Question

show how would you join three resistor , each of resistance 9 Ω so that the equivalent resistance of combination is (a) 13.5Ω (b) 6Ω

Answer 1

Here's answer to your question........

Hope it will help you........

Answer 2

Hey there :).

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 helps you.... :)