# Power consumption when DC motor runs on battery

I have small doubt about power supply selection for a running motor. I have a small dc motor, which is rated for 12V , 3A(rated ). When the motor runs with a load 4000N, the current consumption is 1.5A.

So I have to choose a 12V, 3A = 12 * 3= 36W power supply to run the motor.This is because DC power supply can supply continous 3A current without any disturbance.

Now I wanted to run same motor on battery. I would like to know how much power should be supplied by the battery to run the motor theoretically.

When motor runs on battery, it takes full current from the battery; as per formula ($e= l \frac{di}{dt}$). It said that current required by motor = 3 $\times$ current required while running on starting.

When we run the motor on battery eventually battery voltage got dropped, taking more current.

I would like to know what is the percentage of current it takes.

If we consider above system 12V, 75Ah battery is it efficient?

Can someone explain with proper calculation here?

While running on battery; motor will turn on every 5 minutes and after that it will be turned off. The motor is made to run 11 hours on daily basis.

• 4000N oO you sure about that figure – JonRB May 22 '15 at 8:38
• yes . Its linear actuator which used dc motor – user50949 May 22 '15 at 9:11
• oh right... the motor doesn't see this. the end product of the EMA is 4000N. ok that makes more sense. You need to determine the load as seen at the rotor, not the load at the actuator – JonRB May 22 '15 at 10:47

Before calculations you need to be clear about some fundamental concepts about motors. "It said that current required by motor = 3 × current required while running on starting." This statement can not be true, because motor draws more current at start up than it rotates in operational angular velocity.

"When we run the motor on battery eventually battery voltage got dropped, taking more current."

This statement is also false. Motor speed ----> Voltage, Motor torque -----> current.

My advice is as follows. Your application does not require detailed calculations with motor specs. What you need to do is measure the current when your motor rotates under the load, which you choose. Then compare it with battery capacity.

About voltage drop of battery, it does not draw more current when battery gets low on Voltage. Voltage is about speed(Theoretically). Thus, as long as your load is the same, theoretically, motor draws same current.

NOTE: Of course current change when the Armature voltage changes, but this is negligible.

• as i told when i run on regulated supply with load it takes 1.5 A.My application is solar tracker .if we consider normal load like panel & structure its fine i guess. But assume there is wind load . will the load current remain same. Is info – user50949 May 22 '15 at 9:45
• Current is dependent on motor loading, as Zgrkpnr stated. You need to model your system, take measurements for loading in isolation and add estimates for typical wind loading. Wind is hardly a controlled factor, so you can only ever estimate battery life or bound it to max/min ranges when operating under unpredictable loads. – Oliver May 22 '15 at 10:18

75A/h means "75A for one hour." Maximum. So if your motor, on average, draws 1.5A, and ran continuously, then it should run for 75/1.5 = 50 hours. If it only operated one second out of every minute, then it's duty cycle would be 1/60, so would last up to 50h * 60 = 300 hours. Keep in mind most batteries self-discharge a little. The motor will slow down some as the battery gets depleted, so it's a good idea to de-rate this at least 15%. Some battery types (such as lithium-polymer) can even be destroyed by over-discharging them.

If the DC motor is 12 v and has 3 amps, then is 36watts and a 12 volt battery on 75 amps/h will give you 12*75=900watts/36 = 25 hours. I am interested in spinning a PMG and kick the wind out of equation. The torque demanded would shrink some of that 25 hours. I suggest Varta Silver or gel kind of batteries. Succes !

• Hi, welcome to the site. This is a very old question and it also has several answers. Please try to answer newer and-or unanswered questions – Claudio Avi Chami Jan 30 '17 at 5:16

It's not 75 A/h, it's 75 Ah. That means you can draw at 75 amperes for an hour, then the battery would be drained - theoretically - this represents the amount of charge in a battery. If you multiply that with the (battery) voltage, the resulting value gives you an energy measured in Joules, the SI (IS) unit for energy. J and Wh are both units of energy. $$1kW h = \bigg(\frac{1kJ}{1s}\bigg)*3600s = 3600 kJ$$

For a brief reminder: $P = IV$, where $P$ is power, $V$ is voltage, $I$ is current intensity. and $E=Pt=IVt$ where $E$ is the energy and $t$ is time.

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