Electrical – iv characteristic graph confusion


I am trying to understand how a iv graph for a battery rated at 10V with internal resistance of 10 Ohms would look like. I was thinking it would just be a constant 10V for the graph, however, I am unsure how internal the internal resistance would impact the terminal voltage.

Best Answer


simulate this circuit – 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.