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I'm confused again about electricity. There seems to be two types of resistance, a dissipative and a restrictive one. Everything I've read, however, says there's only one type of resistance.

Let's assume, for instance, you have current supplied to a heating coil (nichrome wire in a coil) in a circuit with a potentiometer in series. The heating coil converts the current energy into heat via the resistance of the electrons. The potentiometer, on the other hand, can equally resist the same amount of current flowing to the coil, yet it doesn't dissipate much heat at all. It appears the heating coil's resistance works by dissipating the excess current into heat, while the potentiometer restricts the current flowing through it. Is there really two types of resistance? Why doesn't the potentiometer heat up as much as the coil?

Notes I made a small but potentially important edit. When I used "coil" in my example, I meant nichrome wire formed into a coil used as a heating element as commonly found in soldering irons, space heaters, etc. A poster clued me in that "coil" might be interpreted as an induction heater.

Btw, I've had nothing but excellent experiences with this community. I am deeply touched by all of your efforts and assistance! While it maybe off topic, I cannot allow such good deeds to go unappreciated. Thanks for all the help!

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  • \$\begingroup\$ What are the resistance of the coil, the resistance of the potentiometer, and the supply voltage? \$\endgroup\$
    – user253751
    Feb 5, 2021 at 3:50
  • \$\begingroup\$ Well, it shouldn't matter, but let's say 120V 10 amps and a 1200watt heater and a 10K pot. \$\endgroup\$
    – user148298
    Feb 5, 2021 at 3:59
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    \$\begingroup\$ You can't have 120V, 10 amps, and a 10K resistance. Those things are not compatible. V=I*R is Ohm's Law. It's as impossible as having a bag with 5 1kg weights in it that weighs 3kg. \$\endgroup\$
    – user253751
    Feb 5, 2021 at 4:00
  • \$\begingroup\$ the pot is variable. I just picked a large upper range. \$\endgroup\$
    – user148298
    Feb 5, 2021 at 4:01
  • \$\begingroup\$ What's its actual resistance though? Pick a number. And the resistance of the heater. Also pick a number. \$\endgroup\$
    – user253751
    Feb 5, 2021 at 4:04

9 Answers 9

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Maybe an actual circuit would help. R1 represents a heating coil and R2 is a potentiometer.

schematic

simulate this circuit – Schematic created using CircuitLab

Here's a graph showing the power dissipated in the coil (R1) and the potentiometer (R2) as the potentiometer resistance varies. Note that the power dissipated by R2 is always much less than the max power that can be delivered to R1.

DC sweep of R2's resistance. The outputs are P(R1) and P(R2).

There's only one kind of resistance. Resistors are devices that obey Ohm's Law. There are two equations at work here for power:

$$V = IR$$ $$P = VI$$

As a consequence of those:

$$P = \frac{V^2}{R} = I^2R$$

That's it. There's no magic here. Two resistors with the same resistance carrying the same current will dissipate the same power. If the power is different, then the resistance and/or the current must also be different.

It might also help to consider the extremes and what has to happen in the middle. When R2 = 0 ohms:

$$I = \frac{V}{R_1 + R_2} = \frac{100\mathrm V}{100\Omega + 0\Omega} = 1 \mathrm A$$ $$P_{R1} = I^2R_1 = (1\mathrm A)^2 \cdot 100\Omega = 100 \mathrm W$$ $$P_{R2} = I^2R_2 = (1\mathrm A)^2 \cdot 0\Omega = 0 \mathrm W$$ $$P_{total} = P_{R1} + P_{R2} = 100\mathrm W + 0 \mathrm W = 100 \mathrm W$$

Now what if we made R2 infinite (an open circuit)?

$$I = \frac{100\mathrm V}{100\Omega + \infty\Omega} = 0\mathrm A$$ $$P_{R1} = P_{R2} = P_{total} = 0 \mathrm W$$

So as we turn up the pot (to infinity!), the total power dissipated goes from 100 W to 0 W. The power in R1 likewise goes from 100 W to 0 W. The power in R2 starts at 0 W and ends at 0 W. But in between it's not zero! This means R2's power has to go up and come back down again.

It turns out R2's power reaches a maximum when its resistance is the same as R1. This is called the maximum power transfer theorem.

The order of the resistors doesn't matter. When R1 is greater than R2, R1 will dissipate most of the power. When R2 is greater than R1, R2 will dissipate most of the power. You can see that in the graph -- at first, the total power line is close to the R1 line, but by the end it's close to the R2 line.

