Analyzing Energy Distribution in a Charging Circuit | Figure 28.19

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The discussion revolves around analyzing energy distribution in a charging circuit involving a battery, resistor, and capacitor, as illustrated in Figure 28.19. It emphasizes that half of the energy supplied by the battery is converted into internal energy in the resistor, while the other half is stored in the capacitor. To demonstrate this, one must derive the current over time and integrate the heat dissipated in the resistor from zero to infinity. This integration should equal the energy stored in a fully charged capacitor. The conversation seeks clarification on these concepts and calculations.
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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