Maximizing Amps in Circuit with 90W Power Supply

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SUMMARY

The discussion centers on maximizing current output from a 90W power supply rated at 19V and 4.74A. The user seeks to draw as many Amps as possible, suggesting a theoretical maximum of 90A, but acknowledges the impracticality of this goal. Experts in the thread caution against attempting to extract 40A, highlighting the significant risks of overheating and potential fire hazards associated with such high currents in amateur setups.

PREREQUISITES
  • Understanding of Ohm's Law (W=V*A)
  • Basic knowledge of electrical circuits and components
  • Familiarity with power supply specifications
  • Safety protocols for handling high current electrical systems
NEXT STEPS
  • Research safe methods for current limiting in power supplies
  • Learn about thermal management techniques in high-current applications
  • Study the design and specifications of power supply units (PSUs)
  • Explore circuit protection mechanisms to prevent overheating and fires
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Electrical engineers, hobbyists working with power supplies, and anyone interested in safely managing high current in electronic circuits.

JoeSalerno
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So, I know the basic equation of W=V*A, but how would I physically do this in a circuit? Say I have 90W=19V*4.74A (http://www.chargerbuy.com/replacement-hp-90w-smart-ac-adapter-19v-474a.html?gclid=CjwKEAiAi-_FBRCZyPm_14CjoyASJAClUigOPligEwQ9DD87xaDL8SQcvxNtWCeo_2nl68GmdozOoxoCbW3w_wcB) for reference. In the project that I am undertaking I need to get as many Amps out of this as possible. Theoretically, I could get 90A out of this but I know it's highly unlikely. If there's at least a way to get 40A out of this circuit I would greatly appreciate knowing how to do so. I plan on cutting the proprietary plug and wring to the +/- leads. If you need any other info just reply, thanks.
 
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The device that is shown in the picture is rarely capable at all to handle such high currents. In addition 40 A are far too dangerous to deal with on an amateur basis. The electronics will likely melt or burn. Plus it cannot be recommended to accidentally get in touch with it. So please stay away from this plan, you could severely get hurt or burn down the house.

Thread remains closed.
 
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I am trying to understand how transferring electric from the powerplant to my house is more effective using high voltage. The suggested explanation that the current is equal to the power supply divided by the voltage, and hence higher voltage leads to lower current and as a result to a lower power loss on the conductives is very confusing me. I know that the current is determined by the voltage and the resistance, and not by a power capability - which defines a limit to the allowable...

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