Wednesday, February 13, 2013

Solar Panel Testing

-->
To understand the variables attached to solar panels, we explored how different distances, positions and surface areas affected the voltage and current captured in the solar panel.  Additionally, we were tasked with finding the resistor that will maximize voltage and current.  We used a 50-watt bulb to serve as the sun, and offering a light source.  In a 100% efficient solar panel, all 50 watts would be captured and converted to useable power.  We measured the current to be 54.9 milliamps, which converts to .055 amps, and the voltage at .5 volts.  To get the wattage produced by this current and voltage, we would multiply the two quantities, .055 * .5 = .0275 watts.  Out of a possible 50 watts, only .0275 watts were captured, yielding a 5.5*10-4% efficiency.  This efficiency would be the case if the wattage the bulb required was exactly equal to the wattage it put out.  However, the bulb its self is not 100% efficient.  We noticed that the bulb generates a significant amount of heat and in conjunction with the amount of useable light given off by the bulb, the light that reaches the solar panel is much less than 50 watts.  The average cheep solar panel is around 10% efficient, and knowing that the wattage of useable light available to the solar panel is much less than 50 for a variety of reasons, we can conclude that the panel is probably around 10% efficient.  The panels are not as horrible as we first calculated, but they are will clearly not be used by NASA.  The next experiment we conducted using the solar panels was to add resistors to the circuit and find the resistor that yielded the highest current and voltage.  We tested a total of 8 resistors and fount that they all were very close to the ideal current and voltage, that measured with no resistor in place.  The resistors ranged from 10 ohms to 100 killi-ohms, however we focused primarily on resistors with orders of magnitude from 1 killi-ohm to 100 killi-ohm.  Within the primary resistors we tested, our average resistance was 26.6 ohms and we found that there was an average current of 51.8 milli amps and average voltage of .58 volts.  Ideally, we would have liked to find a resistor that yielded a current and voltage equal to when there was no resistor present, 54.9 mA and .50 V.  However, because no resistor will be ideal, we will choose the resistor with the current and voltage reading closest to 54.9 an .50 to be our "best" resistor.  The 10 ohm resistor yielded a current and voltage that was significantly different from the other data compiled.  Originally, I suspected that this point was an outlier; however, upon further examination, this may not be the case.  According to the mathematical formula, V = IR ( Voltage = Current * Resistance), a larger resistance will yield a higher voltage, which is consistent with our data, the voltage of the circuit is smallest with the smallest resistor.  Although the formula V=IR can be rearranged to say V/R = I, which would suggest that a smaller resistance should yield a larger current, if you have a small voltage, the current should be small as well.  The last test we conducted was to test the effects of "bird poop" on the current and voltage of the solar panel.  We found that the effect of covering one spot on the solar panel had the same effect on the current and voltage as covering half of the panel.  Each component, current and voltage, decreased dramatically when just a speck of the panel was covered.  We concluded that this is due to the interconnected-nature of the solar panel. 
Resistor
Current
Voltage
No resistor
54.9 mA
.50 V
1.2 k ohms
53.9 mA
.49 V
4.7 k ohms
54.2 mA
.48 V
10 k ohms
49.1 mA
.47 V
15 k ohms
53.9 mA
.48 V
22 k ohms
49.1 mA
.47 V
33 k ohms
50.8 mA
.48 V
100 k ohms
51.8 mA
.48 V
10 ohms
38.8 mA
.38 V

  

1 comment:

  1. The efficiency calculation you took on is tricky ... the bulb uses 50 Watts, but does not emit 50W of usable light (think about how much heat it's generating). And then think about what % of that light that is generated actually reaches the panel, as opposed to how much does not. So the efficiency of your panel is not AS bad as you calculated :)

    Please explain why the 4.7 kohm resistor was bad, and also remember to include units in your axis labels.

    We'll discuss further in class.

    ReplyDelete