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Question
An electronics hobbyist is building a radio which requires 150 Ω in her circuit, but she has only 220 Ω, 79 Ω, and 92 Ω resistors available. How can she connect the available resistors to get the desired value of resistance?
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Solution
Available resistances = 220 Ω, 79 Ω 92 Ω
Case I:
If 3 resistors are connected in series, then
RS = R1 + R2 + R3 = 220 + 79 + 92 = 391
This value is greater than the required resistance so it is not possible.
Case II:
If 3 resistors are connected in parallel, then
`1/"R"_"p" = 1/"R"_1 + 1/"R"_2 + 1/"R"_3`
`1/"R"_"p" = 1/220 + 1/79 + 1/92` = 0.0279
Rp = 35.84 This does not meet the requirement.
Case III:
If R1 and R2 are connected in parallel and R3 in series
`1/"R"_"p" = 1/"R"_1 + 1/"R"_2 = 1/220 + 1/79 = 0.0172`
`=> "R"_"p" = 58.14 Omega`
This meets the requirement.
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