Show how would u join three resistor , each of resistance 9ohm so that the equivalent resistance of the combination is (I) 13.5 (2) 6ohm
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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
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
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