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

Answer

C

Explanation

Resistors `2 Omega` and `6 Omega` are connected in parallel. Their equivalent resistance is
`(1)/(R_(p))=1/2+ 1/6=4/6= 2/3`
`R_(p)=3/2 Omega= 1.5 Omega`
Current flowing through the battery is
`I=(epsilon)/(R_(p)+r)=(2V)/(1.5 Omega+0.5 Omega)=1A`