In the circuit shown, E, F, G and H are cells of emf 2V, 1V, 3V and 1V respectively and their internal resistances are `2 Omega, 1 Omega, 3 Omega` and `1 Omega` respectively.

Explanation

(i) `2/13 V, (ii) V_G=21/13V, V_H=9/13V`
Applying kirchoff's second law in loop BADB, `2-2j_1-j_1-1-2(j_1-j_2)=0`
Similarly applying kirchoff's second law in loop BDCB
`2(j_1-j_2)+3-3j_2-j_2-1=0`………(ii)
Solving Eqs. (i) and (ii) we get
`j_1=5/13,j_2=6/13 and j_1-j_2=-1/13`
(i) Potential difference between B and D,
`V_B+2(j_1-j_2)=V_D`
`therefore V_B-V_D=-2(j_1-j_2)=2/13 V`
(ii) `V_G=E_G-j_2r_G=3-6/13 times 3=21/13 V, V_H=E_H+j_2r_h=1+6/13 times 1=19/13V`