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I have a simple circuit of a DC power source (15 V) powering a DC motor. There is a voltmeter connected across the battery terminals that reads 15 V when the switch is open.

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When the switch is closed, the voltmeter reads 12 V.

I have some questions:

  • What is the significance of the 3 V? If the motor drew more current, could it be more than 3 V?
  • Is the potential across the motor just 3 V and not 12 V?
  • Can we use this to calculate the current drawn by the motor if we knew the power supplied by the power source? Current (I) = (power supplied from source) / 3 V. What's wrong with this assumption?
  • 1
    \$\begingroup\$ how can you be measuring any voltage? ... your schematic shows disconnected batteries \$\endgroup\$
    – jsotola
    Aug 25 at 19:13
  • 5
    \$\begingroup\$ It dawned on me that this question's description is typical of a fully charged, good car battery and the starter motor on a gasoline car engine. \$\endgroup\$
    – Perry Webb
    Aug 26 at 9:26

4 Answers 4


The DC power source isn't "ideal" and has an internal resistance. The larger the current, the more voltage will drop across this internal resistance, so yes, if the motor would pull more current, you would see a larger voltage drop than 3 V.

The internal resistance of a battery depends on chemistry, state of charge, battery size, age, temperature, and probably some other things I forgot. Other non-ideal power sources also have internal resistance.

The voltage across the motor is what you measured across the motor: 12 V.

Power dissipated in the internal resistance of the battery divided by 3 V will give you the current, as will power dissipated in the motor divided by 12 V.


In the leftmost schematic shown below, the OP's original circuit is reproduced, with some resistances added. Each battery has an internal resistance. The motor has an internal resistance as well (Rmotor).


simulate this circuit – Schematic created using CircuitLab

Can we measure all these resistances?
Not easily.
The motor's resistance (Rmotor) is quite low. For a brushed DC motor, its resistance is measured in series with its two brushes. Brush resistance is variable, and for some angles-of-rotation shorts or connects internal windings in parallel. If you try to measure Rmotor resistance with an ohmmeter, you'll likely find that it jumps around - the lowest resistance you see is likely somewhat closer to the truth. Another problem with a different Rmotor measurement method is outlined below.

What about battery resistance? Batteries tend to be non-linear devices, where internal resistance can vary with state-of-discharge. And short-circuit resistance likely differs from this scenario, where less current flows.
The technique of properly measuring battery resistance is a bit awkward. A resistor of known value is placed across the battery. Open-circuit battery voltage \$V_{UNLOADED}\$ is compared with the smaller voltage \$V_{LOADED}\$ measured while the resistor \$R_{load}\$ is attached:
\$ RB1=({V_{UNLOADED}\over{V_{LOADED}}}-1)R_{load}\$

In the second circuit on right, you can combine all the battery resistances into one (3xRB1) since they're all in series.
The model for the motor has been made more complex by adding a voltage source in series with it's static resistance Rmotor. When the motor turns, it generates a voltage (\$V_{bemf}\$) that tends to reduce current flowing from V1. A DC ammeter could measure this current.
We could use this measured current to find RB1...
\$3\times RB1=(V_{UNLOADED}-V_{LOADED}) I_{MEASURED}\$
While the motor is turning, \$V_{bemf}\$ cannot be measured, since it is inside the motor, and is rotating. It will be less than the 12V applied, because of brush losses. So motor resistance Rmotor cannot be determined by simply knowing the current flowing through it.
One could try to grab the motor shaft to stall it (on a weak toy motor) so that \$V_{bemf}\$ falls to zero. Then Rmotor would be simply \$12V\over{I_{MEASURED}}\$. However, since current will be very high, you will no longer have 12V applied...it will have dropped lower due to 3xRB1. So you would need two meters, one to measure current, and one to measure voltage at the motor's terminals.
You would also find that a stalled-motor will give a current that jumps around quite a bit because of brush resistance - same problem as the ohmmeter method.


The "missing" 3 volts is dropped across the internal resistance of the battery. Either the battery is near fully discharged, or the battery is not capable of delivering the current demanded by the motor.


The 3 volts is dropped because of the internal resistance of one of the batteries.

The current can be calculated by dividing the power dissipated by the internal resitance with the 3V.

  • 6
    \$\begingroup\$ One or all of the batteries. Probably all if the batteries are all the same kind. \$\endgroup\$ Aug 25 at 19:48

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