[Disclaimer: I have no idea if any of this is how old toasters worked. But the circuit theory is valid.]

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  • \$\begingroup\$ Boy, that circuit diagram and graph helps. The potentiometer works backwards, but makes sense when you see. \$\endgroup\$
    – user148298
    Feb 5, 2021 at 5:03
  • \$\begingroup\$ LOL! Damned be physics! It gets clearer by the minute. The magic is in the pot. The heater will always get first dibs on the power. \$\endgroup\$
    – user148298
    Feb 5, 2021 at 5:12
  • \$\begingroup\$ @user148298 Kind of. The total resistance determines the total power to both resistors. Turning the potentiometer to zero reduces the total resistance to the minimum, which gives maximum power in the heating coil. Turning it up reduces the total power, but with a non-zero resistance the pot starts dissipating some of the power. I updated my graph to show the total power too; see if that clarifies things. \$\endgroup\$
    – Adam Haun
    Feb 5, 2021 at 6:27
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The potentiometer heats up too, it's just that this isn't its intended purpose.

This is, in fact, the reason you don't use a potentiometer in the way you've described. If you were to actually hook it up like that, it's quite likely that either the pot would overheat, or both the pot and the heater would remain cool. Instead, the proper way to do this would involve using phase angle control with a thyristor for AC, or PWM control with a FET for DC.

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  • \$\begingroup\$ It heats up, but not as much. Where's the energy going? \$\endgroup\$
    – user148298
    Feb 5, 2021 at 3:04
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    \$\begingroup\$ @user148298, if it has the same resistance as the heating coil and you push the same amount of current through it, it produces just as much heat. If it doesn't get as hot that is because it has a more effective heat sink attached to it. \$\endgroup\$
    – The Photon
    Feb 5, 2021 at 3:06
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There is only one kind of resistance and it is both restrictive and dissipative.

You are confusing temperature with heat (wattage). If I dissipate 100W over 1 square kilometer, the temperature is going to rise by almost nothing because all that power is spread out over a huge area. If I dissipate 100W over 1 square nanometer, it is going to reach thousands (or maybe millions?) of degrees because all that power is concentrated in a tiny area.

Temperature is determined not by the amount of power dissipated, but by the concentration or density of the power dissipated.

I could have two resistors that are both 10Ohms with the same current running through them so they are dissipating the same amount of heat. But the giant one will be very cool while the tiny one will be extremely hot, but both are dissipating the same wattage.

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  • \$\begingroup\$ Interesting... So the pot will be hotter if the resistance is higher. \$\endgroup\$
    – user148298
    Feb 5, 2021 at 4:24
  • \$\begingroup\$ @user148298 I didn't say that. For two pots of the SAME PHYSICAL SIZE the one that has higher resistance will only be hotter IF they both have the SAME CURRENT running through them. But, the pot with the higher resistance will require more voltage to push that current through it while the lower resistance pot will require less voltage. \$\endgroup\$
    – DKNguyen
    Feb 5, 2021 at 4:40
  • \$\begingroup\$ That means that if you apply the same voltage across both pots, the higher resistance will actually be COOLER since it will have less current running through it than the lower resistance pot. \$\endgroup\$
    – DKNguyen
    Feb 5, 2021 at 4:42
  • \$\begingroup\$ It is like saying if you push a heavy block and a light block at the same speed, the heavy block will have more kinetic energy. But the heavy block requires more force to to push at the same speed as little block. If you push the both blocks with the same force, the little block ends up moving faster and has more kinetic energy than the larger block. \$\endgroup\$
    – DKNguyen
    Feb 5, 2021 at 4:45
  • \$\begingroup\$ OK. Got it now. Thanks. \$\endgroup\$
    – user148298
    Feb 5, 2021 at 4:49
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Resistance is unique and defined yet has many adjectives depending on how it is used.

"Everything has some resistance", including insulators, also known as dielectrics. Insulators have two components of resistance, the effective series (ESR) and the leakage current parallel equivalent.

Semiconductors have variable resistance controlled by an electric field in a tiny junction by voltage with some exponential factor. The low end of R is limited by fixed bulk resistance due to chip size and thus power dissipation limit.

There are special effects at a higher frequency called Skin Effects from Eddy currents which adds resistance to conductivity as the current circulates to oppose the voltage, below the surface like the force on a paddle in the water.

