A current of 2A is flowing through a cell of e.m.f . 5V and internal resistance 0.5Ω from negative to positive electrode. If the potential of negative electrode is 10V, the potential of positive electrode will be :-
A
5V
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B
14V
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C
15V
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D
16V
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Solution
The correct option is B14V Potential drop across internal resistance is Vr=ir=2×0.5=1V Potential of positive electrode, V+=V−+E−Vr=10+5−1=14V