Question

Two cells of emfs 1.5V and 2.0V and internal resistances 2Ω and 1Ω respectively have their negative terminals joined by a wire of 6Ω and positive terminals by a wire of 7Ω resistance. A third resistance wire of 8Ω connects middle points of those wires. Find the potential difference at the end of the third wire.
a) 2.25V b) 1.36V c) 1.26V d) 2.72V

Answer 1

see the diagram.

1.5 = 5.5 i2 + 8 (i1 + i2) + 3 i2   = 16.5 i2 + 8  i1      -- (1)
2.0 = 4.5 i1 + 8 (i1 + i2) + 3  i1  =  8 i2  +  15.5 i1    --- (2)

       (1) * 4 =>  6 = 66 i2 + 32 i1
       (2) *3  =>  6 = 24 i2  + 46.5  i1 
   subtract:    0 = 42  i2 - 14.5 i1
           i1 = 42/14.5 * i2  = 2.897  i2

   substitute it in (2), we get:
             2.0 = 8 * i2 +  15.5 * 2.897 * i2
             i2 = 0.0378 Amp
             i1 = 0.1095 Amp

Potential difference across the third wire with 8 ohm resistance:
           (i1 + i2) 8 volts = 1.178 volts.