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`) A. 1VB. 2VC. 3VD. 4V

In the network shown the potential difference between `A` and `B` is `

1.	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`) A. 1VB. 2VC. 3VD. 4V
Answer» Correct Answer - B 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`

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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`) A. 1VB. 2VC. 3VD. 4V

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