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The Resistance of Two Resistors Joined in Series is 8ω and in Parallel is 1.5ω. Find the Value of the Two Resistances. - Physics

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Question

The resistance of two resistors joined in series is 8Ω and in parallel is 1.5Ω. Find the value of the two resistances.

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Solution

In series,  R1 + R2 = 8 Ω

In parallel `("R"_1"R"_2)/("R"_1 + "R"_2) = 1.5` Ω

∴ R1R2 = 8 × 1.5 = 12 Ω

Now (R1 - R2)2 = (R1 + R2)2 - 4R1R2 

∴ (R1 - R2)2 = (8)2 - 4 × 12

or (R1 - R2)2 = 64 - 48 = 16

or  R1 - R= 4 Ω

On solving equations (i) and (ii) R1 = 6 Ω and R2 = 2 Ω 

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