




$(2)$ The product of a volt and an ampere is a joule/second.
$(3)$ The product of volt and watt is horse power.
$(4)$ Watt-hour can be measured in terms of electron volt.
State if
$(i) R = r$ $(ii)$ Power in $R$ is $\frac{{{E^2}}}{{4R}}$
$(iii)$ Input power $\frac{{{E^2}}}{{2R}}$ $(iv)$Efficiency is $50\%$