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)=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)