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The ratio of potential gradients is 1:2 of two potentiometers of same length are 2 $Ω$ and 4 $Ω$ respectively. The current flowing through them are in the ratio

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1:3

1:2

1:1

2:1

answer is D.

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Detailed Solution

$PG, \frac{V}{L} = \frac{iR}{L}$ ; L constant; $i \propto \frac{PG}{R}$
$\frac{i_{1}}{i_{2}} = \frac{{PG}_{1}}{{PG}_{2}} \times \frac{R_{2}}{R_{1}}; \frac{i_{1}}{i_{2}} = \frac{1}{2} \times \frac{4}{2} = 1$

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