Two resistors of resistances `2 Omega` and `6 Omega` are connected in parallel. This combination is then connected to a battery of emf 2 V and internal resistance `0.5 Omega` . What is the current flowing through the battery ?

Options

A. `4A`
B. `(4)/(3)A`
C. `(4)/(17)A`
D. 1A

Answer

D

Explanation

`R_(P) = (R_(1)R_(2))/(R_(1) + R_(2))`
`R_(p) = (2 xx 6)/((2+6)) = 1.5 Omega`
`I = (E)/(R_(p) + r) = (2)/(1.5 + 0.5) = 1A`