$(i) R = r$ $(ii)$ Power in $R$ is $\frac{{{E^2}}}{{4R}}$
$(iii)$ Input power $\frac{{{E^2}}}{{2R}}$ $(iv)$Efficiency is $50\%$

Reason : The current flows towards the point of the higher potential, as it does in such a circuit from the negative to the positive terminal.

