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I've a DC motor (brushed) that is driven by a set of 12 Ah 12 V lead-acid batteries. When motor stops moving, I assume that all the current (at least most of it) in the batteries is drained. However, when I check the voltages of each battery they are well above 12 V. The motor has no load current consumption of 2.2 A and peak current consumption ~15 A. So, I would like to know at what voltage all the 12 A of the battery are discharged? Why would motor completely stop moving (i.e. with no load) when there is full battery voltage left? (Is there something wrong with the motor or I am missing something? - BTW motor heats up quite a lot)

*** Motor has fused by now so, this probably is not entirely relevant practically. The problem was that brush board of the motor was loose and got in contact with the coil resulting in shortage. However, this is theoretically still a valid question.

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    \$\begingroup\$ Measure the voltage under load and report back. \$\endgroup\$
    – winny
    Mar 29, 2020 at 14:18
  • \$\begingroup\$ What's the no-load speed of the motor and what speed is it running under your load? Heating sounds like you're bogging it down until it stalls. If so, gear it down. \$\endgroup\$
    – user16324
    Mar 29, 2020 at 14:47
  • \$\begingroup\$ 12.5V without motor, ~11.5 with motor turned on. However, I noticed fumes and smell while motor stopped afterwards. BTW I was giving motor voltages directly from the battery (without mosfet based driver) if that's okay? \$\endgroup\$
    – azad.wolf
    Mar 29, 2020 at 14:59
  • \$\begingroup\$ So just to be clear -- you're connecting a motor directly to a 12V lead-acid battery, and you're letting it run until it stops. Yes? Are you measuring the battery voltage with the motor still connected? \$\endgroup\$
    – TimWescott
    Mar 29, 2020 at 16:23
  • \$\begingroup\$ Please edit your question with the comment about "fumes and smells while motor stopped". Tell us what sort of smells you're smelling, and what (i.e., battery or motor) the fumes and smells came from. How long are you running the motor before it stalls? \$\endgroup\$
    – TimWescott
    Mar 29, 2020 at 16:25

2 Answers 2

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There is no co-relation between a battery's remaining capacity and it's open-circuit voltage.

The discharged battery's no load voltage could be above 12 V but its internal resistance would also be high.

Without the required starting current, the motor would remain stationary and present itself as a dead short across the battery terminals. The battery voltage would be close to zero under this condition.

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12 Ah specifies how much charge is stored in each battery (1 Ah is 3600 Coulombs). It reveals how long the battery can supply a given current.

However, the batteries will also have a maximum power rating (in Watts or VA) that reveals the maximum rate at which it can deliver the charge. If the power rating were unbounded, then the battery could deliver all 12 * 3600 = 43200 Coulombs of charge in an instant (which would be fun to watch for a brief moment until your eyes melted). But the chemical reaction in the battery can only take place at a limited rate.

When the batteries are nearly depleted, the maximum power output drops -it takes longer to build up the same amount of charge on the terminals. If you disconnect the batteries and measure the voltage it can still be 12V because nothing is drawing current from them. If you connect them to a load however, it will drain off the charge more quickly than the battery can supply it, and the voltage will drop.

So it doesn't really make sense to ask at what voltage the batteries will be discharged, because when the batteries are nearly drained the voltage depends on the resistance of the load.

Technically however, when the batteries are completely drained (which won't really happen in practice), they cannot supply any charge, and the voltage will be zero.

For contrast, consider a balloon that you have rubbed against your hair for a bit to build up some static charge. It can actually have hundreds or even thousands of volts relative to ground, but the charge it stores is still so low that it can supply only a negligible power when connected into a circuit.

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  • \$\begingroup\$ According to ohm's law aren't current and voltage directly proportional? Shouldn't voltage drop in accordance with the capacity of the charge present (or the current flow in the circuit)? Or to put it another way; since voltage is potential difference and when current capacity is low (i.e. low potential/charge) then difference is less as well. In this way, shouldn't there be a linear/non-linear relationship between charge present in the battery an d the voltage? \$\endgroup\$
    – azad.wolf
    Mar 30, 2020 at 15:16
  • \$\begingroup\$ @azad.parinda Ohm's Law only applies to ideal resistors. It does not apply to ideal voltage sources. If you talk about a real life battery, then you can only apply Ohm's Law to the assumed internal resistance of the battery. \$\endgroup\$
    – Elliot Alderson
    Mar 30, 2020 at 15:21

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