I have a simple circuit containing a small light bulb, a small DC motor and a 3v DC battery all wired in series. When the motor spins freely the lightbulb is dim, but when I put a load on the motor (by pinching the shaft between my fingers) the light gets brighter.

This is all new to me, but I assume from my observations that the electrical resistance of the motor decreases as the physical load on the motor increases. But I do not understand why?

Can anyone explain?

 

one thing you should remember abt machines is:

Volts = rmp
Current = Torque

By putting a load onto the machines rotor you are causing the machine to draw more current to meet that torque thus more current through the bulb = brighter light

 

Thank you. I am still a little confused. Voltage being constant I thought that current was regulated only by resistance. How does the motor "draw" more current without changing its electrical resistance?

 

When an electric motor spins it generates a voltage. Try spinning the motor with its leads connected to the lightbulb instead of the battery. If you spin it fast enough you should be able to light up the lightbulb.

The same thing happens when the motor is spun by a battery. It generates a voltage that opposes the voltage of the battery. This causes the voltage that gets to the light bulb to be less causing it to dim. This effect is called "Back Electro Motive Force(EMF)" if you want to look it up on Google.

Hope this helps

Brent

 

another thing about machines is the power output is equal to

Tw - where T = Torque and w = angular velocity

Since you are saying you just plugged in the DC machine (ie no control associated with it) it will run at maximum speed (goverend by your battery).

As you load the machine it will draw more current to establish enough field to provide the neccesary torque. More current raw = more resistive volt drop in the leads/winding thus less vots to force current round the windings thus slows down the machine