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.
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Answer:
R₂ = (4 + √15)R₁
Explanation:
Resistance in Series = R₁ + R₂
Voltage = V₀
Current Is= V₀/(R₁ + R₂)
Resistance in Parallel = 1/(1/R₁ + 1/R₂) = R₁*R₂/(R₁ + R₂)
Voltage = V₀
Current Ip= V₀(R₁ + R₂)/R₁*R₂
Ip = 10 Is
=> V₀(R₁ + R₂)/R₁*R₂ = 10 V₀/(R₁ + R₂)
=> (R₁ + R₂)² = 10R₁*R₂
=> R₁² + R₂² + 2R₁*R₂= 10R₁*R₂
=> R₁² + R₂² - 8R₁*R₂= 0
Let Say R₂ = kR₁ where k > 1
=> R₁² + k²R₁² - 8kR₁² = 0
=> R₁² ( 1 + k² - 8k) = 0
=> k² - 8k + 1 = 0
=> k = (8 ±√(64 - 4))/2
=> k = (8 ± 2√15)/2
=> k = 4 ± √15
=> k = 4 + √15 as k > 1
R₂ = (4 + √15)R₁
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