In the network shown the potential difference between `A` and `B` is `(R = r_(1) = r_(2) = r_(3) = 1 Omega, E_(1) = 3V, E_(2) = 2V, E_(3) = 1V`)

Options

A. 1V
B. 2V
C. 3V
D. 4V

Answer

B

Explanation

At junction point E we have,
`I_(1)+I_(2)+I_(3)=0` ...(i)
Apply KVL for the loop `E E_(1) F E` we have,
`I_(1)-3+2-I_(2)=0`
`I_(1)-I_(2)=1`
`:.I_(2)=I_(1)-I` ...(ii)
Apply KVL for the loop `E E_(1) FE_(3)E` we have,
`I_(1)-I_(3)=2`
`:.I_(3)=I_(1)-2` ...(iii)
From equation (i), (ii) and (iii) we have,
`I_(1)=1A, I_(2)=0` and `I_(3)=-1A`
Now, potential difference between point A and B = potential difference between E and F
`:.V_(EF)=E_(2)-I_(2)r_(2)=2-0xx1=2V`