When a path for current flow from the batteries is first applied, a large amount of current flows due to this low resistance. However, as soon as the armature starts moving, a strange thing happens. The Motor actually starts to also act as a generator and creates its own voltage that is referred to as a “counter-electromotive force” (CEMF), which opposes the polarity of the source voltage (battery supply voltage) so the amount of current being drawn from the batteries is drastically reduced. For this example, we are pretending that all of this stuff that goes on while starting the Motor has all occurred and we just have a running Motor producing 5 HP into a 5 HP load connected to an ample source of current (our Battery Pack). We will look at it like this:
We know that A and F each have some small DC resistance, as well as inductive characteristic. But in this example, we will ignore all of that and the CEMF and just treat the Motor as a pure DC resistance, which we will now calculate as a resistor called Rm.
We know that our Source is 36 volts and our Motor is operating at 5 HP.
Back to Ohm’s Law, we get:
P = I x E
5 HP = I x 36
5 x 746 = I x 36 (1 HP is defined as 746 watts)
3730 = I x 36
I = 3730 / 36 = 104 amps (rounded)
So, as the Motor spins along producing 5 HP from a 36 volt source, our Battery Pack is supplying 104 amps. But there is another way to express I and we will use it to calculate Rm, our DC resistive equivalency of the Motor (A and F) as it produces the 5 HP with the 36 volt source.
I = E / R so 104 amps = 36 volts / Rm
Rm = 36 volts / 104 amps = .35 ohms (rounded)
So, when we treat the Motor as Rm, the following would be our drawing.
Now, it gets interesting. Let’s pretend that in wiring our batteries in series and wiring the batteries to the Motor, we have some “less than perfect” connections due to corroded battery posts, frayed wires, corroded battery cable connectors, etc. Let’s say that all of these additional resistances add up to .35 ohms (the same as Rm). Believe me, it doesn’t take much corrosion to add up to .35 ohms. So, let’s call this new accumulation of resistance Ra (a is for accumulated resistance).
Our new diagram would be as follows:
But now, when we calculate the current, we must add Rm and Ra together (they are in series) for what we will call Rt (t is for total of Rm + Ra).
I = E / R = 36 volts / .7 ohms = 52 amps (rounded)
So now, our accumulated resistance (Ra) has caused a reduction of 50% in current flow. Hold on, it gets worse.
If we now calculated the voltage drop across Rm and Ra, we see that:
E = I x R
So
Erm = I x Rm = 52 amps x .35 ohms = 18 volts (rounded)
Era = I x Ra = 52 amps x .35 ohms = 18 volts (rounded)
So, not only has Ra caused a 50% decrease in current flow through the circuit, but it has dropped half of the supply voltage across itself (Ra) and only left half of the supply voltage for Rm (our Motor). Going on to see what effect it has had on the power available to the Motor:
P = I x E
P = 52 amps x 18 volts = 936 watts
If 1 HP equals 746 watts, we’ve just reduced our 5 HP of performance down to:
936 / 746 = 1.25 HP
That’s one quarter of what it was. That’s why I called connections not only important but exponentially important.
Conclusion: Good connections are extremely important in high current situations, as even a very small amount of resistance added to a high current circuit can drastically affect the performance of the cart.
For information about books written by Ron Staley about both electric and gas driven golf carts and their repair, visit the following links.
Electric Golf Cart Repair, both as an eBook and in Hardcopy:
Book: Ronald L Staley: 9780578560557: Amazon.com: Books
Gas Golf Cart Repair, both as an eBook and in Hardcopy:
Gas Golf Cart Repair Book: Ron Staley: 9798987911303: Amazon.com: Books