Two DC motors (Brushed PM Micro) are attached to a voltage source by using a single wire (1mtr). Motors are then connected in parallel (two seperate but same spec. wires, each 1ft in length).For simplicity lets name the motor as A & B and assume that wire resistance is same across A & B. Given following parameters :

  1. DC resistance of motors at rest: A=2.32ohm ; B=4.04 ohm
  2. I @12V A=2.5A & B=0.72A.
  3. Mass of the object in case of A is greater.
  4. CEMF @12V A=6.2V & B=9.09V (as per Vs-IaR)

    The above motors will be run by two different sources: (i) Solar Panel (22V) (ii) 12V Battery

    At low battery, suppose we run the A motor and then the B motor. They'll draw a lower amount of current because of the reduced voltage. But what if they both were running together. Will they show the same current characteristic, i.e., there current reduces by the same amount as it was when they were running individually? In other words, is it not a possibility that the motor under heavy load sucks a lot more current than the motor under light load leading the latter to starve.

    Also given the fact the CEMF of 'A' motor (heavy load) being less than that of 'B' motor, it opposes less to the current and thereby allows more current to flow. Also, what will be the behaviour under low solar radiation?

I need your views on this matter.


is it... a possibility that the motor under heavy load sucks a lot more current than the motor under light load leading the latter to starve.

Yes, depending on the nature of the load on each motor.

At startup each motor draws a stall current determined by its resistance. If both motors are started at the same time this current will be ~60% higher than if only motor A was turned on, or ~270% higher than if only motor B was turned on. If the power source cannot deliver enough current to power both motors then they will both be starved.

Solar panels have a sharp current limit which varies according to the amount of light hitting them. If your panel cannot deliver enough current to power both motors then they may remain stalled. But if only one motor is connected then it will receive all the current, which could then be enough to get it started.

Motor torque is proportional to current. A load such as a reciprocating pump may require a minimum torque to get started, and it if it doesn't get it the motor will remain stalled. A different load such a flywheel may still start up even when torque is limited, only slower than normal. Once the motors get spinning they produce back-emf which reduces current draw, so if they produce enough torque to start the loads spinning they should speed up. However some loads (eg. propellers and fans) have increased torque loading as they speed up, which will cause the motors to run slower when current is limited.

  • \$\begingroup\$ I really like your explanation, let me make it more clear. Motors are running & batt. draining. Ideally both motors should draw that much current that they were drawing at a particular voltage when running indivdly. Ignore the change in collective current whn runing 2gthr. Both motors should get slow and come to their expected speeds. What happens practically heavily loaded motor tries to keep its speed leaving other to starve, is that the prior lower CEMF of the 'A' motor the culprit or the batteries ghostly unpredictable nature. I need a systematic explanation of this situation. \$\endgroup\$ – Akash Feb 4 '16 at 17:50
  • \$\begingroup\$ The more heavily loaded motor will draw more current and get more power than the other one. As both motors are in parallel they get the same voltage, but the more powerful motor may be able to maintain a better speed at low voltage. \$\endgroup\$ – Bruce Abbott Feb 4 '16 at 20:17
  • \$\begingroup\$ And its the last line that needs an explanation as to why that happens???? Observations from test revealed this problem. I am aware of this problem, but what's the cause? \$\endgroup\$ – Akash Feb 5 '16 at 5:33
  • \$\begingroup\$ Hard to say without knowing what load each motor each is driving, but at the same loading a motor with 2.32 Ohms resistance should hold its rpm better than one with 4.04 Ohms, due to lower voltage drop inside the motor. \$\endgroup\$ – Bruce Abbott Feb 5 '16 at 5:41

Since the motors are connected to a voltage source (constant voltage, variable current) and they're in parallel, each motor will draw whatever current its mechanical load demands from the source.

In your example, motor A draws 2.5A and motor B draws 0.72A, so if they're in parallel the total draw will be 3.22 amperes.

If, say, the load on motor B changes and requires an additional ampere from the source, then the [voltage] source will have to output 4.22 amperes while its output voltage remains at 12 volts.

Assuming you've chosen your PV array to supply \$\approx\$ 12 volts under full load, low solar radiation will cause the motors to slow, or stop.

  • \$\begingroup\$ Thanks for your reply. I am very much aware of these theoretical issues.But imagine a battery of 12V 7Ah cap, its voltage dipping as it looses its charge, voltage source is not constant for the whole period of time.Your explanation stands true in case of a bench supply.I totally agree.Things get complicated when the battery is in its discharged phase,inductive loads, reduced CEMF etc. I want to know what would happen at that instance. Please guide me only if you have faced such an issue in any of your practical designs before and what conclusion you'd drawn from the tests. Thanx for your time. \$\endgroup\$ – Akash Feb 4 '16 at 17:29

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