When a current of $0.5 A$ is passed through two resistors in series, the potential difference between the ends of the series arrangement is $12.5 V$ . On connecting them in parallel and passing a current of $1.5 A$ , the potential difference across them is $6 V$ . The two resistances, in ohms, are

5, 20

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5, 15

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5, 10

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15, 20

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Solution

The correct option is B 5, 20
$\Rightarrow 12.5 = 0.5 (R_{1} + R_{2})$

$\Rightarrow R_{1} + R_{2} = 25$ ...........(1)

$\Rightarrow 6 = 1.5 (\frac{R_{1} R_{2}}{R_{1} + R_{2}}$

$4 (R_{1} + R_{2}) = R_{1} R_{2}$

$\Rightarrow R_{1} R_{2} = 100$ ............(2)

Using (1) and (2)..

$R_{1} + \frac{100}{R_{1}} = 25$

$R_{1}^{2} - 25 R_{1} + 100 = 0$

$\Rightarrow (R_{1} - 20) (R_{1} - 5) = 0 \Rightarrow R_{1} = 5; R_{2} = 20$

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