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Home Q & A Board Science-Related Homework Help Physics
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lf1206 lf1206
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11 years ago
here's my question....When R1 and R2 are connected in series, Req is 16 ohms. When R1 and R2 are connected in parallel Req=3.0 ohms. what is the value of R1 and R2? I'm not sure how to do this question any help would be great.
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wrote...
#1
11 years ago
series, R1+R2 = 16
parallel, R1R2/(R1+R2) = 3
R1 = 16?R2

R2(16?R2)/(16?R2+R2) = 3
(16R2?R2²)/(16) = 3
16R2?R2² = 48
R2² ? 16R2 + 48 = 0
(R2?12)(R2?4) = 0

R2 = 4,12
R1 = 12, 4

.
RJW
wrote...
#2
11 years ago
Equation 1 -- resistors connected in series

R1 + R2 = 16


Equation 2 -- resistors connected in parallel

1/R1 + 1/R2 = 1/3

3(R2 + R1) = (R1)(R2)

From Equation 1, R1 = 16 - R2

and substituting this into the above Equation 2,

3[R2 + (16 - R2)] = (16 - R2)(R2)

3R2 + 48 -  3R2 = 16R2 - (R2)^2

48 = 16R2 - (R2)^2

Rewriting the above,

(R2)^2 - 16R2 + 48 - 0

Factoring the above,

R2 = 12 and R2 = 4

If

R2 = 12, then R1 = 4

and if

R2 = 4, then R1 = 12

In other words, R1 and R2 can have two interchangeable values.
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