Electronics > Beginners
how exactly resistor works
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IDEngineer:

--- Quote from: hamster_nz on September 28, 2018, 11:51:42 am ---Unless the pipe is tapered (e.g. a nozzle) the same mass of fluid that enters the pipe exits the pipe, traveling at the same speed, so it is carrying the same energy, regardless of the diameter of the pipe.
--- End quote ---
In a hydraulic system (which is what your pipe and fluid represents) there are two main variables: Pressure and flow. The amount of energy transferred is related to BOTH of them, and you can trade one for the other if you wish. That is, you can drop the flow volume by raising the pressure of the system and still deliver the same amount of energy.

Likewise electronics. Energy transferred ("watts") is the mathematical product (multiplication) of pressure ("voltage") and flow ("current"). This is one solution to Ohm's Law, P = E * I. Like hydraulics, you can halve one variable by doubling the other and the energy/watts remains unchanged.

Now let's "taper" (or constrict) the pipe, as you say. This introduces resistance against the flow. Surprise - we use the same term, "resistance", in electronics. As you introduce resistance, while keeping all else the same, the current (flow) is indeed reduced. But remember, that resistance is "working" to hold back the flow that would otherwise occur. Put yourself in its place... if there were a flow of water coming through a window, and your job was to hold a piece of plywood across the window to resist the flow, would you be "working"? Would you expend effort? Of course. And the more of the window you tried to cover - the more flow you held back - the harder you'd be working.

Your body would start heating up while you worked. Your perspiration system would kick in and you'd start to sweat to cause evaporative cooling to maintain your body temperature. A resistor has no such cooling system to maintain a constant temperature, so that work performed by the resistor raises its temperature as it dissipates the heat. Stop the current flow and the resistor cools down because it takes no effort to hold a piece of plywood against a window with no water flowing through it (nor does it take effort to hold back electrons that aren't trying to move).

Those are the extremes of the situation. Now think about the middle ground. The flow starts up again, a very slight trickle of water through the window (or a very small current through the resistor). How much work is it to resist that? Not much. But as the flow increases, the amount of work required to hold it back goes up. This is, in fact, a linear relationship.

Kick those concepts around in your head for a while and see if it starts to make sense.
Brumby:

--- Quote from: wolframore on September 28, 2018, 01:57:09 pm ---resistance is futile  ;D ;D ;D

--- End quote ---

The collective wants you back.
Brumby:
As with all analogies, the water analogy might struggle a bit here.

The first thing to understand is that energy is lost from the electrical system when current passes through a resistance - and the resistor will heat up.  There is no way around this.

The mental image I have is of a ball rolling down a plane.  With nothing in the way, the potential energy it has at the top is turned into kinetic energy at the bottom - and there is no energy lost (except for a tiny, tiny bit in the rolling action, which we can ignore for this exercise).  This would be the equivalent of a wire.

Now place some posts on the plane and as the ball rolls down, it will hit some of the posts and lose some energy to those posts - sort of like how a cricket ball hitting a bat creates a hotspot:


Add more posts and you are increasing the number of obstacles and the number of impacts which has the result of wanting to reduce the flow of balls down the plane.  This is increasing the resistance.
rstofer:
I'm not a huge fan of the water analogy beyond about 10 minutes the first day of class.

We have Ohm's Law and that tells us all we need to know about the relationship between voltage, current and resistance.  It's a Law, not a suggestion!

You know that if you have a current through a resistor, it drops voltage (E = I * R).  You also know that Power = Current times Voltage (dropped in this case), (P = I * E).  There are other expressions that are algebraic manipulations but, in the end, that current flow creates a voltage drop and the product is heat (power).

https://www.electronics-tutorials.ws/dccircuits/dcp_2.html
ArthurDent:
So far we’ve had resistor analogies involving melons, water, balls, and cricket. I’d humbly like to use frosting, partly because I like food analogies.

If you’ve ever seen someone decorate (embellish) a cake where they fill up a zip-lock bag with frosting and then cut a corner off the bag for the frosting to come out of, they have to supply quite a bit of pressure (intensity) to squeeze the frosting out of the small hole in the bag. The smaller the hole in the corner, the more pressure they have to apply because of the back-pressure (resistance) to force the same volume of the viscous frosting through the hole. If they want to embellish the same number of cakes per hour the decorator either have to increase the pressure (intensity) or decrease the resistance to get the same volume of frosting.

The decorator has to use a formula to determine how to apply these variables and the hole size in the bag will determine the amount of frosting that will go through the hole. They use the equation of Embellishment=Intensity x Resistance or E=IR or as some like to state it, Volume=Intensity x Resistance or V=IR
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