Comparing the powers of electric motors 1
Practical Activity for 14-16
Comparing the power output of two motors.
Apparatus and Materials
- Electric motor, small
- Switch unit
- Mass, 1 kg
- Line shaft unit
- Power supply, 0-12 V
- Fractional horse-power motor (or the largest DC motor available)
- Rubber band or driving belt
- Knife switch
- Metre rule
Health & Safety and Technical Notes
The person controlling the switch must stop the motor before the load reaches the line shaft. (If the load spins round the shaft, the string may break and the load go flying.)
To stop the motor, the current must stop flowing in both field and armature coils. The power supply may incorporate a smoothing capacitor which discharges slowly when the mains switch is operated.
Small electric motor
- Connect the small electric motor to the line shaft using the drive belt. Attach the mass to the line shaft with the cord.
- Connect the motor to the variable DC supply using a switch and set at the appropriate voltage for the motor so that the load rises slowly. (Although the knife switch is not essential, it allows the power to be connected for a more precise time.)
- Time the load being lifted a fixed distance, say 1 metre.
- Connect the field and armature terminals of the fractional horse-power motor in parallel to the DC terminals of the power supply set at 6 volts (or the rated voltage for the motor).
- Attach the mass to the shaft by cord, so that it is lifted when the motor is switched on, as before.
- Note the time for raising the load through 1 metre and compare this with time taken for the small motor.
- The time taken to raise the load is shorter when a higher potential difference is used. The motor is said to be ‘more powerful’. The loads can be changed and the potential difference kept constant; larger loads take longer to raise. There is only so much energy per second that a motor can transfer for a given potential difference, and so it works more slowly with heavier loads.
- Change in energy stored gravitationally = m x g x Δh joules
- Power output to raise the load = m x g x Δh / t watts
- Where m is the mass in kilograms, Δh is the vertical distance in metres, g is the gravitational field strength in newtons per kilogram, and t is time taken in seconds.
- Energy transferred from a power supply to a component, such as a motor = V x Q joules
- Power input due to current in motor = V x Q / t = V x I watts
- Where V is potential difference in volts, Q is charge in coulombs, I is current in amps and t is the time taken in seconds.
- The field coils of the fractional horse-power motor also have energy transferred to them. For a more accurate comparison with other motors, measure the potential difference across the armature and the current passing through it.
- If you compare the small motor and the fractional horse-power motor, both running on 6 volts and hauling up the same load, the larger motor is faster. Measuring the current through the armature will show that the current is also greater through the heavier motor.
- Another variation is to run the two motors at the same voltage, but with different loads, so that the speed with which the load is raised is the same for both motors. The bigger motor will have a larger current passing through it, and will haul up a larger load.
- The motors can also be compared by measuring the braking force to stall the motor. Increase the load until the motor stops turning. (Do not keep it long in a stalled position.)
This experiment was safety-tested in January 2006