Electrical and Mechanical Energy as it relates to Capacitors

[dunn1313]

Homework Statement

In my lab, we are studying the efficiency of a simple DC motor. In order to do this, we related the initial potential energy of the capacitor to the final potential energy of a mass (which remained constant for all three capacitors) that the motor lifted. I was instructed that the motor should have a constant efficiency even when it is powered using different capacitors with a capacitance of 0.21 F, 0.11 F, and 0.068 F respectively, however, in our experiment we found that the efficiency of the motor increased as the capacitor and I am having trouble explaining this. Furthermore, in trying to determine potential sources of error I was curious if the charge of the battery could have an effect on these calculations.

Homework Equations

η= (mgh)/(.5CV^2)

The Attempt at a Solution

I haven't been able to find much information regarding why the efficiency of the motor was experimentally determined to be different using different capacitors expect for that capacitors discharge much faster than batteries so I would assume that an increased capacitance would result in a faster discharge which could generate more torque on the motor and a high torque results in a higher efficiency. I'm not sure if this train of thought is correct but I think I am on the right track... Additionally, I wasn't sure if it would have in impact on the efficiency but I thought that using an old (not fully charged) battery may have an effect on its ability to fully charge the capacitor which could contribute to additional sources of error.

 

[andrevdh]

Did you measure the voltage while the caps were charging or just use the voltage of the battery/power supply?

 

[kuruman]

η= (mgh)/(.5CV^2)

If you use this expression to calculate the efficiency, you have to allow the capacitor to discharge completely, i.e. wait until the motor stops running. Did you do that? Sorry for asking, but your description of what you did is not complete. Also, increased capacitance will result in a slower discharge because the time constant RC will increase assuming that the resistance is constant, which it very likely is. A larger capacitor holds more charge which takes longer to discharge.

http://hyperphysics.phy-astr.gsu.edu/hbase/electric/capdis.html#c2

 

[dunn1313]

Did you measure the voltage while the caps were charging or just use the voltage of the battery/power supply?

We measured the voltage while we were charging the capacitor to determine when our capacitor was fully charged. We used a 6V battery and measured voltages slightly less than that (~5.99-5.8)

 

[dunn1313]

If you use this expression to calculate the efficiency, you have to allow the capacitor to discharge completely, i.e. wait until the motor stops running. Did you do that? Sorry for asking, but your description of what you did is not complete. Also, increased capacitance will result in a slower discharge because the time constant RC will increase assuming that the resistance is constant, which it very likely is. A larger capacitor holds more charge which takes longer to discharge.

http://hyperphysics.phy-astr.gsu.edu/hbase/electric/capdis.html#c2

We did let the capacitor completely discharge and I believe we know this because we measured the maximum height that the mass was lifted and if the capacitor was not fully discharged the mass would continue to rise but I may be mistaken.

 

[andrevdh]

Maybe the tail end of the discharge of the various capacitors gives rise to the efficiency being lower for smaller capacitors - that is the motor succeeds to complete more revolutions for larger caps, but fails to do so for smaller ones?

 

[David Lewis]

...a high torque results in a higher efficiency.

When a motor produces more torque, it consumes more current.
Efficiency = (torque * rotational speed)/(voltage * current).

 

[David Lewis]

If you set torque as the independent variable, and speed and efficiency as dependent variables then (for a given voltage) your graph will look like this. In addition, when the motor is first connected to the cap, current will be high because the motor shaft is accelerating from a standing start, and cap voltage is at its highest.

 

[andrevdh]

You could check experimentally if my hypothesis is correct by checking the voltage over the cap when the motor/mass came to a standstill, that is there will still be some voltage (which will bleed away over time), but the motor (mass) is already stationary.

 

[CWatters]

Its unlikely you were measuring the motor efficiency. Most likely you were measuring the overall energy efficiency.

Larger caps typically have lower equivalent series resistance so lower losses.

 

[andrevdh]

Aaaa, so caps have an internal resistance? Makes sense. So there are energy losses in the internal resistance too.

 

[David Lewis]

There will be loss resistance in the connecting wires, and radiation resistance due to rapidly changing current. I believe these will dissipate half the energy.

 

[andrevdh]

Basically it is an LCR circuit.