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Two resistors R(1) and R(2) of 4 Omega a...

Two resistors `R_(1) and R_(2)` of `4 Omega and 6 Omega` are connected in parallel across a battery. The ratio of power dissipated in them, `P_(1) : P_(2)` will be:

A

`4:9`

B

`3:2`

C

`9:4`

D

`2:3`

Text Solution

Verified by Experts

The correct Answer is:
C
3:2
Explanation: Since resistors are connected in parallel so potential across them will be same. As a result power disslpated in the circult is
given by `P = ((V^(2))/(R))t`
`(P_(1))/(P_(2)) = (R_(2))/(R_(1))`
So, `= (6)/(4)`
`= (3)/(2) = 3:2`
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