What will be the most suitable combination of three resistors A = 2Ω , B = 4Ω , C = 6Ω so that ( 3 22 ) Ω is equivalent resistance of combination?

If the 2Ω resistance and the 4Ω resistance are connected in parallel to each other, then R ′ = 4 + 2 4 × 2 = 3 4 Ω . If the resistance of 6Ω is now connected in series to the previous combination. R e q = 2 + 4 2 × 4 + 6 = 3 22 Ω

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What will be the most suitable combination of three resistors $A = 2Ω, B = 4Ω, C = 6Ω$ so that $(\frac{22}{3}) Ω$ is equivalent resistance of combination?

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Solution:

If the $2Ω$ resistance and the $4Ω$ resistance are connected in parallel to each other, then $R^{^{'}} = \frac{4 \times 2}{4 + 2} = \frac{4}{3} Ω$ . If the resistance of $6Ω$ is now connected in series to the previous combination.

$R_{e q} = \frac{2 \times 4}{2 + 4} + 6 = \frac{22}{3} Ω$

Stream:JEESubject:PhysicsTopic:Current ElectricitySubtopic:Equivalent Resistance

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