When motors are mechanically loaded, their inductance is shunted by a parallel resistance.

Metal conductors have a positive temperature coefficient (PTC) and both conductivity and temp. sensitivity depends on the metal. Some alloys (NiCr) are used as heater wires while different metal oxides are PTC for temp. compensation or used as resettable fuses and others are NTC for inrush current limiting (ICL).

Damping: a low resistance source such as an audio amp (voltage source) and the inductive “kickback” or back EMF of a woofer.

Dissipative: a resistor is a conductor with losses that creates heat energy unlike ideal reactive parts like L, C. Transistors also dissipate heat due to internal resistance which is controlled by bias voltage or current.

Lossy: any non-ideal conductor including inductors and capacitors which have low series resistance.

Leakage resistance: high resistance in parallel with insulators or dielectrics or capacitors or batteries which shared similar properties to store charges.

Restrictive: current limiting resistor

Toaster Pots are not resistors but rather thermostats. They use the bending properties of bimetallic strips with heat with pot-controlled mechanical deflection to represent the temperature rise before switching off.

Often many examples neglect resistance when it is not significant compared to the load, such as a switch for ON/OFF. Electronically fast-switched transistors with pulse width modulation (PWM) have the average resistance which depends on the total load resistance divided by the duty cycle where the switch resistance is selected by computing the temperature rise for some low ratio of the load power much less <<1% when on.

... and many more examples, remember everything has resistance !

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The potentiometer, on the other hand, can equally resist the same amount of current flowing to the coil, yet it doesn't dissipate much heat at all.

It converts electrostatic potential energy into heat the same way the heating coil does.

But the heating coil is designed to pass a large current, and generate a lot of heat, without melting.

If you tried to pass the same amount of heat through the potentiometer as you can through the heating coil, it would heat up so much that it melted or caught fire.

Why doesn't the potentiometer heat up as much as the coil?

Because the user has the wisdom not to push so much current through it that it would melt or catch fire.

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  • \$\begingroup\$ The same current should be flowing through both the heating coil and the pot? Correct? \$\endgroup\$
    – user148298
    Feb 5, 2021 at 3:08
  • \$\begingroup\$ @user148298, no, the typical potentiometer you buy from Digikey will go up in smoke with 1/100 the current you would expect to pass through a typical heating coil. \$\endgroup\$
    – The Photon
    Feb 5, 2021 at 3:17
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If the potentiometer is the same resistance as the coil, and the same current is flowing through both, both will heat up the same.

Sometimes we will use a resistor to limit the current in a circuit. Other times we use a resistor to reduce the voltage applied to other components. Sometimes we use a resistor to produce heat.

Any resistor will do any of these things - or really will do all of these things at once.

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  • \$\begingroup\$ If I put a resistor in series with heating coil and the coil only reaches half of the maximum, the other half of the heat is dissipating through the resistor? \$\endgroup\$
    – user148298
    Feb 5, 2021 at 3:13
  • \$\begingroup\$ If the resistor and the heating coil are the same resistance, they will produce the same amount of heat. \$\endgroup\$ Feb 5, 2021 at 3:18
  • \$\begingroup\$ @user148298 If you put a resistor (with resistance equal to that of the coil) in series with heating coil and plug them to the same voltage, total amount of energy will be half of what it was and amount of energy dissipated in the coil will be a 1/4 of what it was. \$\endgroup\$
    – AlexVB
    Feb 5, 2021 at 8:07
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As you mentioned, the coil heats up, but have you ever thought why the wire is looped around and made into a coil? to include a longer length of wire in a smaller space. so why longer length? to have higher resistance in the wire. taking the power equation in its simplest form:

$$P = I^2R$$

So one thing that I believe you did not understand while looking at this question, is that if you put a pot with a coil in series, you are just adding a fixed resistance (since there is no connection to the movable bit of the pot) in series with your coil. The pot would REALLY get hot if you put it in series with a coil, say if you are connecting it to mains (230 Vrms) and your pot is 10K and your coil is, say 15 Ohms, you are getting 5.27W of loss from the original 5.28W!! That would make the resistor really hot! (ofc depending on your type of resistor, but a normal chip SMD or a THT used in dev kits would start to get so hot that it'll affect it's resistance)

There is no such thing as a restrictive resistance and a dissipative resistance. The fact that a resistor "resists" current flow is in fact because of dissipating their energy into heat. Now you can make use of this fact in a coil, generating heat, and it would be a problem for resistance in your wires and other stuff, resulting in power loss.

btw, if you want to control how much heat is generated through your coil, you will need a bit more complicated of a system to control how much power is delivered to your coil.

