Understanding Amps: How Can They Be Determined Without Knowing the Circuit?

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Amperage ratings on electrical devices can be determined based on known voltage sources and the design of the circuit, particularly for devices plugged into AC mains. These devices are typically connected in parallel, allowing them to operate independently without affecting each other's performance. For configurable devices like computer motherboards, manufacturers provide maximum power ratings based on potential configurations, which can vary depending on the components used. Adding more components, such as RAM or video cards, can increase the total current draw because they reduce the overall resistance seen by the power supply. Understanding these principles clarifies how amperage is calculated and why adding devices can lead to circuit overloads.
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I often see on electrical devices, they say how many amps the device uses. But I wonder how can that be? if you do not know the voltage and the resistance (im guessing they do know the resistance) how can they tell you how many amps it pulls? Also doesn't it matter what components are in the circut before the device? What if the circut divides into 100 parallel resistors right before the device, won't the amperage be significantly less? And what if these 100 resistors are in series, won't the voltage then be considerably less, causing a lowering in amperage?

The question I am asking, is how can they make a determination of the amps used without knowing the circuit.
 
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Wishbone said:
The question I am asking, is how can they make a determination of the amps used without knowing the circuit.

"They" designed the circuit so they do know what's inside.
 
And the voltage source for the type of circuits you are talking about is known, and has low source impedance. I'm assuming the devices you are asking about are ones that plug into the AC Mains power outlets in the walls, etc. All power consuming devices are plugged in parallel into this power source, so one device doesn't affect another by being in series with it. And the source resistance (or impedance) of the AC Mains is very low, so its voltage level does not droop much when a maximum load is placed on it.
 
Well I an see what you mean for things like TVs or Vacuum cleaners. However, for something like computer hardware, you cannot determine the amperage for a motherboard, or RAM, unless you know all other components of the computer and power supply, no? Or for a more general example, you simply cannot make any determination about the amperage of a device when you move it from one circuit to another, correct?
 
Wishbone said:
Well I an see what you mean for things like TVs or Vacuum cleaners. However, for something like computer hardware, you cannot determine the amperage for a motherboard, or RAM, unless you know all other components of the computer and power supply, no? Or for a more general example, you simply cannot make any determination about the amperage of a device when you move it from one circuit to another, correct?

Ah, that's a better question. When the designers put a power rating on a configurable device like a computer motherboard with empty memory sockets, they are listing the maximum power that can be consumed by the device, when fully populated with options. If you open up your PC (with the power off) and look at the sticker on the power supply, it will list the maximum output currents that are available from the supply for its different output voltage wire harnesses. The input power label on the outside of the PC and in the PC's Owner's Manual correspond to the maximum total output power of the PC's power supply, plus some fraction due to the lost heat because the power supply is not 100% efficient in converting the input AC Mains power to its output supply rails.
 
Ah excellent explanation, my final question is: why would adding greater resistance (more RAM chips, video cards, or other stuff that draws power) increase the amperage, and thus bringing it to the maximum levels? How come pugging more resistance (ie. vacuum cleaner, tv) into a outlet can blow a fuse? I figure the more resistance you add to a circuit with a constant voltage source, the lower the current will be.
 
Because the added resistances are in parallel, thus REDUCING the total resistance as seen by the power supply.
 
I understand now, thanks.
 
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