Show how would you join three resistors each of resistance 9 ohm so that the equivalent resistance of the combination is 13.5 ohm and 6 Ohm
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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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Answer:
1) Join two resistors of 9 ohms in parallel with third resistor in series
1/Rp = 1/9 + 1/9
Rp = 4.5 ohms
Now in series,
Rs= 4.5 ohms + 9 ohms
= 13.5 ohms
2) Now, Equivalent resistance is 6 ohms
Req = R'Re/(R'+ R3) [R'and R3 are in parallel
Req=18×9/(18+9)
= 18×9/27
= 6 ohms
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