Analyzing Energy Distribution in a Charging Circuit | Figure 28.19

[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.

 

[joe_andersen]

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.



Joe Andersen

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[kyle8921]

Thank you for reaching out for assistance with this question. I can provide a response to help you better understand the energy distribution in this charging circuit.

Firstly, it is important to understand that energy is conserved in a closed system, meaning that the total amount of energy supplied by the battery must be equal to the total amount of energy stored in the circuit. In this case, the battery supplies energy in the form of electrical potential energy, which is then converted into internal energy in the resistor and stored as potential energy in the capacitor.

To show that half the energy supplied by the battery is stored as internal energy in the resistor, we can use the equation for electrical power, P = IV, where I is the current flowing through the circuit and V is the voltage across the resistor. We can also use the equation for electrical energy, E = PΔt, where Δt is the time the battery is connected to the circuit.

Since the resistor and the capacitor are connected in series, the current flowing through them is the same. Therefore, the energy supplied by the battery is split between the two components. Using the equations above, we can calculate the energy stored in the resistor, ER, and the energy stored in the capacitor, EC, as follows:

ER = PΔt = IVΔt
EC = ½CV² = ½(QV) = ½(IVΔt)V = ½IVΔtV

Since the current and time are the same for both components, we can see that the energy stored in the resistor and the capacitor is equal, each being half of the total energy supplied by the battery.

In conclusion, half of the energy supplied by the battery is stored as internal energy in the resistor and the other half is stored as potential energy in the capacitor. This is due to the fact that the resistor and capacitor are connected in series, causing the same current to flow through both components and splitting the energy equally between them.

I hope this explanation helps you better understand the energy distribution in this charging circuit. If you have any further questions, please don't hesitate to ask. Best of luck with your studies.