Show how would you on three resistors, each ofresistance 9 Ω so that

1.	Show how would you on three resistors, each ofresistance 9 Ω so that the equivalent resistance of the combination is(i) 13.5Ω, (ii) 6Ω ?
Answer» 1. equivalent resistance = 13.5 ΩJoin two resistors in parallel and one in series as shown in 1st figure We know, in parallel , 1/R' = 1/R₁ + 1/R₂ Here , R₁ = R₂ = R₃ = 9Ω Now, 1/R' = 1/9Ω + 1/9Ω ⇒ R' = 9/2Ω = 4.5 Ω So, equivalent resistance , Req = R' + R₃ [ ∵ R' and R₃ in series ] ∴ Req = 4.5Ω + 9Ω = 13.5Ω 2. Equivalent resistance = 6Ω two resistors are in series which is parallel with 3rd resistor as shown in 2nd figure .now, R' = R₁ + R₂ = 9Ω + 9Ω = 18Ω [ R₁ and R₂ are in series ] now, equivalent resistance, Req = R'R₃/(R' + R₃) [ R' and R₃ are in parallel ] ∴ Req = 18 × 9/(18 + 9) = 18 × 9/27 = 6Ω

Show how would you on three resistors, each ofresistance 9 Ω so that the equivalent resistance of the combination is(i) 13.5Ω, (ii) 6Ω ?

Answer»

1. equivalent resistance = 13.5 ΩJoin two resistors in parallel and one in series as shown in 1st figure We know, in parallel , 1/R' = 1/R₁ + 1/R₂ Here , R₁ = R₂ = R₃ = 9Ω Now, 1/R' = 1/9Ω + 1/9Ω ⇒ R' = 9/2Ω = 4.5 Ω So, equivalent resistance , Req = R' + R₃ [ ∵ R' and R₃ in series ] ∴ Req = 4.5Ω + 9Ω = 13.5Ω

2. Equivalent resistance = 6Ω two resistors are in series which is parallel with 3rd resistor as shown in 2nd figure .now, R' = R₁ + R₂ = 9Ω + 9Ω = 18Ω [ R₁ and R₂ are in series ] now, equivalent resistance, Req = R'R₃/(R' + R₃) [ R' and R₃ are in parallel ] ∴ Req = 18 × 9/(18 + 9) = 18 × 9/27 = 6Ω

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