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

i. Find the potential difference between B and D.
(ii) Calculate the potential differences across the terminals of each of the cells G and H.

Explanation

The circuit with the currents shown is redrawn in Applying the loop law to BADB, we get

`-2i_1+2 -1 -1i_1 - 2(i_1-i_2) = 0` or `- 5i_1 +2i_2 = -1 (i)`
Applying the same law to loop DCBD, we get
`-3 - 3i_2 - 3i_2 +1+(i_1 - i_2) = 0` or `2i_1 - 6i_2 = 2 (ii)`
From Eqs. (i) and (ii) we get
`i_1 = (1)/(12)A, i_2 = -(4)/(13)A`
`:. i_1 - i_2 = (5)/(13) A`
i. `V_D - V_B = (2Omega)((5)/(13))A = (10)/(13)V`
ii. Potential differences across the cell G is
`V_D - V_C = 3 +3i_2 = 3+3(-4//13) = 2713V`
Potential difference across the cell H is
`V_B -V_C =1 -i_2 = 1+ 4//13 = 17//13V.`
`(:. V = E + ir)`