Question 1146599: The resistance R of two resistors in parallel is given. R is found to be 283.0 Ω, but if R1 is doubled and R2 halved, then R is found to be 291.0 Ω. Find R1
and R2
1/R = 1/R1 + 1/R2
Answer by Alan3354(69443)   (Show Source):
You can put this solution on YOUR website! The resistance R of two resistors in parallel is given. R is found to be 283.0 Ω, but if R1 is doubled and R2 halved, then R is found to be 291.0 Ω. Find R1
and R2
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R1*R2/(R1+R2) = 283 ------------- Eqn 1
R1*R2/(2R1 + R2/2) = 291
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R1*R2 = 283(R1+R2)
R1*R2 = 291*(2R1 + R2/2)
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283(R1+R2) = 291*(2R1 + R2/2)
283R1 + 283R2 = 582R1 + 145.5R2
137.5R2 = 299R1
R1 = 137.5R2/299
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Sub into Eqn 1
(137.5R2^2/299)/(137.5R2/299 + R2) = 283
(137.5R2^2/299)/(137.5R2/299 + 299R2/299) = 283
(137.5R2^2)/(137.5R2 + 299R2) = 283
(137.5R2^2)/(137.5R2 + 299R2) = 283
(137.5R2)/(137.5 + 299) = 283
137.5R2/436.5 = 283
137.5R2 = 123529.5
R2 = 898.4 ohms
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Solve for R1, take a break
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PS R2 is not a standard value. 910 ohms is the closest standard value.
It has to be special ordered.
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