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Upper secondary extension · Circuits · Forces and motion · Energy transfer

A stopped motor uses no electricity. Really?

Follow energy from the source and series resistor through a motor and drum to a lifted load.

The problem

A 12 V source supplies a motor through a closed switch and a rheostat with 1 Ω effective resistance. The motor has a 1 Ω winding, rated voltage 12 V and ideal no-load angular speed 60 rad/s. It drives a light drum of radius 5 cm with an anti-rollback ratchet, lifting a 1 kg load from rest for 2 s.

Assumptions and boundaries
  • Gravity is 9.80665 m/s². Ignore inductance, friction, rotor inertia and drum inertia.
  • The ideal linear DC motor has equal back-EMF and torque constants of 0.2 in SI units.
  • The ratchet prevents downward motion only. Lift distance is scaled to fit the drawing.

Questions to explore

  1. What is the initial current, and why does it then fall?
  2. How far does the load rise in 2 s? Is all input energy stored as potential energy?
  3. What happens if the load is changed to 3 kg?
Edit the code in Studio →
Electric crane Scientific model diagram + + - - 12 V 12 V S closed S closed control 25% control 25% Shaft: torque and speed Shaft: torque and speed M M DC motor DC motor D r = 5cm D r = 5cm Anti-rollback ratchet Anti-rollback ratchet 1 kg 1 kg cargo cargo As the load rises, back EMF reduces current As the load rises, back EMF reduces current Electrical input = potential energy gained + kinetic energy + resistive heat Electrical input = potential energy gained + kinetic energy + resistive heat 2 s: h = 3.327 m | v = 1.774 m/s | Q = 29.97 J 2 s: h = 3.327 m | v = 1.774 m/s | Q = 29.97 J Starts at rest; no inductance, rotor inertia or friction; motion scaled to fit Starts at rest; no inductance, rotor inertia or friction; motion scaled to fit
time 0s displacement from equilibrium 0m speed 0m/s current 6A electrical input 0J potential energy gained 0J kinetic energy 0J resistive heat 0J Playback speed is adjusted; time readouts still show physical time.
Diagram description

Animated physics diagram: electric_hoist; showing time, displacement from equilibrium, speed, current, electrical input, potential energy gained, kinetic energy, resistive heat.

Unpack one connected process

Motor torque drives the drum and lifts the load. Load motion determines motor speed and back EMF in return. Stalling, switching off and normal lifting are distinct states, even when the picture looks still.

U = IR + kω; τ = kI; Eelectrical = ΔUgravity + K + Q

1. What is the initial current, and why does it then fall?

Initially the motor is stationary, so back EMF is zero. With 2 Ω total resistance, current is 6 A. As the motor speeds up, back EMF opposes the supply voltage and current falls. At 2 s, it is about 2.452 A.

2. How far does the load rise in 2 s? Is all input energy stored as potential energy?

The load rises about 3.327 m and reaches 1.774 m/s. Electrical input also becomes kinetic energy and heat in the winding and series resistor, so mgh alone does not account for all supplied energy.

3. What happens if the load is changed to 3 kg?

Maximum initial lifting force is 24 N, less than the weight of 3 kg. The ratchet holds the load and the motor stalls, but current remains 6 A. All 144 J supplied in 2 s becomes resistive heat, including 72 J in the motor winding.

Test your prediction in code

Change the load to 3 kg. Does a stalled motor still consume energy and produce heat?

scene electric_hoist

source battery 12V
switch S closed
rheostat control 4ohm position=25%
motor M rated_voltage=12V no_load_speed=60rad/s resistance=1ohm
drum D radius=5cm brake=ratchet
load cargo 1kg

battery <-> S <-> control <-> M <-> battery
M drives D
D lifts cargo

simulate 2s playback=4s
show displacement velocity current energy heat
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