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  • \$\begingroup\$ I've seen toasters with simple pots to control the heat. So if they can drop the heat by a 400 degrees, does it mean it is dissipating 400 degrees as heat too? If so, then the knob should melt too. \$\endgroup\$
    – user148298
    Feb 5, 2021 at 3:51
  • \$\begingroup\$ @user148298 you should understand that those pots are not connected to the coils, everything is not just what you see, same thing with the pots on your oven! They are an input into the control mechanism, a more complicated thing than just connecting the pot to the coil, and again connecting a pot in series with a coil does not even mean a variable resistance! it's fixed, a pot has 3 terminals, and if you connect it's - and + in series with the coil your just adding fixed resistance, I suggest you look deeper into what a pot actually is. \$\endgroup\$
    – NeuroEng
    Feb 5, 2021 at 3:58
  • \$\begingroup\$ What about those antique toasters and ovens? No pwm, no triacs or diacs. A pot is fixed resistance? If you don't touch the dial? \$\endgroup\$
    – user148298
    Feb 5, 2021 at 4:13
  • \$\begingroup\$ @user148298 dude... I appreciate this out of box thinking I really do! but imagine this, your pot has 3 terminals, + supply, - supply and a third one which changes resistance w.r.t either of the ends. but forget about the pot, the knobs on the old toasters were either not pots, or were variable inductors. hope this helps \$\endgroup\$
    – NeuroEng
    Feb 5, 2021 at 4:24
  • \$\begingroup\$ @user148298 in an antique toaster you would expect dial to switch coils. For example, connect extra coil in parallel, or switch two coils from parallel to series connection. \$\endgroup\$
    – AlexVB
    Feb 5, 2021 at 8:16
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Are there two types of resistance?

In a way there are, but "type of resistance" and "restrictive" are not the terms that are normally used. Resistance dissipates energy as heat. Inductive and capacitive reactance present forms of restriction to current flow without dissipating very much power.

The assumption that you presented does not describe "two types of resistance at all. Other answers explain that.

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  • \$\begingroup\$ What about dissipating energy as light instead of heat as in LEDs? Gotcha! \$\endgroup\$
    – user148298
    Feb 5, 2021 at 3:18
  • \$\begingroup\$ An LED is not a type of resistance and would not be likely to be described as such. Are you playing a game or asking a serious question? \$\endgroup\$
    – user80875
    Feb 5, 2021 at 3:21
  • \$\begingroup\$ Serious question. The LED and an incandescent is dissipating the energy into light and heat. Correct? I've seen countless toasters and ovens with a simple potentiometer controlling the heat. If they can drop the heat from 500 to 100 degrees, then either it restricts the current or it dissipates the excess (400 degrees) of heat. If the latter, the plastic knob should melt. Correct? \$\endgroup\$
    – user148298
    Feb 5, 2021 at 3:43
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    \$\begingroup\$ Older heater or toaster controls used a bimetalic strip that bent when heated. The adjustment knob would move a contact relative to the bimetalic strip - when the strip heated enough, it would bend away from the adjustable contact to open the circuit and turn the heater off. When it cooled a bit, it would bend back to make contact and turn the heater on again. \$\endgroup\$ Feb 5, 2021 at 4:52
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    \$\begingroup\$ @user148298: Those toaster "pots" were thermal timers, not pots. The toaster ran on full power for more or less time. In any case a rheostat (variable resistor) would be used if you had to reduce power, not a potentiometer. Rheostats were used in theatre dimmers. \$\endgroup\$
    – Transistor
    Feb 5, 2021 at 9:01
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All resistors have (only one kind of) a resistance R across which a voltage drop V = I*R is lost. This drop is subtracted from the input voltage and (if the circuit is powered by a constant voltage source) the current I decreases. This is accompanied by power dissipation in the form of heat.

If the heat cannot be dissipated, the temperature of the resistor begins to rise and it turns into a heater. This is a desirable property in heat engineering.

In electronic circuits, however, heating is an undesirable property, and therefore it is done so that the temperature does not rise. For this purpose, the applied power must be less than the resistor nominal power.

The desired property here is only the voltage loss which is used to decrease the input voltage. Typical examples of this technique are the rheostat and voltage divider.

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