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Tamil Nadu Board of Secondary EducationHSC Science Class 12

An electronics hobbyist is building a radio which requires 150Ω in her circuit, but she has only 220Ω, 79Ω, and 92Ω resistors available. - Physics

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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?

Numerical

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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Chapter 2: Current Electricity - Evaluation [Page 122]

APPEARS IN

Samacheer Kalvi Physics - Volume 1 and 2 [English] Class 12 TN Board
Chapter 2 Current Electricity
Evaluation | Q IV. 7. | Page 122
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