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What causes a mobile phone charger or any adapter to increase the current on same voltage. I saw that chargers are rated with different amp on same volt ( 5v.1A 5v,500ma 5v,250ma).

Ohm law says (V=IR) I am confused....!!!

Tony Stewart EE75
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    Long story short, the current rating is the maximum value that the charger can supply. The load will determine how much current is drawn. – Tom Carpenter Nov 14 '17 at 17:09
  • Ohms law says that a certain voltage is required to drive a certain current through a certain resistor. It doesn't say you can take any random voltage and random current and divide the two. – Tom Carpenter Nov 14 '17 at 17:15
  • Look up Thevenin and Norton equivalent circuits. That might help you understand why ohm's law is not the sole governing factor for the current draw in this situation. Although, in reality, USB chargers are much more complicated that. – kjgregory Nov 14 '17 at 19:19

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To translate ohm's law to weight for easier understanding. \$V=I×R\$ => \$g = cm³×(g/cm³)\$, one is weight, one is volume, and one is density. You can't choose to have 1000 cm³ water weighing 30 kg. If you choose to have 1000 cm³ then it will weigh 1 kg, if you want 30 kg water then you will have 30000 cm³.

This is assuming water weighs 1 kg with a volume of 1000 cm³.

I know this question will be closed within the hour, but meh, thought I'd at least elaborate on ohm's law.

Harry Svensson
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  • @KingDuken You can edit if you want, I won't retract the edit. - But you at least said '*"cubed" symbol*', so you did see what it was. But whatever floats your boat. – Harry Svensson Nov 14 '17 at 17:26
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Ohm's Law applies to things that draw current, not things that supply voltage. If you connect a load to a 5V power supply and the current drawn is 1A at 5V then Ohm's Law says the load has a resistance of 5Ω.

Your 'chargers' are actually power supplies rated to deliver a maximum of 500mA or 1A. The charger in the phone etc. will draw whatever current it wants, up to the maximum the power supply can deliver.

Maximum current output of a power supply is determined by several components (transformer, rectifier diodes, filter capacitors, transistors etc.) all of which have to be matched to the current rating. Switch-mode power supplies often have a current-limiting circuit in them which prevents drawing more current than the supply can safely deliver. If the current limit is exceeded the power supply may shut down until the load is removed.

Bruce Abbott
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Ohm's Law applies to the unknown equivalent load resistance, Req.

I=V/Req

Although charging devices are dynamic, the current they draw, depends on hard limits imposed in the design and the battery condition and if the device display is also on as both increase demand current.

Since the target device usually has a smart battery charger, the current is limited by the target before it is limited by a voltage source, capable of more current.

I recently bought a 6 port hub 2.4A 5V adapter for about $35Cdn at Walmart. It avoids the "octupus outlets". It supports 2.4A per port but only 12A max = 60W.

  • Devices such as Ipads and iPhone X are capable of accepting 12W charging or 2.4A when needed.
  • Smaller devices with smaller batteries must use less current to reduce heat & ageing on the battery, so the "smart" load determines current.

Here is just an example of a cheap 60W USB hub. (12A/6port) Not a testimonial but $0.50/W is typical for USB hubs and this will come down with volume. Cheapest is not the best.

https://www.ebay.com/itm/Multi-Port-60W-USB-Charger-12A-Rapid-Charging-Station-Desktop-Travel-Hub-iPhone-/162596367150?_trksid=p2349526.m4383.l4275.c10&var=461598838749

If you just have a passive heater resistor, then it is linear and again I=V/R for example 5V/2.4A=2.08 ohms for 12 watts.

Tony Stewart EE75
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