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

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2015-04-02T05:29:42+05:30

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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.


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