In the circuit shown in fig. E, F, G and H are cells of emf 2, 1, 3 and 1 volts and their internal resistances are 2, 1, 3 and 1 ohm respectively. Calculate.
(i) The potential difference between B and D is given by 
Volt then valu of n will be.
(ii) The ratio of potential difference across the terminals of the cell G to cell H is given by 
the value of n will be.
Text Solution
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[(i) 11, (ii) 19]
Sol Applying Kirchhoff ’ s second law in loop ADBA:
2 – 2i 1 – i 1 – 1 – 2(i 1 – i 2 ) = 0 .........
Similarly applying Kirchhoff ’ s second law in loop BDCB
2(i 1 – i 2 ) + 3 – 3i 2 – i 2 – 1 = 0.........
Solving Equation and we get,
i 1 = 
, i 2 = 
and i 1 – i 2 = – 
(i) Potential difference between B and D.
V B + 2(i 1 – i 2 ) = V D
V B – V D = – 2(i 1 – i 2 ) = 
volt
(ii) V G = E G – i 2 r G = 3 – 
× 3 = 
volt
V H = E H + i 2 r H = 1 + 
× 1 = 
volt
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