Group members

Hadi, Fernan, Zhihua (Dominic)

Introduction:  

The purpose of this experiment is to measure the efficiency of different types of light bulbs used in households. This experiment will also demonstrate the concept of efficiency by the use of hand-cranked generator. In addition, it will include the use of photoresistors in measuring the light intensity of the system.

 

Tools and equipments:

In this experiment, we are going to use the following:

-3 types of light bulbs: Halogen, CFL, and LED.

-Light bulb socket.

-Hand-cranked generator.

-Multimeter.

-Desk lamp.

-Breadboard.

-Wires.

-1k ohm resistor.

-Photoresistor.

-Power outlet.

 

 

Procedure:

Step 1:

First we set up the experiment tools and components. First, we are going to measure the voltage supplied by the hand-cranked generator using the multimeter. So we connect the multimeter to the output of the generator, then we spent the lever and measure the voltage generated by it. The voltage that we got was about 140V AC. Now we can start testing the light bulbs each at a time. We connect the first type of light bulbs (Halogen) to the socket and then to the hand cranked generator. We then start spinning the lever until we get the light to turn on. Notice the effort that it takes to turn it on. Repeat this step 2 times with different types of light bulbs (CFL, LED). Which of the 3 types was easier to light up, and which was harder? What does that have to do with the efficiency?

If the power for each of these type is given as in Table1, calculate the current flowing through each of them if the average voltage generated is 140V.

 

 

 

 

 

Light Bulb Type	Input Power (W)	I (mW)
Halogen	43	0.31
CFL	13	0.093
LED	3.5	0.025

Table1.

Remember that Power=V*I

 

 

Step 2:

Connect the circuit as shown in Fig.1.

 

Fig.1

 

We will use a lamp connected to the outlet of the wall that supplies 120V AC. Then we will build the circuit on the left on a breadboard. We use a DC power supply of 5V, and connect it to a photoresistor in series with 1k ohm resistor. Photoresistors are resistors that change their values depending on the light spotted on them. The more the light, the less resistance it has and vise versa. We are going to observe the voltage across the 1k ohm resistor. The voltage across the 1k resistor depends on the value of the photoresistor. According to the voltage divider rule, Vout=Vs*(Rout/(Rout+Rp). Vs is the source voltage. We start with the Halogen bulb and insert it into the lamp. We turn it on and place it so it is shining on the photoresistor. We measure the Vout. We repeat this 2 times using CFL and LED lights.

Then fill the table below:

 

Light type	Vout (V)	I (mA)	Output Power (mW)	Input Power (W)	Efficiency
Halogen	3.9	3.9	15.21	43	0.35m
CFL	3.96	3.96	15.68	13	1.2m
LED	3.56	3.56	12.67	3.5	3.62m

 

The input power is given from the manufacturer of these products.

We need to use these formulas to acquire the values.

V=I*R, we use this equation to calculate the current through the 1k resistor.

The output power= I*Vout.

The efficiency=Output Power/Input power.

Determine which of these lightbulbs is the most efficient and which is least efficient. Do they agree with your results in part 1?

 

Conclusion:

At the end of this experiment, we know that the LED light is the most efficient types of light bulbs, then CFL, and lastly the Halogen. Our results from the first step agreed with our results in the second step. We had some difficulties in our experiment that restrained us from applying some measurements such as the current flowing through the lamp due to the danger of working with 120V. We also had to change our plan from using DC power supplies to the 120V AC outlet because we couldn’t acquire enough voltage to turn the lights on. However, we were lucky to find the photoresistors which made it very easy to conduct our experiment and measure the light intensity of each of the bulbs.