Steele Taylor
April 12, 2010
Lab Practicum Report
1. Minimum Voltage Required to Light LED
A) Finding the resistance of the LED light. A circuit was created containing he following in series: the tangent galvanometer, the LED light, and 2 1.5 Volt Batteries. With the coils of the tangent galvanometer parallel to the horizontal component of the geomagnetic field, the angle of deflection of the compass needle towards the normal of the galvanometer loops was measured.
The galvanometer works because it measures the extent to which the magnetic field induced by the current of the coil displaces the compass needles from pointing towards the horizontal component of the geomagnetic field. This component was previously calculated:
The dip angle in VT is 26o and the accepted BVT is 5.4x10-5 T,
SO: Bhorizontal = (BVT)(sin 26) = 2.37x10-5 Tesla
THUS: Bcoil = (2.37x10-5 )(tan Ф)
For a solenoid, Bcoil = (µoNI)/(d) = (2.37x10-5 )(tan Ф)
This can be rearranged to solve for I:
I = [(d)(2.37x10-5 )(tan Ф)] / (µoN)
Diameter of galvanometer loops: d = 20.2cm = .202meters
Trial 1: Loops = n = 70
Angle Deflected = 21o
Calculated I = 0.02089 Amps
Trial 2: Loops = n = 30
Angle Deflected = 10o
Calculated I = 0.02239 Amps
Trial 3: Loops = n = 60
Angle Deflected = 19o
Calculated I = 0.02186 Amps
Average I = (.02089 + .02239 + .02186)/3 = 0.0217 Amps
Resistance = (V/I) = (3 Volts / 0.0217 Amps) = RLED = 138.25 Ohms
B) Finding the minimum voltage required to light the LED:
1. The LED light was hooked up to a single 1.5 volt battery, and this was insufficient to light the bulb. Therefore it was immediately estimated that the minimum voltage to light the LED is between 1.5 and 3 Volts.
2. A series circuit was set up that included the rheostat, the galvanometer, and the LED light hooked up through the 2x1.5 Volt batteries. The rheostat could not deliver sufficient resistance to cause the LED to go out. A similar set up was then attempted, also with no success, that included the fixed resistors. Although the LED light was very dim, no amount of added resistance seemed to cause it to go out completely. AND, at this level of dimness the deflection of the compass in the galvanometer was infinitesimally small, thus an accurate reading would not be obtained anyway.
3. It can be concluded here that an erroneous experimental design led to the inadequate results. The rheostat SHOULD HAVE INSTEAD been included in parallel with the LED light. When the rheostat is at zero resistance, all of the voltage would be dissipated across the rheostat (thus the LED would not be lit), and as the resistance is slowly increased, more of the voltage would diverted across the LED and ultimately it would light. The rheostat setting could then be left untouched and rewired as a simple series containing the hampden power supply and the galvanometer. The current passing thru the rheostat could then be calculated, and subsequently the resistance could be calculated as R = V/I.
4. This value for the resistance of the rheostat could then be used in combination with the previously determined resistance of the LED in order to find an equivalent resistance for the circuit: (1/Req) = (1/RLED) + (1/RRheostat). Then the total current in the circuit could be calculated: Itotal = Voltage / Req, and Voltagetotal = ILEDRLED + IrheoRrheo. The current across the individual resistors can then be calculated and subsequently the voltage across each can be calculated.
2. Power Consumed by LED (Watts): P = IV
A) Using the current and the voltage for the LED found in part 1A: V = 3volts and I = 0.0217 amps.
Power = IcircuitVbattery = (0.0217 Amps)(3 Volts)
Power = 0.0651 Watts
3. Energy Saving / Efficiency of LED vs. Household 15 Watt Bulb
A) The resistance of the 15W bulb was determined using method described in part 1A.
B) The photometer was placed exactly between the LED light (hooked up to the 2 1.5 Volt batteries) and the 15W household light (hooked up in series with theHampden Power Supply, the rheostat, and the galvanometer).
C) The resistance in the rheostat was adjusted until the bulbs had equal brightness (according to the photometer placed exactly ½ way between the bulbs). The new current was calculated based on galvanometer readings. This current is the same across all resistors in series, so it could be multiplied by the Resistance of the 15 Watt bulb already calculated in order to find the voltage across the LED resistor.
D) The Power can now be calculated as P = IV, and compared to the power used by the LED to achieve the same brightness. The power supplied to the LED to achieve the brightness has already been calculated above in part 2.
Calculation of R15 Watt Bulb
Trial 1: Loops = n = 10
Voltage = 16 Volts
Angle Deflected = 77o
Calculated I = [(d)(2.37x10-5 )(tan Ф)] / (µoN)
= [(.202)( 2.37x10-5 )(tan 77)] / [(µo)(10)]
= 1.65 Amps
Resistance of 15Watt Bulb = 9.7 Ohms
Calculation of Power15 Watt Bulb at equivalent brightness to LED
Loops = n = 10
Angle Deflected = 62o
Calculated I = [(d)(2.37x10-5 )(tan Ф)] / (µoN)
= [(.202)( 2.37x10-5 )(tan 66)] / [(µo)(10)]
= 0.85 Amps
Voltage Across 15 Watt Bulb at equivalent brightness = V = IR
V = (9.7 ohms)(0.85) = 8.245 Volts
Power = IV = (0.85)(8.245) = 7.008 Watts
Comparison of LED and 15 Watt Bulbs
In order to achieve the same brightness, the conventional light bulb must consume 7.01 Watts, while the LED only needs to consume 0.0651 Watts. Therefore the conventional 15 Watt light bulb uses (7.01/.0651) times more energy, or 107.7 times more energy. In other words, the LED is 108 times more energy efficient than the 15 Watt Bulb.
Conclusion:
LED is AWESOME!