Question

In which branch of the circuit shown in figure a `11 V` battery be inserted so that it dissipates minimum power. What will be the current through the `2Omega` resistance for this potential of the battery?

Answer 1

Suppose we insert the battery with `2Omega` resistance. Then we can take `2Omega` as the internal resistance (`R`) of the battery and combined resistance of the other two as the external resistance (`R`). The circuit in that cae is shown in figure,

Now,` P=E^2/(R+r)`
This power will be minimum where `R+r` is maximum and we can see that `(R+r)` will be maximum when the battery inserted with `6Omega` resistance as shown in figure.

Net resistance in this case is
`6+(2xx4)/(2+4)=22/3 Omega`
`=i=(11)/(22/3)=1.5A`
This current will be distributed in `2Omega and 4Omega` in the inverse ratio of their resistance.
`:. i_1/i_2=4/2=2`
` i_1=(2/(2+1))(1.5)=1.0A`