Correct Answer - D
Given, `R_(1)=2Omega, R_(2)=6 Omega`
E = 2 V and `r=0.5 Omega`
`R_(1), R_(2)` are in parallel series.
So, `(1)/(R )=(1)/(R_(1))+(1)/(R_(2))rArr (1)/(R )=(1)/(2)+(1)/(6)`
`rArr " " (1)/(R )=(3+1)/(6)`
`R = (6)/(4)Omega = 1.5 Omega`
Then, the current in the circuit,
`l=(E )/(r+R)=(2)/(0.5+1.5)=(2)/(2.0)=1 A`
(as internal resistance, r is in series with other resistant)