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If the power dissipated in 5Omega is 20W...

If the power dissipated in `5Omega` is 20W then power dissipated in `4Omega` is:-

A

4W

B

6W

C

10W

D

20W

Text Solution

Verified by Experts

`P=VI=V^(2)//R`, voltage constant
`Pprop1//R`

then power in ` 10Omega` will be 20W
when I constant then
`P=I^(2)RrArrPpropR`
`(P)/(10)=(4)/(10)rArrP=4W`
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