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

Options

A. `10Omega`
B. `30Omega`
C. `20Omega`
D. `15Omega`

Explanation

`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`