Two cells of emf 4.5 V and 6.0 V and internal resistance `6 Omega` and `3 Omega` respectively have their negative terminals joined by a wire of `18 Omega` and positve terminals by a wire of `12 Omega` resistance. A third resistance wire of `24 Omega` connects middle points of these wires. Using Kirchhoff's find the potential difference at the ends of this third wire.

Explanation

(1) Let the currents through the various arms of the network be as shown in fig.
(2) Applying KVL to closed mesh ABCDA, we have
`4.5 - 6 I_(1) - 18 I_(1) - 24 (I_(1) + I_(2)) = 0` ltbRgt `implies 48 I_(1) + 24 I_(2) = 4.5` `to` (i)
(3) For a closed mesh CDEFC, we have
`- 24 (I_(1) + I_(2)) - 12I_(2) + 6 - 3I_(2) = 0`
`24 I_(1) + 39 I_(2) = 6`
(4) `(ii) xx 2 - (i) implies 48 I_(!) + 78 I_(2) = 12`
`48 I_(1) + 24 I_(2) = 4.5`
`54 I_(2) = 7.5`
`:. I_(2) = (7.5)/(54) = 0.139 A to (iii)`
Substituting (iii) in (i), we get
`48 I_(1) + 78 xx 0.139 = 12`
`48 I_(1) = 12 - 10.84 = 1.158`
`I_(1) = (1.158)/(48) = 0.024 A`
(5) Potential difference across third wire
`= (I_(1) + I_(2)) xx 24 = 0.163 xx 24 = 3.912` volt.