In the circuit as shown in figure,
`E_(1)=2V , E_(2)=2V, E_(3)=1V, and R=r_(1)=r_(2)=r_(3)Omega`.
The potential difference between points A and B will be

Options

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

Answer

B

Explanation

Applying junction law at point E
`I_(1)+I_(2)+I_(3)=0 " "...(i)`
Apply KVL for loop `E E_(1)FE_(2)E`
`I_(1)-3+2-I_(2)=0`
or `I_(1)-I_(2)=1`
`:. I_(2)=I_(1)-1" "...(ii)`
Apply KVL for the loop `E E_(1) FE_(3) E`, we have
`I_(1)-I_(3)=2`
`:. I_(3)=I_(1)--2`
From Eqa. (i),(ii), (iii) we, have M
`I_(1)=1A,I_(2)=0 and I_(3)=-1A`
Now, potntial difference between point A and B= potential difference between E and F+ potential difference between A and E.
`:. V_(EF)=E_(2)-I_(2)r_(2)+IR`
`=2-0xx1+0=2V`