A current of two amperes is flowing through a cell of e.m.f. 5 volts and internal resistance 0.5 ohm from negative to positive electrode. If the potential of negative electrode is 10V, the potential of positive electrode will be
V2−V1=E−ir=5−2× 0.5=4 volt
⇒ V2=4+V1=4+10=14 volt