A battery has internal resistance $\displaystyle R_i$

and open circuit terminal voltage $\displaystyle V_b$

Show that the power supplied to a resistive load cannot exceed $\displaystyle \frac{({\frac{V_b}{2})^2}}{R_i}$

well I only changed this to

$\displaystyle \frac{{\frac{V_b}{2}^2}}{R_i}

\Rightarrow

({\frac{V_b}{2})^2}\times\frac{1}{R_i}

\Rightarrow

\frac{V_b^2}{4R_i}

$

but not sure how you determine what cannot be exceeded

any suggest...

mod.... this should of been put in "other topics" its not an advanced topic.