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

`0.29 V,3.143 V,3.34 V`

Explanation

The distribution of current in the network following kircbboff's first rule has been
According to kirchbboff's second rule in the closed loop `ABDA`

`-1+2=1(I-I_(1))-2I_(1)+(I-I_(1))2`
or `1 = 3I - 5I_(1)` ....(i)
In the closed loop `BCDB`
`4 - 2 = 1 xx l + 2 xx I + 2 I_(1) or 2 = 3I + 2 I_(1) `.....(ii)
Subtricting (i) and (ii) we get
`1 = 7 l_(1) or l_(1) = (1)/(7) A`
From (i) `1 = 3I - 5 xx (1)/(7)`
or `3I = 1 + (5)/(7) = (12)/(7) or l = (12)/(7 xx 3) = (4)/(7) A`
(i) Potential difference between D and B
`= I_(1) xx 2 = (1)/(7) xx 2 = (2)/(7) V = 0.29 V`
(ii) potential difference across the terminals of `G` (receiving current for charging)
`V_(C) = epsilon + lr = 2 + (4)/(7) xx 2 = (22)/.(7) = 3.143 V`
potential difference across the terminals of `H`
`V_(H) = epsilon ' - lr' = 4 - (4)/(7) xx 1 (24)/(7) = 3.34 V`