Find Ratio r of R1/R2 in Two-Resistor Circuit

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SUMMARY

The discussion focuses on finding the ratio \( r \) of two resistors \( R1 \) and \( R2 \) in a circuit where \( R2 > R1 \). The relationship between the currents in series and parallel configurations is established, leading to the equation \( I_s(R1 + R2) = 10I_s\left(\frac{R1R2}{R1 + R2}\right) \). Through algebraic manipulation, the quadratic equation \( r^2R2^2 - 8rR2^2 + R2^2 = 0 \) is derived, resulting in the solutions \( r = 7.873 \) and \( r = 0.127 \). Given the condition \( R2 > R1 \), the valid ratio is confirmed as \( r = 0.127 \).

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Homework Statement



Two resistors of resistances R1 and R2 , with R2>R1 , are connected to a voltage source with voltage V0 . When the resistors are connected in series, the current is Is . When the resistors are connected in parallel, the current Ip from the source is equal to 10Is

Let r be the ratio R1/R2

Find r

Homework Equations


The Attempt at a Solution



First I found an expression for voltage in each circuit and then equated them to give me

Is*(R1+R2) = 10Is*((R1*R2)/(R1+R2))

Although from here I'm kind of stuck. I'm not sure how to, or even if I am able to manipulate my equation to get R1/R2. Unfortunately it seems my algebra is letting me down in a lot of physics work.
 
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Hi Yaaaldi! :smile:

(Remember, they're also both equal to V0)

There's various ways of solving this, but one is to replace R1 by rR2, to get a quadratic equation in R2. :wink:
 
r^2.R2^2 + rR2^2 + R2^2 = 0

For the quadratic equation in terms of r equation I got a=1, b=1, c=1

if I try to solve this I'll get complex roots..

Have I done something wrong?
 
Hi Yaaaldi! :smile:

(try using the X2 tag just above the Reply box :wink:)
Yaaaldi said:
r^2.R2^2 + rR2^2 + R2^2 = 0

How did you get that? :confused:
 
Nevermind.. somehow managed to forget to write the 10 infront of the rR2^2 on the RHS earlier.

now have a=1 b=-8 and c=1

Solved to get 7.873 and 0.127 for r

and as I know R2>R1

r=0.127

Thanks!
 

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