Reason : An ideal voltmeter draws almost no current due to very large resistance, and hence $(V)$ and $(a)$ will read zero.


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