The given situation is shown in figure

VA – VB = V \(\Rightarrow\) Terminal voltage of the battery
\(I = \frac{E}{R+ r} \) ...(i)
(I) Variation of V with R
\(V= E- Ir \Rightarrow V = E - \left(\frac{E}{R+r}\right) r\)
\(V= \frac{ER}{r +R}\)

(II) Variation of V with I
V = E – Ir
