Four resistance 40Ω  , 60Ω  , 90Ω   and 110Ω   make the arms of a quadrilateral ABCD. Across AC is a battery of emf 40 V and internal resistance negligible. The potential difference across BD is V is ______

In upper branch, ABCA​+40−40i1−60i1=0⇒40=100i1​⇒i1=410AIn lower branch,ADCA ​+40−90i2−110i2=0⇒40=200i2​⇒i2=420ALet say potential of A = 40V​⇒40−40i1−VB=0​⇒VB=40−40×410and​⇒40−90i2−VD=0​VD=40−90×420∴VD−VB=40×410−90×420=−2 V​⇒VB−VD=+2 V