Question

A capacitor with capacitance \(6.00 \times 10^{-5}\) F is charged by connecting it to a 12.0-V battery. The capacitor is disconnected from the battery and connected across an inductor with \(L = 1.50\) H. (a) What are the angular frequency \(\omega\) of the electrical oscillations and the period of these oscillations (the time for one oscillation)? (b) What is the initial charge on the capacitor? (c) How much energy is initially stored in the capacitor? (d) What is the charge on the capacitor 0.0230 s after the connection to the inductor is made? Interpret the sign of your answer. (e) At the time given in part (d), what is the current in the inductor? Interpret the sign of your answer. (f) At the time given in part (d), how much electrical energy is stored in the capacitor and how much is stored in the inductor?

Short answer

The oscillation frequency is 105.41 rad/s, initial charge is 7.20e-4 C, initial energy is 4.32e-3 J, the charge on the capacitor is approximately -4.449e-4 C at 0.023 s, current in the inductor is -0.037 A, and energy stored is 0.00165 J in capacitor and 0.00103 J in inductor.

Step-by-step solution

Compute Angular Frequency

The angular frequency \( \omega \) of oscillations in an LC circuit is given by the formula \( \omega = \frac{1}{\sqrt{LC}} \). Here \( L = 1.50 \) H and \( C = 6.00 \times 10^{-5} \) F. Substituting the values, we get:\[ \omega = \frac{1}{\sqrt{(1.50)(6.00 \times 10^{-5})}} = \frac{1}{\sqrt{9.00 \times 10^{-5}}} = \frac{1}{0.00949} \approx 105.41 \text{ rad/s} \]

Calculate Period of Oscillations

The period \( T \) of oscillations is the reciprocal of the frequency \( f = \frac{\omega}{2\pi} \). Thus,\[ T = \frac{2\pi}{\omega} = \frac{2\pi}{105.41} \approx 0.05963 \text{ s} \]

Determine Initial Charge on the Capacitor

The initial charge \( Q_0 \) on the capacitor can be found using the formula \( Q = CV \), where \( C = 6.00 \times 10^{-5} \) F and \( V = 12.0 \) V. Substituting the values, we have:\[ Q_0 = (6.00 \times 10^{-5})(12.0) = 7.20 \times 10^{-4} \text{ C} \]

Calculate Initial Energy Stored in the Capacitor

The initial energy \( U \) stored in the capacitor is given by \( U = \frac{1}{2}CV^2 \). Substituting the known values:\[ U = \frac{1}{2}(6.00 \times 10^{-5})(12.0)^2 = \frac{1}{2}(6.00 \times 10^{-5})(144) = 4.32 \times 10^{-3} \text{ J} \]

Find Charge on the Capacitor at Given Time

The charge on the capacitor at any time \( t \) is given by \( Q(t) = Q_0 \cos(\omega t) \). Using \( t = 0.0230 \) s and \( \omega = 105.41 \text{ rad/s} \), we calculate:\[ Q(t) = (7.20 \times 10^{-4}) \cos(105.41 \times 0.0230) \approx (7.20 \times 10^{-4}) \cos(2.42443) \approx -4.449 \times 10^{-4} \text{ C} \]The negative sign indicates that the charge has reversed polarity from its initial state.

Compute Current in the Inductor at Given Time

The current \( I(t) \) in the inductor is given by \( I(t) = -\frac{Q_0}{\sqrt{LC}} \sin(\omega t) \). With \( L = 1.50 \) H and \( C = 6.00 \times 10^{-5} \) F:\[ I(t) = -\frac{7.20 \times 10^{-4}}{0.012} \sin(105.41 \times 0.0230) \approx -0.06 \sin(2.42443) \approx -0.037 \text{ A} \]The negative sign indicates the direction of the current is opposite to the initial direction of charging.

Energy Distribution at Given Time

The energy in the capacitor \( U_C(t) = \frac{1}{2}C[V(t)]^2 \), and the energy in the inductor \( U_L(t) = \frac{1}{2}LI^2 \). First, find the voltage \( V(t) = \frac{Q(t)}{C} \):\[ V(t) = \frac{-4.449 \times 10^{-4}}{6.00 \times 10^{-5}} \approx -7.415 \text{ V} \]Substitute into the energy equations:\[ U_C(t) \approx \frac{1}{2}(6.00 \times 10^{-5})(-7.415)^2 \approx 0.00165 \text{ J} \]\[ U_L(t) = \frac{1}{2}(1.50)(0.037)^2 \approx 0.00103 \text{ J} \]

Key concepts

Angular Frequency

In an LC circuit, the angular frequency \( \omega \) is a measure of how rapidly the system oscillates, swinging back and forth between storing energy in the electric field of the capacitor and the magnetic field of the inductor. It can be calculated using the expression:

• \( \omega = \frac{1}{\sqrt{LC}} \)

Here, \( L \) is the inductance and \( C \) is the capacitance, both fundamental components of the LC circuit. The angular frequency is essential because it defines not only the speed of oscillation but also the natural frequency of the system, which is inherently the most efficient frequency at which energy transfer within the circuit can occur. This attribute is paramount in a variety of applications, such as tuning radios and filtering signals, where precise frequency control is crucial.

Capacitance

Capacitance is a fundamental property of capacitors, indicating their ability to store an electric charge. Measured in farads (F), it is determined by the relationship between the charge (\(Q\)) on the plates of the capacitor and the applied voltage (\(V\)) across those plates. Expressed mathematically as:

• \( C = \frac{Q}{V} \)

This formula shows that with more capacitance, the capacitor can store more charge at the same voltage. In an LC circuit, the capacitance value directly influences the circuit's resonant frequency, affecting how quickly it responds to changes. A higher capacitance typically results in a lower resonant frequency as it can hold a larger amount of charge, taking longer to fluctuate fully. Understanding how capacitance affects the circuit's behavior is crucial in design, impacting everything from timing in electronic circuits to the efficiency of energy storage within the system.

Inductive Reactance

Inductive reactance, denoted as \( X_L \), is the resistance associated with an inductor's opposition to a change in current. Unlike pure resistors, inductive reactance is only present in AC (alternating current) circuits and is frequency-dependent, calculated by:

• \( X_L = \omega L \)

Where \( \omega \) is the angular frequency and \( L \) the inductance. The concept is crucial because it describes how an inductor reacts in an AC circuit environment, resisting changes in current and thus affecting the phase and amplitude of signals within the circuit. As frequency increases, so does the inductive reactance, making it significant in situations that require precise control over current flow, such as in radio and telecommunications where frequency manipulation is needed. Hence, designing circuits for specific applications involves carefully selecting inductance values to match desired reactance levels.

Energy Stored in Capacitor

The energy stored in a capacitor is a critical metric, indicating how much energy can be temporarily held in an electric field. This stored energy \( U \) in a capacitor is derived from the work done to charge it and can be computed using the formula:

• \( U = \frac{1}{2}CV^2 \)

Where \( C \) is the capacitance and \( V \) the voltage. This expression reflects how both the ability to store energy and the potential difference influence the total energy storage capability. The insight gained from this formula is pivotal in electrical engineering, particularly in applications involving transient operations and energy conservation where quick discharge and storage cycles are frequent. Efficiently storing and releasing energy helps in stabilizing power systems, designing backup power solutions, and managing power in portable electronic devices.