To find the relationship between the weight of an object and the efficiency of an electric motor

1. An electric motor
2. Some wires
3. A power pack
4. An ammeter
5. A voltmeter
6. A piece of cotton
7. Some weights
8. A stop clock
9. Safety goggles
10. A metre ruler

Method:

1. Place an electric motor on the edge of a table and tie a piece of cotton around its spindle so that the cotton is a few centimetres off the floor. Set up the motor circuit containing a voltmeter and an ammeter.
2. Tie a chosen weight at the end of the cotton so that the apparatus looks the one shown in the diagram.
3. Switch on the motor and immediately start the time. The motor should pull the string upwards.
4. Quickly stop the clock once the weight has reached the top of the cotton and take the reading from the ammeter and voltmeter.
5. You can now work out the output and the input and thus work out the efficiency of the motor for the chosen weight and length of the cotton. You can do this by using the equations:

Input = Power x Time (electricity into motor)

Output = Mass x Gravity x Height

6. Once you have found out the input and output you can then work out the efficiency of the motor by using the equation:

Efficiency = output/ input x 100

7. Repeat the experiment with different weights and record the results in a table. The weights I have chosen to use are: 10g, 20g, 30g, 40g, and 50g.                

Results:

• Total Input Energy(Joules) = power(Watts) x time(s)
• Total Output Energy(Joules)= mass(kg)x gravity(m/s/s)x height(m)
• Efficiency = Output Energy / Input Energy x 100%

                                   

        

Conclusion:

When a current is passed through a DC motor, a torque is generated by magnetic reaction, and the motor revolves. However, the revolution of the motor induces a voltage in the windings which is opposite in direction to the voltage applied by the battery, and hence is known as back emf (electromotive force).

total voltage = battery voltage – back e.m.f.

Vtotal = V – Eb

If the motor resistance is R and the current is I, then by Ohms Law,

Vtotal         = V – Eb = IR

V = Eb + IR

        The total input power (taken from the battery) is voltage x current and is:

VI = (Eb + IR) x I

VI = Eb I   +  I2R

Eb I is the useful power and is thus equal to the output power.

I2R is the power dissipated in the motor as heat and is wasted.

Efficiency        = energy in / energy out

        = power out / power in

This equation shows that as I increases, the efficiency decreases, so a greater fraction of the energy will be wasted as heat.

As the current increases, the efficiency decreases, so a greater fraction of the power will be wasted as heat in the resistance of the motor.

The output power is also proportional to the load lifted by the motor, so as the load increases, the efficiency decreases.

This experiment has shown that my prediction was correct and that as the input increases, the efficiency of the motor decreases, so a greater fraction of the power will be wasted as heat in the resistance of the motor.

Evaluation:

Although my results were mainly correct and did support my hypothesis, my efficiency results were surprisingly low. I believe this may have been due to the inaccuracy of some of my readings, especially time. This was because it was very difficult to stop the clock at exactly the right time, as soon as the weight reached the top. Also, the readings taken from the ammeter and voltmeter may have had a slight inaccuracy. This was a result of not knowing when to take a reading off the metre, as the numbers did not tend to settle at a particular reading.

All in all, I think this was a successful experiment and my results from it prove my hypothesis.