# 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.

## 1 Answer

Your reasoning is correct.

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

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