Calculating Current and Resistance: "17V, 3C, 25J

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SUMMARY

The discussion focuses on calculating current and resistance using the formula Current = (emf)/(resistance) and understanding the relationship between capacitance, voltage, and stored energy. The user attempts to derive voltage from stored energy using the equation Stored energy = (Capacitance)(Volt ^ 2) /(2), ultimately determining that the voltage is 17V when using 3C of charge. The conversation clarifies that the energy extracted from the battery is related to the emf and charge, emphasizing the importance of understanding these electrical concepts for accurate calculations.

PREREQUISITES
  • Understanding of Ohm's Law and its applications
  • Familiarity with the concepts of capacitance and stored energy
  • Knowledge of electrical units such as volts, coulombs, and joules
  • Basic proficiency in algebra for manipulating equations
NEXT STEPS
  • Study the relationship between emf, current, and resistance in electrical circuits
  • Learn about the principles of capacitance and its role in energy storage
  • Explore the derivation and applications of the energy stored in capacitors
  • Investigate practical examples of calculating energy in electrical systems
USEFUL FOR

Students in physics or electrical engineering, educators teaching circuit theory, and anyone seeking to deepen their understanding of electrical energy calculations.

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Homework Statement


http://i33.photobucket.com/albums/d86/warnexus/emf_zpsbff89890.jpg


Homework Equations


Current = (emf)/(resistance)

Capacitance = (Charge)/(voltage)

Stored energy = (Capacitance)(Volt ^ 2) /(2)

The Attempt at a Solution



I need to find current and resistance so I can find emf.

capacitance = (charge)/(volt)

stored energy = (charge(voltage)/(2)

25 J (2) /(3C) = 50/3

voltage is 50/3 or 17 V.
 
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This has nothing to do with either resistance or capacitance.

You have a battery rated at "emf" volts and you have sucked 3 C (coulombs) of charge out of it. How much energy is that?
 
rude man said:
This has nothing to do with either resistance or capacitance.

You have a battery rated at "emf" volts and you have sucked 3 C (coulombs) of charge out of it. How much energy is that?

thats right! volt is measured in joules per coulomb! got it thanks!
 

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