Three resistors of values $2 Ω, 4 Ω, 6 Ω$ are connected is series. The equivalent resistance of combination of resistors is............

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Solution

$R_{e q} = R_{1} + R_{2} + R_{3}$ (In series)
$= 2 Ω + 4 Ω + 6 Ω = 12 Ω$

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