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I am in the middle of trying to wrap my head around basic electronics. I understand the concepts of voltage, amperage, and resistance, however I'm confused about how heat is generated and related to these three things. These are my assumptions and questions:

  1. When you plug an appliance into a 120 V wall outlet, some amount of amperage will run through the wire depending on the amount of resistance in the circuit (I=V/R, where the voltage is a constant 120 V). Resistance in the circuit depends on the material resistivity and the length/shape of both the wire and appliance.

  2. The difference between a high amperage appliance (like a refrigerator or space heater) and a low amperage appliance (like a light bulb) is the resistance in the circuit. A light bulb draws less amperage because there is more resistance. In other words, resistance is used to control the amperage drawn by an appliance. The appliance is intentionally made to have the right amount of resistance so as to draw the right amperage.

    • Is this assumption correct?
  3. More resistance creates more heat. This is due to the electrons bumping into the atoms in the material they are moving through. I kind of envision this as being like having more friction, so therefore more heat. This is the reason a frayed wire can heat up and cause an electrical fire.

    • Does this mean that since a light bulb has more resistance than a space heater, it is more likely that it can heat up and cause an electrical fire? Are small appliances therefore more dangerous than large appliances due to their higher resistance?
    • Does current in itself create heat? So when you reduce resistance and therefore increase current, does more heat get produced (although heat due to resistance decreases)? Conversely, does increasing resistance (e.g. fraying a wire) also help it cool down since current is reduced?

Any explanations would be greatly appreciated.

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    \$\begingroup\$ "More resistance creates more heat" - It can, BUT - only if the current remains constant. If your supply voltage is fixed, then increasing the resistance will cause the current to decrease, resulting in less power dissipated (as heat). \$\endgroup\$ – brhans Feb 15 '18 at 19:32
  • \$\begingroup\$ What "more resistance" does depends what's providing power. If the power source has fixed voltage, more resistance reduces current drawn from it, and that means less power, less heat. So your light bulb produces less heat than an electric stove. If the power source produces a fixed current (unusual but possible) more resistance can't decrease the current, so teh voltage increases instead. THEN you get more power and more heat. \$\endgroup\$ – Brian Drummond Feb 15 '18 at 21:35
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1.When you plug an appliance into a 120 V wall outlet, some amount of amperage will run through the wire depending on the amount of resistance in the circuit (I=V/R, where the voltage is a constant 120 V). Resistance in the circuit depends on the material resistivity and the length/shape of both the wire and appliance.

Correct.

2.The difference between a high amperage appliance (like a refrigerator or space heater) and a low amperage appliance (like a light bulb) is the resistance in the circuit. A light bulb draws less amperage because there is more resistance. In other words, resistance is used to control the amperage drawn by an appliance. The appliance is intentionally made to have the right amount of resistance so as to draw the right amperage. •Is this assumption correct?

Yes. A little more complex than that, but essentially correct.

3.More resistance creates more heat.

Incorrect. For a fixed input voltage, LESS resistance creates more heat.

\$Power = Volts^2 / Resistance\$

As the resistance falls you consume more current and hence more power.

This is due to the electrons bumping into the atoms in the material they are moving through. I kind of envision this as being like having more friction, so therefore more heat. This is the reason a frayed wire can heat up and cause an electrical fire.

True, but again, it is dependent on how fast and how many electrons are moving. Higher current = more collisions = more heat.

•Does this mean that since a light bulb has more resistance than a space heater, it is more likely that it can heat up and cause an electrical fire? Are small appliances therefore more dangerous than large appliances due to their higher resistance?

Your invalid assumptions make this a little invalid.

Further, temperature change is also dependent on geometry. The filament in a 100W light bulb gets orders of magnitudes hotter than your 750W refrigerator because the heat is concentrated in a small area. It is important to separate hot from heat here. Your fridge puts out more heat than the bulb, but does not get so hot.

•Does current in itself create heat?

We already covered that.

