Question

Show how would you join three resistors, each of resistance 9Ω so that the

equivalent resistance of the combination is (i) 13.5 Ω (ii) 6 Ω?



plz solve it step by step.​

Answer 1

Answer

1. equivalent resistance = 13.5 Ω

Join 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

hope this will help you....

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Answer 2

Answer:

1. equivalent resistance = 13.5 Ω

Join 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