I've been thinking about this conundrum.

Given we use joules to move a metal block over a surface, we do work.

So imagine a motor, that spins and pulls a string, moving said metal block over one meter toward itself.

We supply it with 1 volt, 1 amp and it needs 5 seconds of time to do the work of pulling the block for 1 meter.

Given volts are J/C and amps are C/s, we can say volts are how much work, certain amount of electrons will do, over time. And volts * amps are work over time or J/s.

So, 1 V × 1 A × 5 s = 5 Ws = 5 J

Now the question is:

How much energy of 5 J was used to move the block, and what amount was spent on heating up the wires?

Am I right to think, that since 99% of resistance is taken by the motor, it leads to firstly meaning 1% of all energy spend goes to heating up the wires, and secondly (imagining a 80% efficiency motor) ~80% of those 5 joules are spend on moving the block and ~20% of those 5 J are heating up the motor's insides?

In retrospect, we can ponder about how much energy gets spent on friction between the block and it's supporting surface, but I digress.

Your input on this, is greatly appreciated. Thank you.

  • \$\begingroup\$ You are mixing up energy and work/power. Voltage times current is instantaneous power in watts. Power integrated over time is energy in joules. 1V \$\times\$ 1A \$\times\$ 5s is not 5W/s it is 5W\$\cdot\$s or 5J. You are also mixing the symbol for a unit of something (s for seconds) and the symbol for an amount of that quantity (t for time). \$\endgroup\$ – Elliot Alderson Jul 16 '19 at 20:05
  • \$\begingroup\$ 1 V * 1 A * 5 s = 5 W/s No, W * s = Ws, not W/s. Note that SI units named after a person have their symbols capitalised but are lowercase when spelled out. 'V' for volt, 'A' for ampere, 'K' for kelvin, 'Ω' (capital omega) for ohm, etc. Meanwhile 'k' is for kilo. Also a space should separate the number from the unit. See here. (We don't write "5apples" and we don't write '5V'.) Welcome to EE.SE. \$\endgroup\$ – Transistor Jul 16 '19 at 20:07
  • \$\begingroup\$ "What amount of those 5J were used to move the block, and what amount was spent on heating up the wires?", I think you must know the friction and the weight of the metal block if you want to know exactly. Otherwise you just have too many unknowns. \$\endgroup\$ – Harry Svensson Jul 16 '19 at 20:07
  • \$\begingroup\$ oh my, you make me blush. You're all correct of course. Do I correct the question or leave it as it is? \$\endgroup\$ – DayDreamer Jul 16 '19 at 20:09
  • \$\begingroup\$ Tidy it up and we'll upvote you! \$\endgroup\$ – Transistor Jul 16 '19 at 20:10

If the motor is 80% efficient, 20% of the electrical input is used to heat up the motor (internal wire and iron) and 80% is used to supply mechanical energy to whatever is connected to the shaft. That is mostly correct. A small amount of the lost energy is used to move air through the motor to carry the heat away. An even smaller amount is used to make the humming noise that you hear.

A certain amount of energy is lost in the wire between the power supply and the motor. That could be very little if the supply is close and the wires are large. The losses between a power generation station and motors in a factory average on the order of 7%.


"What amount of those 5J were used to move the block, and what amount was spent on heating up the wires?"

How thick and how long were the wires? If kept short, and high enough AWG (gage), then I suspect next to none would be used. You can easily find sites that will determine resistance of various AWG wires, and their temperature rise given a current.


since 99% of resistance is taken by the motor... and secondly (imagining a 80% efficiency motor)

and assuming no mechanical loss in pulling the string...

~20% of those 5J are heating up the motor's insides?

Yes. But you are making a lot of assumptions that may not be true in practice. Motor efficiency is not a constant - it varies depending on voltage, rpm, and current. If the block is stationary at the start then the motor has to accelerate from 'stall' (0 rpm) to its normal operating rpm. Power out = torque x rpm, so at stall the motor's efficiency (defined as mechanical output power / electrical input power) must be 0%.

At stall the motor may draw much higher current than normal, so losses in the wiring will be higher. As the motor speeds up its efficiency improves, and it also stores some energy in the rotational mass of the armature or rotor.

So your conclusion is correct if the motor is pulling the block at constant speed and torque, and the energy used to get it started (and stopped?) is ignored, and the wiring loss really is only 1%, and the motor really is 80% efficient at that speed and torque. However with all those assumptions the conclusion is not very useful.

The graph below shows typical characteristics of a shunt-wound or permanent magnet brushed DC motor or brushless DC (BLDC) motor. The dotted lines at high current indicate that the motor cannot be operated continuously in this region because it would burn out due to excessive resistive losses. Not shown on this graph is that efficiency at maximum power output is always 50% or less - ie. half the input power is wasted (as heat, vibration, sound, light, etc) inside the motor.

enter image description here

  • \$\begingroup\$ Appreciate your comment, learned much new information. The question is aimed to clarify the ubiquitous saying "You're circuit draws n watts, because V*I and, watts are dissipated as heat", which made me wonder, if it's all heat, how do we count energy used to create light in a LED or soundwaves in a speaker. So, it's more a thought problem, rather than a practical one :) My hope is - this thread will help other confused souls like me. \$\endgroup\$ – DayDreamer Jul 16 '19 at 21:00
  • \$\begingroup\$ Given that energy is conserved, it is trivially obvious that any energy put into a device (whatever it may be) that does not appear at the output must have been absorbed or dissipated. This is basic physics. However questions here are supposed to be about electrical engineering, so... \$\endgroup\$ – Bruce Abbott Jul 16 '19 at 21:30
  • \$\begingroup\$ @DayDreamer At the end of the day (end of the universe?) light from an LED is absorbed and becomes heat, sound waves become heat, sliding a mass causes friction which causes heat, ... \$\endgroup\$ – Elliot Alderson Jul 16 '19 at 23:19

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