Source transformations: confusing problem wording

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SUMMARY

The discussion revolves around calculating the open-circuit voltage of a practical DC voltage source that can deliver 10.5 A when short-circuited and 30 W to a 20 Ω load. The key equation used is Ohm's Law, V=IR. Participants clarify that the 10.5 A refers to the current limited by the internal source resistance when shorted. Understanding these parameters is crucial for solving the problem accurately.

PREREQUISITES
  • Understanding of Ohm's Law (V=IR)
  • Knowledge of electrical power calculations (P=IV)
  • Familiarity with circuit analysis concepts
  • Basic understanding of internal resistance in voltage sources
NEXT STEPS
  • Research how to calculate open-circuit voltage in practical voltage sources
  • Learn about internal resistance and its impact on circuit performance
  • Study power calculations in electrical circuits
  • Explore equivalent circuit models for voltage sources
USEFUL FOR

Electrical engineering students, circuit designers, and anyone involved in analyzing or designing voltage sources will benefit from this discussion.

EngnrMatt
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Homework Statement



A certain practical dc voltage source can provide a current of 10.5 A when it is (momentarily) short- circuited, and can provide of 30 W to a 20 Ω load. Find the open-circuit voltage

Homework Equations



V=IR

The Attempt at a Solution



Well I'm pretty sure that the current entering the resistor load would have to be √(3/2), but I'm not sure exactly what to do from there. I don't quite understand what is meant by the 10.5 A when short-circuited statement. Like would that be the current through the source resistance?
 
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EngnrMatt said:

Homework Statement



A certain practical dc voltage source can provide a current of 10.5 A when it is (momentarily) short- circuited, and can provide of 30 W to a 20 Ω load. Find the open-circuit voltage

Homework Equations



V=IR

The Attempt at a Solution



Well I'm pretty sure that the current entering the resistor load would have to be √(3/2), but I'm not sure exactly what to do from there. I don't quite understand what is meant by the 10.5 A when short-circuited statement. Like would that be the current through the source resistance?
Yes on both counts. If the power source's output is shorted, the internal source resistance will be the only thing limiting the current.
 

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