Two resistors of resistances `2Omega` and `6Omega` 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

Given, `R_(1)=2 Omega, R_(2)=6 Omega, E = 2 V, r = 0.5 Omega, l = ?`
`R_(1), R_(2)` are in parallel combination
So, `(1)/(R )=(1)/(R_(1))+(1)/(R_(2))=(1)/(2)+(1)/(6)`
`R =(6)/(4)Omega = 1.5 Omega`
The current flowing through the battery
`l = (E )/(r+R)=(2)/(0.5+1.5)=1A`