In the circuit shown (figure), calculate the current through 6 V battery.

Equivalent circuit:
$\mathrm{I}=\frac{5-5}{8}=0\text{ or }{\mathrm{V}}_{\mathrm{A}}-{\mathrm{V}}_{\mathrm{B}}=0$
So, current will be only in the smaller loop containing the 6 V and 4 V battery. So, current through the 6 V battery is 
${\mathrm{I}}^{\mathrm{\prime }}=\frac{6-4}{4}=\frac{1}{2}\mathrm{A}$