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When two resistances are connected in series. The effective resistance is found to be 48 $Ω$ . Their resistances if their values are in the ratio 3 : 1.

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$36 Ω and 12 Ω$

$36 Ω and 36 Ω$

$24 Ω and 24 Ω$

$12 Ω and 48 Ω$

answer is B.

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Detailed Solution

$\begin{array}{l} R_{1} + R_{2} = 48 as \frac{R_{1}}{R_{2}} = \frac{3}{1} \Rightarrow R_{1} = 3 R_{2} \\ as R_{1} + R_{2} = 48; 3 R_{2} + R_{2} = 48; 4 R_{2} = 48 \\ R_{2} = 12 Ω; R_{1} = 3 R_{2} = 3 \times 12 = 36 Ω \end{array}$

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