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?

Answer 1

Available resistances = 220 Ω, 79 Ω 92 Ω

Case I:

If 3 resistors are connected in series, then

R_S = 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

R_p = 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.