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Judge the equivalent resistance when the following are connected in parallel − (a) 1 Ω and 106Ω, (b) 1 Ω and 103Ω and 106Ω. - Science

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Question

Judge the equivalent resistance when the following are connected in parallel − (a) 1 Ω and 106Ω, (b) 1 Ω and 103Ω and 106Ω.

Numerical

Solution

(a) When 1 Ω and 106 Ω are connected in parallel:
Let R be the equivalent resistance.

R2 = 106 Ω    .....[given]

`1/"R" = 1/1 + 1/10^6`

`1/"R" = (10^6 + 1)/10^6`

R = `10^6/(1 + 10^6)` ≈ 1 Ω

R ≈ 1 Ω

Therefore, equivalent resistance ≈ 1 Ω

(b) When 1Ω, 103 Ω and 106 Ω are connected in parallel:

Let R be the equivalent resistance.

`1/"R" = 1/"R"_1 + 1/"R"_2 + 1/"R"_3`

R1 = 1 Ω    .....[given]

R2 = 103 Ω

R3 = 106 Ω

`1/"R" = 1/1 + 1/10^3 + 1/10^6`

`1/"R" = (10^6 + 10^3 + 1)/10^6`

`1/"R" = 1001001/1000000`

R = `1000000/1001001` ≈ 0.999 Ω ≈ 1 Ω

R ≈ 1 Ω

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Chapter 12: Electricity - Intext Questions [Page 216]

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NCERT Science [English] Class 10
Chapter 12 Electricity
Intext Questions | Q 1 | Page 216

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