My son was building circuits with a hobby kit. He tried connecting a small DC motor and incandescent bulb in series with two batteries. The motor is not marked, probably is rated at no more than 6V. The lamp is marked 3V.

The motor shaft spun, but the bulb did not illuminate. When he added a fan blade to the motor, we were both surprised to see that the lamp turned on.

I know that neither the lamp nor the motor can be explained with Ohm's law, but beyond wildly guessing that the motor increases the circuit's current as it has to work harder I couldn't get very far.

We are both happy to figure this out on our own but need a nod in the right direction. I have a multimeter and a good selection of resistors if that would help us. What kind of an experiment or demonstration could we try to help us understand?

Schematic of the circuit

  • \$\begingroup\$ If you have the rated power spec for motor and bulb using Ohm's Law and a frequency nomograph I could estimate exactly why it turns on the bulb \$\endgroup\$ Apr 23, 2022 at 12:21
  • \$\begingroup\$ Counter-electromotive force \$\endgroup\$
    – J...
    Apr 24, 2022 at 10:32

5 Answers 5


For an ideal motor, speed and voltage are proportional to each other, as are current and torque.

A small low voltage motor will be far from ideal, but the general relationship still holds.

With no load on the shaft, the torque and hence current will be very low (zero in an ideal motor), with little voltage dropped across the bulb, and a high voltage across the motor, reflecting its high speed.

As you increase the load on the shaft, the current increases as the torque output increases.

As you slow the motor down, the voltage across the motor decreases, resulting in more voltage across the bulb, which increases the current.

If you stall the motor, safe to do as you have a current-limiting bulb in series, the voltage across the motor will be small (zero in an ideal motor), and the current through it high.


but beyond wildly guessing that the motor increases the circuit's current as it has to work harder I couldn't get very far.

It's a good guess because it is correct.

The motor draws more current because it is shifting air with the fan. In other words, it is producing a mechanical output power and, that power is supplied by the battery and, that means it has to take more current.

Another good test is to hold the shaft to prevent it from rotating. No mechanical output power is being produced (because the motor shaft is stalled) but, the motor will act a bit like a short circuit under these circumstances and the lamp will glow to near-full brightness.

  • \$\begingroup\$ In any of these situations (no load, fan, pressing with finger), how are the three volts divided between the devices? \$\endgroup\$
    – nuggethead
    Apr 23, 2022 at 10:34
  • 1
    \$\begingroup\$ @nuggethead the more you load the motor shaft the lower the voltage will be across the motor. This then means that the more you load the motor shaft the higher the voltage will be across the lamp (up to the 3 volt limit imposed by the battery). Hence, the more you load the shaft, the brighter the lamp will be. \$\endgroup\$
    – Andy aka
    Apr 23, 2022 at 10:36

To add to the other responses. When the motor is starting up (no fan) it accelerates until the motors back EMF equals the battery voltage. At this point the motor spins at steady state speed and consumes very little current. Only enough current to overcome the bearing friction. When the back EMF and battery voltage are approximately equal there is only a small voltage drop across the lightbulb (hence small current flow and a very dimly lit light).

When you put a load (connect a fan) on the motor. The speed is reduced. Thus, the back EMF, which is proportional to speed) is now less than the battery voltage. The rest of the voltage drop (which is bigger now) occurs across the lightbulb. A bigger voltage drop means more current is flowing. Hence the light will be brighter.


The motor behavior can be also explained in terms of "armature reaction".

Think of the electric motor also as an electric "generator" producing a voltage proportional to the speed and opposing the input voltage. So at normal speed, the current is lower and the lamp will glow less.

If the speed slows down (due to increased load), this "anti voltage" will decrease, the current will increase and the lamp will glow brighter.

This mechanism reminds the so-called "bootstrapping" technique in electronic circuits where the output voltage is subtracted from the input voltage. As a result, the current enormously decreases (as though the input resistance has increased).

If you are meticulous enough, you can also take into account the non-linear behavior of the lamp - the more it heats up, the more its resistance increases. Thus, it acts as a current-stabilizing nonlinear resistor.


Congratulations on asking a great question.

The discussion point here is Kirchhoff's Current Law, KCL. In context, the current through the motor is the same as in the lamp. More current in the motor: more current through the lamp. More current in the lamp: more better brighter the bulb.

As you begin to experiment further, a lamp wired this way is pretty good idea to limit current through your test circuit. Search on "smoke stopper" or "load bulb" for more info!


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