Two cells of emfs 1.5 V and 2.0V internal resistance `2 Omega` and `1 Omega` respectively have their negative terminals joined by a wire of `6 Omega` and positive terminals by another wire of `4 Omega`. A third wire of `8 Omega` connects the mid points of these two wires. Find the current through `8 Omega` and the potential difference at the ends of the third wire.

Explanation

Let the distribution of current in the various arms be as shown in figure

According to Kirchoff's second rule, in closed circuit ABEFA
`2 I_(1) + 8 (I_(1)+I_(2)) + 3 I_(1) + 2I_(1)=1.5`
or `15 I_(1) + 8I_(2) = 1.5`
In closed circuit BCDEB
` - 2 I_(2) - 1 I_(2) - 3 I_(2) - 8 (I_(1) + I_(2)) = - 2`
or `8 I_(1) + 14 I_(2) = 2 or 4 I_(1) + 7 I_(2) = 1`
Multiplying (i) by 7 and (ii) by 8 and substracting, we have, `73 I_(1) = 2.5 or I_(1) = 2.5/73 = 5/146 A`
Putting the value of `I_(1)` in (ii), we get
`4 xx 5/146 + 7 I_(2) = 1 or I_(2) = 9/73 A`
Current flowing through `8 Omega` resistance
`= I_(1) + I_(2) = 5/146 + 9/73 = 23/146 A`
Potential difference across `8 Omega` resistance
`= (I_(1) +I_(2)) R = 23/146 xx 8 = 1.26 V`