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The power dissipated in the circuit show...

The power dissipated in the circuit shown in the figure is 30 watts. The value of R is:

A

`10Omega`

B

`30Omega`

C

`20Omega`

D

`15Omega`

Text Solution

Verified by Experts

`P_("Total")=(V^(2))/(R_(eq))` So `R_(eq)=(V^ (2))/( P_("total"))=(10xx10)/(30)=(10)/(3)`
As `R_(eq)=(R_(1)R_(2))/(R_(1)+R_(2))`
`rArr (10)/(3)=(5R)/(5+R)=10Omega`
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