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.

Answer

A, B, C

Explanation

Let `I_2` current flow through the branch DCB
`:. By Kirchoff's junction law, current in branch DB will be
`I_2-I_1` as shown in figure.

Applying Kirchoff's law in loop BDAB
`+2(I_2 - I_1) + 1 + 1 xx I_2 - 2 + 2I_2 = 0`
`rArr 2I_1 - 5I_2 = -1` .........(i)
Applying Kirchoff's law in loop BCDB, we get
`-2 (I_2-I_1) + 1+I_1 -3 + 3I_1 = 0`
`rArr 3I_1 - I_2 = 1` .........(ii)
Solving (i) and(ii, we get `I_1 = 6//13 amp`
and `I_2 = 5/13 amp`
(i) To find the p.d. between B and D, we move from B to D.
`V_(B) + [5/13 - 6/13] xx 2 = V_(D) :. V_(B)-B_(D) = 2/13 volt`
(ii) p.d. across`G = 3- 6/13 xx 3 = (39-18)/13 = 21/13 volt `
`[:. the cell is in discharging mode ] `
p.d. across H = 1+ 1 xx 6/13 = 19 / 13 volt `
`[:. cell is in charging mode] ` .