Show that only half the total energy drawn from a battery in charging an \(R C\) circuit ends up stored in the capacitor. (Hint: What happens to the rest? You'll need to integrate.)

Understanding energy conservation in capacitors is crucial when analyzing electrical circuits, particularly when dealing with charging and discharging processes. Despite being an essential component in storing energy, a capacitor in an RC circuit does not retain all the energy supplied to it by the battery during charging.

In the context of the exercise, we observe that the energy supplied by a battery is not completely stored in the capacitor. When the capacitor charges, it accumulates potential energy in its electric field, represented by the formula \( U = \frac{1}{2}C V^2 \). However, due to the presence of a resistor in the circuit, only half of the energy drawn from the battery ends up as stored energy in the capacitor. This is because energy conservation, in this case, also involves the conversion of a portion of energy into heat within the resistor.

The process of integrating power over time is a method used to calculate the total energy consumed by an electrical component or a circuit. Power, expressed as the product of voltage and current \(P(t) = V(t) \times I(t)\), varies over time as the voltage and current change. In an RC circuit during the initial charging phase, both these quantities are time-dependent.

Using the supplied formulas, \(V(t) = V_0(1 - e^{-t/RC})\) and \(I(t) = \frac{V_0}{R}e^{-t/RC}\), we can compute the total energy supplied to the circuit by integrating the product of these functions from time zero to infinity. This process of integration is what quantifies the total energy used by the circuit, and it's crucial to understand this process to appreciate energy flow and transformation in electrical systems.

Resistors in an electrical circuit play a critical role in managing current flow and are also the sites of energy dissipation. When the RC circuit is being charged, some of the energy provided by the battery is inevitably dissipated in the form of heat across the resistor, which is a manifestation of the energy conservation principle.

The heat dissipated by the resistor over time is equal to the energy that's not stored in the capacitor. Given that our calculated stored energy in the capacitor is \(U = \frac{1}{2}C V_0^2\), and the total energy supplied by the battery is twice that, it is evident that the other half of the battery's energy is dissipated in the resistor. The laws of thermodynamics confirm that in this closed system, energy is not lost but merely transformed from electrical energy into heat.