In the circuit shown in figure E,F, G and H are cell of emf 2,1,3, and `1 V` respectively. The resistances 2,1,3 and 1(Omega)are their respective internal resistance .Calculate (a)the potential difference between B and D and (b) the potential differences across the terminals of each of each of the cells G and H.

Explanation

Suppose a current i_(1) goes in the branch BAD and current i_(2) in the branch DCB. The current in DB will be i_(1)-i_(2) from the junction law. The circuit with the currents shown is redrawn in figure. Applying the loop law to BADB we ge,
`(2(Omega))i_(1)-2V+1V+(1(Omega))i_(1)+(2(Omega))(i_(1)-i_(2))=0`
`(5(Omega))i_(1)-(2(Omega))i_(2)=1V. ...(i)`
Applying the same law to the loop DCBD,we get
`-3V+(3(Omega)i_(2)+(1(Omega))i_(2)+1V-(2(Omega))(i_(1)-i_(2))=0`
`or, -(2(Omega))i_(1)+(6(Omega))i_(2)=2V. ...(ii)`
From (i)and (ii),
`i_(1)=5/13A,i_(2)=6/13A.`
`so that `i_(1)-i_(2)=1/13A.`
the current in BD is from B to D .
(a)`V_(B)-V_(D)=(2(Omega)((1)/(13)A)=(2)/(13)V.`
`The potential difference across the cell G is
`V_(c)-V_(d)=-(3(Omega))i_(2)=3V.`
`=(3V-(18)/(13)V)=(21)/(13)V.`
`The potential difference across the cell H is
`V_(C)-V_(B)=(1(Omega))i_(2)+1V=(1(Omega))((6)/(13)A+1V=(19)/(13)V.`