Two resistors R₁ and R₂ of 4 $Ω$ and 6 $Ω$ are connected in parallel across a battery. The ratio of power dissipated in them, P₁: P₂ will be.

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4 : 9

3 : 2

9 : 4

2 : 3

answer is B.

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Detailed Solution

In parallel combination potential differences(V) remains same
so, power dissipated in resistance(R)=[(V)²] /R
here potential is same so
P₁ = [(V)²] /4
P₂ = [(V)²] /6
P₁:P₂ = 3:2

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