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I'm trying to understand the basic concept of how such electronic equipment works.

For example, if I have 220V 10A power line, when I plug a heating element into it, will it take all available power?

Without external tool to limit power being drawn, what decides how much power will be drawn by heating element? Is it its resistance? Or is it the voltage and amperage of power supply? If so, how is the formula?

Additional questions, what is usually being used to limit power drawn by such equipment, and how does it works?

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  • \$\begingroup\$ What kind of heating element? Is it a heating system, a peltier or a resistor? \$\endgroup\$ Commented May 10, 2022 at 11:47
  • \$\begingroup\$ Resistor maybe? It's heating coil \$\endgroup\$
    – Bramble
    Commented May 10, 2022 at 12:11
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    \$\begingroup\$ Footnote to all the answers about resistors: while a heating element is a resistor, like all resistors its resistance will vary depending on temperature! \$\endgroup\$
    – pjc50
    Commented May 10, 2022 at 12:11
  • \$\begingroup\$ How does it normally vary? Goes up or down as temperature rises? Is there some exception to that normal? @pjc50 \$\endgroup\$
    – Bramble
    Commented May 10, 2022 at 12:20
  • \$\begingroup\$ For a heating element? It's a resistor, so just learn about Ohm's Law. \$\endgroup\$
    – user16324
    Commented May 10, 2022 at 13:42

3 Answers 3

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For example, if I have 220V 10A power line, when I plug a heating element into it, will it take all available power?

  1. All power supplies/power lines have a maximum current that can deliver. If your heating element tries to draw 11A and the maximum current output of your power supply/power line is 10A, then the output voltage of your power supply/line will drop (for example from 220V to 200V) until it delivers 10A, since this is its maximum current output.

If you want to make the heating element draw all the available power, its resistance should be \$ R=\dfrac{V}{I}=\dfrac{220V}{10A}=22Ω.\$ With a heating element of 22Ω, you will manage to draw 10 Amp. If the resistance is more that 22Ω, it will draw less current. If the resistance is less than 22Ω, It will try to draw more current (see 2.), but it wont be able to. As a result, it will draw 10A and the source voltage will start to drop.

Without external tool to limit power being drawn, what decides how much power will be drawn by heating element? Is it it's resistance? Or is it the voltage and amperage of power supply? If so, how is the formula?

  1. Both. The heating element's resistance will determine the current that it TRIES to draw, and it also follows Ohm's law (\$ I=V/R \$ ), so both input voltage and resistance determine the current draw. I say "try" to draw since due to (1.) it is limited to the power supply's current capability.

Additional questions, what is usually being used to limit power drawn by such equipment, and how does it works?

Depends on what you are trying to do, the application, the type of input power (AC/DC).

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  • \$\begingroup\$ The input is AC power, and it's for those heating coil. \$\endgroup\$
    – Bramble
    Commented May 10, 2022 at 12:17
  • \$\begingroup\$ @Bramble the only thing "limiting" power in AC mains power is the device's resistance, or the circuit breaker which breaks the circuit if that doesn't work. \$\endgroup\$ Commented May 10, 2022 at 13:49
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    \$\begingroup\$ Note that the maximum current of house wiring (which is likely the source of the 220 V) is never related to voltage drop--it's related to heating in the wires. Breakers are sized as they are to prevent fires, not because the wires physically can't provide more than that much current. This is why bypassing fuses is so dangerous. \$\endgroup\$
    – Hearth
    Commented May 11, 2022 at 3:16
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Assuming the heating element is a pure resistor, the power drawn will be defined by the voltage of the source (220V in your case), and the resistance of the heating element. The power can be calculated as the product of voltage and current, and the current can be computed using Ohm's law: $$P = I \times V = \frac{V}{R} \times V = \frac{V^2}{R}$$

A space heater might have a 30 ohm heating element. When connected to a 220V supply, this would result in $$\frac{(220V)^2}{30 Ω} \approx 1613W$$

As for your additional question, there are a couple common ways of controlling this power.

A cheap space heater might simply have two heating elements of different resistances, and a switch allowing you to connect power to just the low-power element, just the high-power element, or both elements, giving you four (including off) power settings.

Another method, commonly used in stovetops, is to turn the heating element on and off with a given duty-cycle. If the element is outputting 1 kW, 70% of the time, and is off the other 30% of the time, the average power output will be 700W. In these applications the thermal load will normally be high enough to smooth out the fluctuations, so that the switching on and off can be done quite slowly (several seconds between switchings).

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  • \$\begingroup\$ With those value, I calculated the current would be 7.33 A. What will happen if the power line can only supply 5 A at most? Will it blow some fuse? If so, what can I do so it won't trigger fuse? \$\endgroup\$
    – Bramble
    Commented May 10, 2022 at 12:16
  • \$\begingroup\$ If the power line is a normal house-circuit with a 5A circuit-breaker, then yes, it will trip the breaker. Note that it might take several seconds, up to a few minutes, for the breaker to trip with only ~50% overcurrent. To avoid triggering the breaker, you would need to increase the resistance, or use a more sophisticated switching converter. \$\endgroup\$
    – sondre99v
    Commented May 10, 2022 at 12:55
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Power 'P' drawn by a heating element is determined by its operating voltage 'V' and operating current 'I'.

P = V * I

Operating current 'I' is determined by the element's operating voltage and its resistance 'R'.

I = V / R

The electrical design parameters of a heating element i.e. its operating voltage and power is marked on the name plate for the user to make a choice based on requirement.

A heating element marked 240V ~ 1.5 kW, for example, would draw only 1.5 kW from a 240V ~ source to which it is connected. The current drawn by the element would be = 1.5 * 1000 / 240 = 6.25 A.

However, it would be fed using 14 AWG cable, protected by a 15 A fuse or circuit breaker, to minimise voltage drop and cable heating.

Heater temperature may be varied through on/off control using an 'energy regulator' commonly known as a 'Simmerstat'.

Closed-loop on/off or PWM control could also be effected with feedback from a temperature sensor.

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