# The shape of the graph of the equation: $V= -r I + E$

I have recently collected data (for a school experiment) in order to measure the EMF and the internal resistance of a solar cell. The data complied with the equation: $V = -rI + E$, i.e. the voltage decreased as the current increased and I have a negative gradient. However, I am having some trouble as to explaining why the graph is the way it is, that is, why is there a negative gradient?

Here's my rationale: Using Ohm's law, we can see that decreasing the (external) resistance increases the current flowing throughout the whole circuit (it is a simple series circuit with just a solar cell, ammeter, variable resistor and a voltmeter (in parallel, obviously)). Now, the voltage drop across the solar cell should increase as well (assuming its internal resistance in constant) because $v = V_r = Ir$, where $v$ is the 'lost volts'. For this reason the voltage drop across the load must decrease(which is what we see from the graph) because $E = V + v$. Thus, the voltage drop across the resistor decreases as current increases.

I don't know to what extent this is correct or if it is correct at all but I would appreciate any and all help given. Also, as you can probably tell, the level of physics used here is very basic, so I would also appreciate it if the answers explained it in this way as well, although, where possible, do not sacrifice accuracy for the sake of simplicity.

If the solar cell is modelled as a voltage source $E$ in series with an internal resistance $r$ and the cell is connected to a load resistance $R_L$, the series current is given by Ohm's law:

$$I = \frac{E}{r + R_L}$$

or

$$E = (r + R_L)\cdot I$$

The output voltage $V$ of the solar cell is the voltage across the load resistance which is, by Ohm's law, $V = R_L\cdot I$.

Thus

$$V = R_L\cdot I = E - r\cdot I$$