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. part a let r be the ratio r1/r2. find r. round your answer to the nearest thousandth.

Since R₂>R₁ The ratio ( r ) R₁/R₂ = 0.127

when resistors are connected in series

Current = I

Req = R₁ + R₂

Vo = I ( R₁ + R₂ ) ( 1 )

When resistors are connected in parallel

Current = 10 I

Req =  R₁R₂ / (R₁ + R₂)

Vo = 10 I ( R₁R₂ / (R₁ + R₂) ) ( 2 )

Equating equation (1) and (2)

10 I ( R₁R₂ / (R₁ + R₂) )  = I ( R₁ + R₂ )

10 ( R₁R₂ ) = 2 (R₁R₂) + R₁² + R₂²

8 R₁R₂ = R₁² + R₂²   ( 3 )

Divide equation ( 3 ) by  R₂²

8 R₁ / R₂  = ( R₁/R₂)² +  1   ( 4 )

Given that ; r = R₁ / R₂

Equation ( 4 ) becomes

8 r = r ² + 1

 r ² - 8 r + 1 =0  

∴ Resolving the equation;  r = 0.127

since R₂>R₁

The ratio of  R₁ / R₂ = 0.127.

Hence we can conclude that the ratio ( r ) R₁/R₂ = 0.127  given that R₂>R₁

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r=0.127

Explanation:

When  connected in series

Current = I

When connected in parallel

Current = 10 I

We know that equivalent resistance

In series  R = R₁+R₂

in parallel  R= R₁R₂/(R₂+ R₁)

Given that voltage is constant (Vo)

V = I R

Vo = I (R₁+R₂)  ------------1

Vo = 10 I (R₁R₂/(R₂+ R₁)) -------2

From above equations

10 I (R₁R₂/(R₂+ R₁)) = I (R₁+R₂)  

10  R₁R₂ =  (R₁+R₂) (R₂+ R₁)

10  R₁R₂  = 2  R₁R₂  + R₁² + R₂²

8 R₁R₂  =     R₁² + R₂²

Given that

r =  R₁/R₂

Divides by R₂²

8R₁/R₂  = ( R₁/R₂)²+ 1

8 r = r ² + 1

r ² - 8 r+ 1 =0  

r= 0.127 and r= 7.87

But given that R₂>R₁  It means that r<1 only.

So the answer is r=0.127

i think its c if i'm wrong