1. Home
  2. Hobby Electronics Basics
  3. Amperage not sensible

Amperage not sensible

I have tried a few times over the years to understand amps, but beyond a superficial grasp, it eludes me.

1) In a river analogy, if voltage is like the action of gravity on water resulting from the slope of the earth's surface, amperage is the amount of current or water passing down the river, and resistance is like a partial dam placed in the river. If gravity (voltage) is increased, current (amperage) increases. If the river is damed to a trickle, current DECREASES. So far so good for Ohm's law.

2) But, if you have an electrically-powered propeller-driven airplane, or an electrically-powered automobile, the larger the propeller or the larger the tires the MORE current is drawn through the system (up to limits of the power source and impedance of the rest of the system).

A larger driven transducer device does more work, so must present increased load and resistance.

So, is amperage (current) increased or decreased by resistance and/or load?

Thanks, sorry for the dumb question.

Reply to
Sarah
Loading thread data ...

Yes, that is pretty much the case. And if you had a water wheel in the stream and you added more resistance (closed the spillway), it would slow to a stop.

Yes...

No. If you increase the load on a motor, and /lowers/ its impedance (AC resistance), and it draws more current.

Resistance and load are not the same thing. In fact, they are nearly the opposite. Increasing the load reduces the resistance and increases the current. Conversely, reducing the resistance increases the load and the current.

Reply to
Chronic Philharmonic

It's helpful to think of voltage like water *pressure*.

As the load (work or power) increases the current rises. A large hyroelectric dam won't generate any power with no water flow will it ?

Ohm's law gives us V = I.R which re = arranged is I = V/R so if resistance reduces, current increases for a given V

Graham

Reply to
Eeyore

The river analogy is a useful one. Think of voltage as the speed of the river, and current as the amount of water flowing, and it may help you visualize it. In this model: Low voltage, low current: a small stream in relatively flat terrain. Low voltage, high current: the Mississippi River. High voltage, low current: a mountain stream. High voltage, high current: the Niagara River. Very high voltage, very low current: a fire hose.

--
Regards,
        Doug Miller (alphageek at milmac dot com)

It\'s time to throw all their damned tea in the harbor again.
Reply to
Doug Miller

Think of a load like a waterwheel.

The work done by the waterwheel (V*I) depends on the amount of water passing it, *and* the height difference (drop) of the water. A fast-running but rather flat river can't be used for much work, nor can a waterfall at a trickle. A fast waterfall gives the most work.

So, consider the waterwheel. Putting the wheel in the water will slow it down but also cause the water level before it to rise, and the level after it to fall. To get the most work, you want to balance the change in water level with the reduction in speed of the water.

Now, a small waterwheel doesn't slow down the water as much, but neither does it change the levels much. A larger waterwheel slows the water down a lot, but also changes the level a lot. It's producing more work for less water speed.

A motor-type electrical load is like that. When lightly loaded, it has less effective resistance[*] so the current is higher, but the voltage drop is less also, so less net energy is used. A higher load increases resistance, reduces current, but increases the voltage drop

- net result, more energy used.

[*] An unloaded motor is like an inductor, not a resistor. The impedance is all imaginary, so no real energy is used. Putting a load on it changes the impedance angle, making more of the impedance real and thus increasing real energy used and real work done.

This is again like a waterwheel - a big wheel with no load will just "go with the flow" and not slow down the river much at all, but as soon as you engage the wheel (say, to grind flour), it starts slowing down the river and letting the water pile up in order to extract useful work from it.

Reply to
DJ Delorie

No. Increased load means less resistance.

Current increases when the resistance is reduced. Resistance is reduced when load is increased.

--
RoRo
Reply to
Robert Roland

And this is where making 'folksy science' comparisons falls down of course.

Graham

Reply to
Eeyore

Current flow _is_ sensible.

  1. Correct.

  1. If you have a simple waterwheel in the river (think undershot, if you're that familiar with waterwheels) with no load, the paddles will dip into the water but won't disturb the flow. Now put a big load on the wheel (like a propellor). Now the wheel will go slower than the river, and the paddle _will_ disturb the flow.

At some point you'll find that you've stretched your analogy to breaking. When that happens it's time to dispense with it and just pay attention to the actual current and voltage.

--

Tim Wescott
Wescott Design Services
http://www.wescottdesign.com

Do you need to implement control loops in software?
"Applied Control Theory for Embedded Systems" gives you just what it says.
See details at http://www.wescottdesign.com/actfes/actfes.html
Reply to
Tim Wescott

Yes. Although the water stream analogy works well to visualize electric current, the waterwheel does not provide a good analog to the electric motor.

Maybe it helps to emphasize the difference between electrical resistance and mechanical resistance:

When the mechanical resistance on the motor's shaft increases, the motor's electrical resistance decreases.

--
RoRo
Reply to
Robert Roland

An electric motor has a back voltage (EMF) proportional to speed. If you put 100 volts to a motor and let it run load-free at high speed, the back EMF could be 95 volts, so only 5 volts net is left for the motor. With a load that would drag it to half speed, the back voltage would be only 50 volts and now there are 50 volts left for the motor. With

50 volts the motor will draw much more current than with 5. John Polasek
Reply to
John C. Polasek

---------------------- Sorry, John, I wrote the following before seeing your reply so take this as an addition to what you have said rather than a contradiction.

In the case of an electric motor, as in the airplane or automobile, Ohm's Law does not apply- a motor is an active device which includes a voltage source as well as an Ohmic resistance.

as an example for a DC motor:

V=R*I +E where V is the applied voltage and E is a back emf dependent on speed. I is the current and R is the resistance. The power converted to mechanical power is E*I =Torque*angular velocity (or force*velocity) In such a motor the torque is proportional to I and the speed proportional to E. Actual mechanical uoutput is lower because of mechanical losses The input power is V*I =R*I^2 (electrical loss) +E*I and if R*I^2 is greater than E*I you have a lousy and inefficient motor.

Another part of the problem is that increased resistance doesn't necessarily mean an increased load. In discussion of motors, power output (as determined by the mechanical conditions) is the "load"

--

Don Kelly snipped-for-privacy@shawcross.ca remove the X to answer

Reply to
Don Kelly

ElectronDepot website is not affiliated with any of the manufacturers or service providers discussed here. All logos and trade names are the property of their respective owners.