The emf of the batteries E, F, G and H are 2 V, 1 V, 3 V and 1 V respectively. Their internal resistance are respectively `2 Omega, 1 Omega, 3 Omega and 1 Omega ` respectively. Calculate potential difference between B and D.

Explanation

Circuit with internal resistances of batteries is as under. Currents on different paths are shown in figure.
Near point D according to Kirchoff.s first law,
I = `I_(1) + I_(2) " "` .... (1)
For loop DBAD using Kirchoff.s second law,
`2I_(1) - 2 (I - I_(1)) - (I - I_(1)) = -2 + 1 `
`therefore 2I_(1) - 2I + 2I_(1) - I + I_(1) = - 1`
`therefore 5I_(1) -3I = -1 " "` ... (2)

For loop D-C-B-D using Kirchoff.s second law,
`therefore - 3I - I - 2I_(1) = - 3 + 1 `
`therefore - 4I - 4I_(1) = - 2 `
` therefore 2 I + I_(1) = 1 " " ` .... (3)
Multiplying equ. (2) by 2 and equ. (3) by 3 and eliminating l ,
` therefore 10I_(1) - 6I = -2 `
`6I + 3I_(1) = 3 `
`therefore 13 I_(1) = 1 rArr I_(1) = (1)/(13) `A
p.d. between points B and D =` V_(BD) = 2I_(1)`
` therefore V_(BD) = (2)/(13) `V