The maximum number of inputs an output can drive reliably is called fan- out.
Added by Edward F.
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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 $\frac{1}{2}=50 \%$
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