Analyzing Energy Distribution in a Charging Circuit | Figure 28.19

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SUMMARY

The discussion focuses on analyzing energy distribution in a charging circuit involving a battery, a resistor, and a capacitor, as depicted in Figure 28.19. It is established that half of the energy supplied by the battery is converted into internal energy within the resistor, while the other half is stored in the capacitor. To demonstrate this, one must derive the current function i(t) from fundamental equations and integrate the heat dissipated in the resistor over time, confirming that this energy equals the energy stored in a fully charged capacitor.

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Lewis
Hello, I'd really appreciate a hand with this question. The text of the question follows and there is an attachment with the referenced figure at the end. Thanks very much.

A battery is used to charge a capacitor through a resistor, as shown in Figure 28.19. Show that half the energy supplied by the battery is stored as internal energy in the resistor, and that half is stored in the capacitor.
 

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you can get i(t) from the basic equations. then, integrate the heat dissipated in the resistor (that's what they mean by internal energy, if I understand correctly) - which depends upon the current - from time zero to time infinity. This should end up being equal to the energy stored in a fully charged capacitor.



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