ENERGY AND LIGHTING
UM Physics Demo Lab 07/2013
Pre-Lab Question
Why do people say you should turn the lights off when you leave the room?
EXPLORATION: Lighting
Materials
Regular (incandescent) 40 W light bulb
Regular (incandescent) 60 W light bulb
40W equivalent compact fluorescent bulb
LED light bulb
Digital voltmeter with alligator leads
Solar panel
Lamp base with cord
Induction ammeter – clamp-on
Light/lux meter
Calculator
Clear ruler
Meter stick
Rag (for handling warm light bulbs)
One third of all the energy used in the USA goes for lighting. In this experiment we
will explore three different kinds of lighting and their efficiency (that is, how much
light each produces for a certain amount of electrical power). We will also explore the
use of solar energy panels which convert energy from light into electrical energy.
1. Screw the 40W incandescent (regular old fashioned hot filament) bulb in the
lamp base and turn it on. Bring your hand to within a few inches of the bulb.
DO NOT TOUCH the bulb. What do you feel? Why?
2. We can measure the power supplied to each bulb, by measuring the current
and knowing the voltage (110 volts). Since the wall voltage (110 Volts) is
dangerous, we need a non-contact way to measure the current. Do NOT
TOUCH the bare contacts of the meter to the light bulb, or base or wall. We
will instead measure the current using the induction ammeter provided.
This meter clips around the wires attached to the lamp holder. Wrap one
lead of the lamp cord through the induction meter 10 times.
What is the relationship between the actual current flowing in the lamp cord and
the current reported on the meter, given that the cord passes through the meter
10 times?
Induction Meter Reading: __________ amps
Actual current Supplied to the Incandescent bulb: ___________ amps
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Power supplied to incandescent bulb:
Power = Current (amps) x Voltage (Volts) = ____________ watts
3. Does the power supplied to the bulb match the 40 Watts rating printed on the
bulb? Why or why not?
We will explain how the induction ammeter works based on coils and generators in
a few weeks. Metal detectors at the airport work in very much the same way.
4. Unplug the lamp base, unscrew the incandescent bulb (careful it is hot) and
insert the compact fluorescent bulb. Plug the lamp base back into the wall outlet
and turn on the light. Again bring your hand within a few inches of the bulb. Is it
giving off less or more heat than the incandescent bulb?
5. Measure the current for the compact fluorescent bulb and determine the power
supplied to it: (also record the power in the table in question 7 below)
Current = ____________ amps
Power = Current x Voltage = ____________ watts
6. Repeat steps 4 and 5 for the LED bulb.
Power supplied to LED bulb = Current x Voltage = ___________ watts
Which of the three bulbs (incandescent, compact fluorescent, or LED) produces the
most heat? And which appears to give the most light (is the brightest)?
Most Heat ___________________________
Most Light ___________________________
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7. Using the figures from parts 2, 5 and 6, calculate the cost per hour to operate
each of the bulbs: (May 2013 Cost of power = $0.13/kWh) Show your
calculations below the table.
Type of bulb
Incandescent
Power
Supplied
to Bulb
(kW)
Lifetime
Hours
(hr)
Energy
Consumed
During
Lifetime
(kW-hr)
Cost of
Energy
Consumed
During
lifetime
($)
Purchase
Cost Per
Bulb
($)
1000
$0.98
Compact
Fluorescent
10,000
$2.74
LED
50,000
$12.86
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Cost Per
Hour of
Operation
($)
3
Assuming 6 hours per day of lighting, what is the cost for each light bulb over a 1 year period?
Total hours used = 6 hours per day x 365 days = 2,190 hours. Show your calculations
below the table. Remember to make sure you factor in the number of bulbs you
need for a full year.
Type of bulb
Cost per hour of
Operation
($)
Cost to Run
One Bulb
for 1 year
($)
Purchase Cost Per
Bulb x
#of Additional
Bulbs Required for
1 Year
($)
Total Cost
for One
Year.
($)
Incandescent
Compact
Fluorescent
LED
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Which bulb is the cheapest to operate over a year?
8. Assuming they perform equally well for their task (lighting), which is the most
efficient in terms of energy usage? (Check with your instructors, but move on for
now if they are busy)
9. Now we will quantify this a bit by measuring the brightness. We do that using a
photocell. This is a device that converts light energy into electrical energy. Set the
digital multimeter on DC volts and clip the multimeter leads across the photocell
output wires (located on the back of the photocell).
