrical resistance "R" using an equivalent treatment: I = U/R // *U ~ I * U = R
I agree with Dr.Peterson. I would like to add, though, that your notation is very difficult to understand. Once you get the hang of it things work out all right but I'd simply use more than one line and use a couple of words:
I = U/R where I is in A, U is in V and R is in [imath]\Ohm[/imath]

I = U/R

I*U = R
(This is a typo... I*R = U.)

-Dan
 
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