iv characteristic graph confusion
simulate this circuit &ndash; Schematic created using CircuitLab

Figure 1. The 10 V battery can be modelled as an ideal 10 V source with a 10 Ω series resistance.


With no load connected the output voltage will be 10 V. This is marked with an 'X' on the upper left of the VI graph.
If we short circuit the battery - seldom a good idea with real batteries - we will have 10 V across R1 and \$ I = \frac {V}{R} = \frac {10}{10} = 1\ \text A \$. This also is marked.


Now you should be able to fill in the rest of the points. Calculate the voltage drop across R1 at 0.2, 0.4, 0.6 and 0.8 A and plot the resulting terminal voltage on the chart.