When three resistors of resistances 3,4,5 are connected in parallel, the currents through them are in the ratio

3:4:5

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5:4:3

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20:15:12

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12:15:20

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Solution

The correct option is B 20:15:12
Current flowing in one resistor is proportional to the effective value of other two resistance.
$R_{1} = 3; R_{2} = 4; R_{3} = 5$

$I_{1} : I_{2} : I_{3} = \frac{\frac{R_{2} R_{3}}{R_{2} + R_{3}}}{\frac{R_{1} + R_{2} R_{3}}{R_{2} + R_{3}}} : \frac{\frac{R_{3} R_{1}}{R_{1} + R_{3}}}{\frac{R_{2} + R_{1} R_{3}}{R_{1} + R_{3}}} : \frac{\frac{R_{1} R_{2}}{R_{1} + R_{2}}}{\frac{R_{3} + R_{1} R_{2}}{R_{1} + R_{2}}}$

$= \frac{\frac{20}{9}}{\frac{3 + 20}{9}} : \frac{\frac{15}{8}}{\frac{4 + 15}{8}} : \frac{\frac{12}{7}}{\frac{5 + 12}{7}}$

$= \frac{20}{47} : \frac{15}{47} : \frac{12}{47}$

$= 20 : 15 : 12$

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