Heat Generation in a circuit due to moving charges

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SUMMARY

The discussion focuses on calculating the heat generated in a circuit involving a capacitor with capacitance C, charged by a battery with emf E and internal resistance r, while connected to an additional resistance of 2r. The correct formula for the heat liberated inside the battery when the capacitor is 50% charged is established as E²/C. Participants noted the necessity of using calculus to account for the changing voltage and current during the charging process, emphasizing the integral of power over time to determine energy.

PREREQUISITES
  • Understanding of capacitor charging equations, specifically q=Q(1-e-t/RC)
  • Familiarity with electrical concepts such as emf, internal resistance, and resistance in series
  • Basic knowledge of calculus, particularly integration for energy calculations
  • Comprehension of energy units and their application in electrical circuits
NEXT STEPS
  • Study the derivation of the capacitor charging equation q=Q(1-e-t/RC)
  • Learn about the relationship between power, energy, and heat in electrical circuits
  • Explore the application of calculus in electrical engineering, focusing on integration techniques
  • Investigate the implications of internal resistance on circuit performance and energy loss
USEFUL FOR

This discussion is beneficial for electrical engineering students, circuit designers, and anyone involved in analyzing energy efficiency in capacitor circuits.

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1. Homework Statement
A capacitor of capacitance C id charged by a battery of emf E and inernal resistance r, A resistance 2r is also connected in series with the capacitor. The amount of heat liberated inside the battery by the time capacitor is 50% charged is? Answer is E2/C



2. Homework Equations
q=Q(1-e-t/RC)
btw, RC means the time constant and Q is the charge at steady state



3. The Attempt at a Solution
tried to find time when charge in capacitor is 50% charged and then find the heat using V2/R but had no success
 
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The voltage and current are changing with time as the capacitor charges. This suggests that some calculus will be involved.

Hint: ∫ watts*dt = Energy
 
You might also note that the answer suggested in the problem statement is clearly incorrect, as the units of E2/C are not energy.
 

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