For each of the three bulbs, hold the bulb 30 cm above the photocell, using the lamp
base and cord to power the bulbs. For each bulb, measure the photocell voltage
using the digital multimeter. Repeat the measurements using the light meter.
Record all your data in the table below.
Note: The light meter measures light output in lux or lumens/m2 (equivalent to
power/area or w/m2). Compare each measured photocell voltage with the light
meter reading. The light meter is just a photocell with its output voltage calibrated
in different units (the photocell voltage is calibrated to correspond to w/m2 of light
output).
Type of bulb
Photocell Voltage (V)
Light Meter Output
(Lumens/m2 ≡ lux ≡ w/m2)
Incandescent
Compact Fluorescent
LED
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10. Are the photocell and light meter measurements consistent? Are these results
what you expected?
Everyday Applications
-
Street lighting (some streets in Ann Arbor are lit by LED’s)
LED traffic lights (long lifetime gives more reliability)
LED automotive tail lights
Home lighting. Compact fluorescents are now common and LEDs will eventually
become the lighting of choice
- Adjustable color lighting (mix red, green, and blue LEDs)
- Efficient flashlights powered by energy stored in capacitors and generated by
mechanical work done when shaking the flashlight (Faraday induction)-no
batteries required.
Solar Power Everyday Applications
-
Power for watches and calculators
Solar powered traffic signs on highways (needs a battery)
Solar powered residential outdoor lighting (needs a battery)
Providing power in remote areas far from the electrical supply grid
Powering electric fences on farms
Recharging batteries for cell phones and lap-top computers
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APPLICATION: SOLAR ENERGY CONVERSION
1. The photocell you are using for this experiment can also be used to convert solar
energy to electrical power. Scaled up and placed on the roofs of buildings, such
panels can generate enough electrical power to supply your daily needs for heating,
lighting and cooking, etc. At the present time they are quite expensive, but costs
are coming down as more are produced. To approximate exposure to bright
sunlight, shine the 60W light bulb on the solar panel from a distance of 10 cm.
Measure the voltage Vsun
Vsun
=
___________volts
Now, with the multimeter on the current setting (mA) measure the current
I=
___________ milliamps = ______________ amps
Psun = I x Vsun
=
____________watts
In Ann Arbor, the sun provides about 700 W of power to an area of 1 square meter
at the brightest part of the day in midsummer. The area of your solar cell is only
about 7 cm x 4 cm or 0.0028 m2, so you would only expect to collect 700 (w/m2) x
0.0028 (m2) = 1.96 watts with this small photocell in bright sunlight. How does this
compare with what you actually measured above (Psun)?
The efficiency of solar cells is an important factor in their usefulness as an alternate
source of energy. Calculate the efficiency of your solar cell as:
Efficiency = Psun/ 1.96 watts =____________ x 100 = ___________%
Manufacturers strive to make solar cells that are as efficient as possible. The cell
you are using, based on Polycrystalline Silicon, is one of the first versions available.
It is still far from 100% efficient, as you can see. There will be a very big effort
worldwide to improve solar energy cells over the next few years. Typical commercial
solar cells are currently 18 – 20% efficient.
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Challenge Work:
1. What would be the area in square meters of an array of solar cells which
generates 5,000 Megawatts of power (typical for conventional power
generating plant burning coal)? Remember that 1 Megawatt (Mw) is one
million watts = 1,000,000 w = 106 w. Use the power per unit area you
determined for the actual solar cells we used in class today = Psun/(area of
solar cell). Assuming this area corresponds to a square array of cells
calculate the length of one side of the array by taking the square root of the
area you calculated. Finally, divide this length by 1000m to express it in
kilometers (1 km = 1000 m).
Summary
1. Engineering trade studies are a tool for making informed technical
decisions.
2. The Kilowatt-hour is a unit of energy. 1kw-hr = 3.6 x 106 Joules.
3. Lighting energy efficiency can be greatly improved by replacing
incandescent bulbs with new technologies such as compact
fluorescent tubes and LEDs.
4. Photovoltaic cells produce electricity directly from sunlight and
can be used for practical power generation. The major impediment to
their more widespread use is cost, not technical feasibility.
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