Problem 20 - Fundamentals of Physics

Question

In an oscillating \(L C\) circuit in which \(C=4.00 \mu \mathrm{F},\) the maximum potential difference across the capacitor during the oscillations is \(1.50 \mathrm{~V}\) and the maximum current through the inductor is \(50.0 \mathrm{~mA}\). What are (a) the inductance \(L\) and (b) the frequency of the oscillations? (c) How much time is required for the charge on the capacitor to rise from zero to its maximum value?

Step-by-Step Solution

Answer

The inductance \(L\) is \(3.60 \times 10^{-3} \text{H}\), frequency \(f\) is approximately \(1,333 \text{Hz}\), and time for maximum charge is about \(1.88 \times 10^{-4} \text{s}\).

Step 1: Understand the given parameters

We are given that the capacitance of the capacitor is \(C = 4.00 \mu \text{F} = 4.00 \times 10^{-6} \text{F}\). The maximum potential difference across the capacitor is \(V_{max} = 1.50 \text{V}\), and the maximum current through the inductor is \(I_{max} = 50.0 \text{mA} = 50.0 \times 10^{-3} \text{A}\). Our goal is to find the inductance \(L\), frequency of oscillation \(f\), and the time required for the charge on the capacitor to rise from zero to its maximum value.

Step 2: Use the formula for maximum energy

The maximum energy stored in the capacitor at the maximum potential difference is given by \(E_C = \frac{1}{2} C V_{max}^2\). Plug the values:\[E_C = \frac{1}{2} (4.00 \times 10^{-6}) (1.50)^2\]After calculation, \(E_C = 4.50 \times 10^{-6} \text{J}\).

Step 3: Use the formula for maximum energy in the inductor

Similarly, the maximum energy in the inductor is \(E_L = \frac{1}{2} L I_{max}^2\). Set \(E_L = E_C\) since energy is exchanged between the capacitor and inductor without loss:\[E_C = \frac{1}{2} L (50.0 \times 10^{-3})^2\]Solve for \(L\):\[\4.50 \times 10^{-6} = \frac{1}{2} L (50.0 \times 10^{-3})^2\]\[\L = \frac{4.50 \times 10^{-6}}{0.5 \times (0.050)^2} = 3.60 \times 10^{-3} \text{H}\].

Step 4: Calculate the frequency of oscillation

The frequency of oscillation in an LC circuit is given by \(f = \frac{1}{2 \pi \sqrt{LC}}\). Use the values for \(L\) and \(C\):\[L = 3.60 \times 10^{-3} \text{H}, \quad C = 4.00 \times 10^{-6} \text{F}\]\[\f = \frac{1}{2 \pi \sqrt{(3.60 \times 10^{-3})(4.00 \times 10^{-6})}} \]Solve to get \(f \approx 1,333 \text{Hz}\).

Step 5: Calculate the time for charge to rise to maximum

The time for the charge to rise from zero to its maximum value is a quarter of a complete cycle of oscillation. This is given by \(t = \frac{1}{4f}\).Using \(f = 1,333 \text{Hz}\):\[t = \frac{1}{4 \times 1,333} \]Calculate: \(t \approx 1.88 \times 10^{-4} \text{s}\).

Key Concepts

Inductance

In an LC circuit, inductance is a key parameter that characterizes the inductor's ability to store energy. Inductance, commonly denoted by the letter \(L\), is measured in henries (H). It relates to how effectively an inductor can create a magnetic field from a given amount of current flowing through it.
In the given exercise, we want to find the inductance \(L\) based on the energy exchange happening between the capacitor and the inductor. We calculate this by equating the energy stored in the capacitor (\(E_C\)) with that in the inductor (\(E_L\)). Once we know the energy for one, the other is determined as they are equal in a lossless LC circuit.

• The formula for energy in a capacitor is \( E_C = \frac{1}{2} C V_{max}^2 \).
• The energy in an inductor follows the relation \( E_L = \frac{1}{2} L I_{max}^2 \).

By comparing these energies and solving for \(L\), we determine the inductance needed for such oscillations to occur.
In the provided problem, \(L\) was found to be \(3.60 \times 10^{-3} \text{H}\), which tells us how much inductance is involved in maintaining the oscillation between the capacitor and inductor.

Oscillation Frequency

The oscillation frequency in an LC circuit refers to how fast the energy circulates between the inductor and capacitor, causing the current and voltage to oscillate. It's a crucial aspect as it defines the time period of these oscillations.
This frequency, denoted as \(f\), is derived from the capacitance \(C\) and the inductance \(L\) using the formula:

• \( f = \frac{1}{2 \pi \sqrt{LC}} \)

In this equation, \(\pi\) is a constant approximately equal to 3.14159, and \(\sqrt{}\) represents the square root function.
By substituting the values of \(L\) and \(C\), we calculate the frequency. This tells us how many cycles of oscillation occur in one second.
From the given exercise, the frequency was determined to be approximately \(1,333 \text{Hz}\), indicating the energy oscillates quite rapidly in this circuit configuration.

Maximum Energy

Maximum energy in an LC circuit is a fundamental concept as it describes the peak energy levels stored in either the capacitor or the inductor during oscillations.
At these maximum levels, energy is fully either magnetic (in the inductor) or electric (in the capacitor), and the values are always equal due to conservation of energy, assuming no losses.

• For a capacitor, maximum energy is \( E_C = \frac{1}{2} C V_{max}^2 \).
• For an inductor, it is \( E_L = \frac{1}{2} L I_{max}^2 \).

In oscillating circuits, these maximums play a crucial role, as they dictate the limits of oscillation power. When the energy reaches its peak in the capacitor, it's fully discharged and then the cycle shifts, allowing the current to build in the inductor.
In the problem, we calculated \(E_C\) to be \(4.50 \times 10^{-6} \text{J}\), and with the circuit's conservation principle, this is also the maximum energy in the inductor.

Capacitor Charge Timing

Capacitor charge timing is a pivotal aspect of analyzing LC circuits because it relates to the dynamics of how quickly a capacitor reaches its maximum charge from an uncharged state.
In an LC circuit, the time it takes for this to happen corresponds to one-quarter of the period of one full oscillation cycle. The period (\(T\)) is the inverse of the frequency (\(f\)), so it follows:

• Full cycle: \( T = \frac{1}{f} \)
• Quarter cycle: \( t = \frac{1}{4f} \)

Given the frequency, you can easily find how long it takes for the charge to rise to its peak value.
This is effectively the time taken for the system to move from a state with no charge on the capacitor to its maximum charge.
For the exercise at hand, this time was computed as approximately \(1.88 \times 10^{-4} \text{s}\), signifying a brief period due to the relatively high frequency of oscillation.