As far as motor is concerned the power delivered is dissipated and can. be represented by a load, $R_{0}$. Thus
$$
I=\frac{V}{R+R_{0}}
$$
and $\quad P=I^{2} R_{0}=\frac{V^{2} R_{0}}{\left(R_{0}+R\right)^{2}}$
This is maximum when $R_{0}=R$ and the current
$I$ is then
$$
I=\frac{V}{2 R}
$$
The maximum power delivered is
$$
\frac{V^{2}}{4 R}=P_{\max }
$$
The power input is $\frac{V^{2}}{R+R_{0}}$ and its value when $P$ is maximum is $\frac{V^{2}}{2 R}$ The efficiency then is $\frac{1}{2}=50 \%$
