Question

A \(12.0-V\) battery is attached to a \(2.00-\mathrm{mF}\) capacitor and a \(100 .-\Omega\) resistor. Once the capacitor is fully charged, what is the energy stored in it? What is the energy dissipated as heat by the resistor as the capacitor is charging?

Short answer

Answer: The energy stored in the capacitor when it is fully charged is 0.144 J, and the energy dissipated as heat by the resistor as the capacitor is charging is also 0.144 J.

Step-by-step solution

Calculate the energy stored in the fully charged capacitor

We are given that the voltage (\(V=12.0 V\)) and capacitance (\(C=2.00 mF\)). Using the formula for the energy stored in a capacitor \(W_c = \frac{1}{2}CV^2\), we can calculate the energy stored: \(W_c = \frac{1}{2}(2.00 \times 10^{-3} F)(12.0 V)^2\) \(W_c = 0.144 J\) So the energy stored in the capacitor when it is fully charged is \(0.144 J\).

Calculate the total energy supplied by the battery

To calculate the energy supplied by the battery, we can use the formula \(W_b = QV\), where \(Q\) is the charge transferred. Since the charge on the capacitor is given by \(Q = CV\), we can rewrite the formula as \(W_b = CV^2\). Now, substituting the given values: \(W_b = (2.00 \times 10^{-3} F)(12.0 V)^2\) \(W_b = 0.288 J\) The total energy supplied by the battery is \(0.288 J\).

Find the energy dissipated as heat by the resistor

Now that we have the energy supplied by the battery (\(W_b = 0.288 J\)) and the energy stored in the capacitor (\(W_c = 0.144 J\)), we can find the energy dissipated as heat by the resistor using the conservation of energy principle: \(W_b = W_c + W_r\). Rearranging the equation to solve for \(W_r\): \(W_r = W_b - W_c\) \(W_r = 0.288 J - 0.144 J\) \(W_r = 0.144 J\) The energy dissipated as heat by the resistor as the capacitor is charging is \(0.144 J\).

Key concepts

Capacitor Energy

In electrical circuits, capacitors are crucial components that store energy. A capacitor is essentially an electronic device consisting of two conductors separated by an insulator. When a voltage is applied across the conductors, an electric field is created. This field allows the capacitor to store energy for later use.

The amount of energy a capacitor can store depends on two main factors:

• The capacitance (\(C\)): This is a measure of a capacitor's ability to store charge. It is often given in Farads (F). In our problem, the capacitance is \(2.00 \text{ mF}\), or \(2.00 \times 10^{-3} \text{ F}\).
• The voltage across the capacitor (\(V\)): This is the electrical potential difference applied, in volts. Here, the voltage is \(12.0 \text{ V}\).

The formula to calculate the energy stored in a capacitor is:\[W_c = \frac{1}{2} C V^2\]Substituting our given values:\[W_c = \frac{1}{2}(2.00 \times 10^{-3} \text{ F})(12.0 \text{ V})^2\]Calculating this gives \(0.144 \text{ J}\). So, when fully charged, the capacitor stores \(0.144 \text{ J}\) of energy.

Resistor Power Dissipation

When charging a capacitor through a resistor, it is crucial to understand the concept of energy dissipation. Resistors regulate the flow of electric current and convert electrical energy into heat, causing a certain amount of energy to be lost in the process.

Based on the conservation of energy principle:

• Total energy supplied by the battery (\(W_b\)) is equal to the sum of energy stored in the capacitor (\(W_c\)) and the energy dissipated as heat by the resistor (\(W_r\)).

Using the battery's supplied energy calculation:\[W_b = CV^2 \]which yields \(0.288 \text{ J}\) in our exercise. Given that \(0.144 \text{ J}\) is stored in the capacitor, the remaining energy is the energy dissipated by the resistor:\[W_r = W_b - W_c\]Thus, \(W_r = 0.288 \text{ J} - 0.144 \text{ J} = 0.144 \text{ J}\). Consequently, the resistor dissipates \(0.144 \text{ J}\) as heat during the charging of the capacitor.

Battery Circuits

Battery circuits are foundational in understanding how electric power is supplied and distributed in a circuit. A battery provides a constant voltage source that drives current through devices like capacitors and resistors.

In the given circuit, a \(12.0 \text{ V}\) battery is connected to a capacitor and a resistor. This setup is typical for energy storage and regulation in electrical engineering:

• The battery's role is to supply energy. In our exercise, it provides a total energy of \(0.288 \text{ J}\), as calculated using \(W_b = CV^2\).
• The capacitor stores some of this energy, which we've calculated to be \(0.144 \text{ J}\).
• The resistor dissipates the remaining energy as heat, equal to \(0.144 \text{ J}\).

In a real-world application, understanding how energy flows and is managed in such a circuit helps in designing efficient electronic devices that optimize power usage while minimizing unwanted energy loss.