So when you reduce resistance and therefore increase current, does more heat get produced (although heat due to resistance decreases)? Conversely, does increasing resistance (e.g. fraying a wire) also help it cool down since current is reduced?

Again you got the resistance part backwards.

Fraying a wire is a little more complex. What you end up doing here is increasing the voltage drop across that frayed part in the wire, which will have a higher resistance than the rest of the wire, adding another load1 in series with the appliance. This new load steals some voltage from the appliance. The total current is reduced a little. That voltage drop times whatever current the combination load continues to take generates heat in the frayed part. Since the frayed part is small, that heat turns into HOT. If it is frayed enough it can actually start a fire.

schematic

simulate this circuit – Schematic created using CircuitLab


ADDENDUM

1 The term "LOAD" can be confusing in EE. A "LOAD" is generally defined as something that consumes power. However, when you add resistive loads in series to a fixed voltage supply, the "load" on the supply goes down not up. Only when you add loads in parallel does the load on the supply go up.

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  • \$\begingroup\$ The part I still don't totally get is the frayed wire. My interpretation of what you said is that when a wire is frayed, the frayed strands can touch (hot & neutral), creating another circuit. This is the "load" you mentioned, which is basically just the small piece of frayed wire. The 120V is now split between the appliance and frayed wire circuits (the voltage drop is the amount taken from appliance for frayed wire circuit?). What determines how big this V drop is? Shouldn't the high resistance of this small load prevent the current from getting too high/too hot? \$\endgroup\$ – Jahlon Feb 19 '18 at 18:35
  • \$\begingroup\$ @Jahlon no.. let me expand the answer a little... \$\endgroup\$ – Trevor_G Feb 19 '18 at 18:36
  • \$\begingroup\$ @Jahlon there..does the image help? 10R is a bit of an exaggeration though, but you should get the point. \$\endgroup\$ – Trevor_G Feb 19 '18 at 18:46
  • \$\begingroup\$ This is probably just because I'm inexperienced with electrical diagrams, but why is the fray marked 10 ohms while the appliance is marked 100 ohms (shouldn't the fray have higher resistance than the appliance)? \$\endgroup\$ – Jahlon Feb 19 '18 at 20:30
  • \$\begingroup\$ @Jahlon wire has a specified resistance per foot. But that resistance is based on the wire being whole, that is, no broken strands or cuts in the metal. When you fray a wire you either break some of those strands of thin down the copper or aluminum, or whatever the wire is made of. The resistance at the fray is then higher than the rest of the wire. The values in the drawing are just off the top of my head. The relative sizes will depend on the wattage of the appliance and how badly the wire is frayed. \$\endgroup\$ – Trevor_G Feb 19 '18 at 20:47
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More resistance creates more heat.

No, more power is more heat.

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

For a constant voltage, like the 120V in your walls, increasing the resistance decreases the power.

Power is heat, but heat is not temperature. A computer and a lightbulb may both produce 60W of heat, but the lightbulb is very hot - so hot the filament glows.

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For your first point: pretty much correct, with the proviso that 120V outlet isn't really a constant voltage. It's alternating current, which means the voltage varies constantly, but 120 Volts is basically the average voltage for any time scale longer than a small fraction of a second.

Second point about appliances: in quite a few cases (motors, for example) it's usually more about reactance than resistance per se (but yes, the basic idea is essentially correct).

Third point about resistance: no, it's not a simple matter of more resistance producing more heat. In fact, mostly rather the opposite is true--less resistance results in greater current, which results in more heat.

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  • \$\begingroup\$ Maybe it isn't really a constant voltage, but a "constant" 120 V RMS AC power supply will deliver exactly the same power to a purely resistive load as a constant 120V DC supply would deliver. \$\endgroup\$ – Solomon Slow Feb 15 '18 at 20:01
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    \$\begingroup\$ @jameslarge: as long as you integrate it over a time scale greater than a small fraction of a second... :-) \$\endgroup\$ – Jerry Coffin Feb 15 '18 at 20:48

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