# Internal Resistance of a Cell

by mayan
Tags: cell, internal, resistance
 P: 87 I can't answer to the second question, but i can try to answer to the first: Let's assume that you have a battery with some internal resistance $$R_{i}$$ and some $$R_{load} = 50\Omega$$ and they are coupled in series. Now, let's say that the battery voltage is constant and is $$V_{b} = 10V$$. Since you want to know what happens to the internal resistance we'll write a loop equation, that describes the circuit: $$V_{b} = I\cdot (R_{i}+R_{load})$$ Re-writing the equation such that the current is a function of resistance gives: $$R_{i}(I) = \frac{V_{b}}{I} - R_{load}$$, now we can plot the function Looking at the function you can easily see that as the current drops, i.e you're using your battery for a long time, the internal resistance will increase. Hence the result: Resistance will increase when the current drops, that could be found by using ohm's law, but i just wanted to show a more practical example. P.S Y-axes is at - since the function by itself goes to infinity pretty quickly, so I had to go below zero to show the effect. Attached Thumbnails