Answer:
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
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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