The power dissipated (in watt) in `3Omega` resistance in the following circuit is

Answer

0.75

Explanation

From circuit `(1)/(R)=(1)/(3)+(1)/(6)=(2+1)/(6)=(3)/(6) Omega`
`rArr R_(1)=2 Omega`

The internal resistance of battery `=1 Omega`
So, the equivalent resistance of circuit
`=2 Omega+1 Omega=3 Omega`
The current in the battery,
`I=(V)/(R)=(4.5)/(3)=1.5A`
Current in `3 Omega` is `(1.5)/(2)xx(2)/(3)A=(1)/(2)A`
The power dissipated in `3 Omega` resistance,
`P=I^(2)xxR`
`=((1)/(2))^(2)xx3`
`=(1)/(4)xx3`
`=0.75W`