Solving for R2 in v1/v2 = R2/R1+R2 Equation

  • Thread starter simpComp
  • Start date
In summary, the conversation discusses how to solve for R2 in the equation v1/v2 = R2/(R1+R2). The process involves factoring out R2 on the right hand side to get (1/R1 + 1)R2, and then using cross multiplication to isolate R2 on one side of the equation. The final solution is R2 = v1R1/(v2-v1).
  • #1
simpComp
44
0
Hello,

I have the following equation:

v1/v2 = R2/R1+R2

How do we solve for R2?

I don't know how to isolate R2 on one side of the equation. All help is appreciated?

thanks
 
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  • #2
In the right hand side, you can factor out the R2:

(1/R1 + 1)R2.

Now divide both sides by (1/R1+1).
 
  • #3
Is your equation

$$\frac{v_1}{v_2} = \frac{R_2}{R_1+R_2}$$

or

$$\frac{v_1}{v_2} = \frac{R_2}{R_1}+R_2?$$

The way you wrote it, it looks like the second one. You need parentheses if you mean the first one.

You also need to show an attempt before anyone can help you with this question.
 
  • #4
v1/v2 = R2/(R1+R2)

I don't understand how you factor R2? I gather when you say to factor R2 you mean to factor R2/R2 ?

Therefore:

How many times does R2/R2 go into >>>> (R2/(R1+R2))

Here below, R2/R2 goes once into R2/(R1+R2) leaving:
v1/v2 = 1/(R1 + 1) R2

right?
 
  • #5
How can R2/R1+R2 be factored out to (1/R1 + 1)R2. ? This is not true!

Lets apply numbers and see... let R1 be 2 and R2 be 10:

10/(2+10) which equals to 0.83

factoring the way it is suggested we get:
10(1/(2+1)) which equals to 3.333 ?

Bof ! confused?
 
  • #6
If you use the standard "order of operations" and look at your notation from the first post, then I think you will see that you were missing parentheses. I now realize what you were actually trying to write, so what I suggest is that you "cross multiply".
 
  • #7
Yes I am sorry I was missing parenthesis!

v1/v2 = R2/(R1+R2)

so when you say cross multiply... do you mean like this:
(R1 + R2) (v1/v2) = (R2/(R1+R2)) (v2)
v1(R1 + R2) = R2v2
v1R1 + v1R2 = R2v2

But now R2 is on both sides, what I really need is to isolate R2 so its on one side only!
Then what is the next step?

.. I am really no good at this...

thanks
 
Last edited:
  • #8
Okay I think I got it:

v1/v2 = R2/(R1 + R2)
(R1 + R2) (v1/v2) = (R2/(R1+R2)) (v2)
v1(R1 + R2) = R2v2
v1R1 + v1R2 = R2v2
v1R1 = R2v2 - v1R2
v1R1 = R2(v2 - v1)

v1R1/(v2-v1) = R2(v2 - v1)/ (v2 - v1)

R2 = v1R1/(v2-v1)

Thanks!
 
  • #9
simpComp said:
Okay I think I got it:

v1/v2 = R2/(R1 + R2)
(R1 + R2) (v1/v2) = (R2/(R1+R2)) (v2)
v1(R1 + R2) = R2v2
v1R1 + v1R2 = R2v2
v1R1 = R2v2 - v1R2
v1R1 = R2(v2 - v1)

v1R1/(v2-v1) = R2(v2 - v1)/ (v2 - v1)

R2 = v1R1/(v2-v1)

Thanks!

Yes, that looks right.

I hope the confusion between you and Vargo will help you remember to be more careful with parentheses in the future! :smile:
 
  • #10
I hope the confusion between you and Vargo will help you remember to be more careful with parentheses in the future!

yes... thanks!

sc
 

1. What does the equation "v1/v2 = R2/R1+R2" represent?

The equation represents the relationship between two voltages (v1 and v2) and two resistances (R1 and R2) in a series circuit. It is known as the voltage divider equation and is commonly used in electronic circuits.

2. How is the equation derived?

The equation is derived using Ohm's Law, which states that the current through a conductor between two points is directly proportional to the voltage across the two points. By rearranging the formula for voltage (V = IR), we can derive the voltage divider equation.

3. What is the significance of the ratio between the resistances in the equation?

The ratio between the resistances (R2/R1+R2) determines the proportion of voltage dropped across each resistor in the circuit. It determines the output voltage (v2) relative to the input voltage (v1).

4. Can the equation be applied to parallel circuits?

No, the voltage divider equation is only applicable to series circuits. In a parallel circuit, the voltage across each branch is the same, regardless of the resistance, so the ratio between the resistances does not affect the output voltage.

5. What are some practical applications of the voltage divider equation?

The voltage divider equation is commonly used in electronic circuits for various purposes, such as dividing a voltage for measurement, adjusting the output voltage of a power supply, and creating reference voltages for sensors and other electronic components.

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