Problem 85 - Fundamentals of Physics

Question

an \(L C\) circuit oscillates at a frequency of \(10.4 \mathrm{kHz}\). (a) If the capacitance is \(340 \mu \mathrm{F},\) what is the inductance? (b) If the maximum current is \(7.20 \mathrm{~mA},\) what is the total energy in the circuit? (c) What is the maximum charge on the capacitor?

Step-by-Step Solution

Answer

(a) Inductance \( L \approx 1.40 \mathrm{mH} \); (b) Energy \( E \approx 3.63 \times 10^{-8} \mathrm{J} \); (c) Maximum Charge \( Q \approx 3.60 \times 10^{-3} \mathrm{C} \).

Step 1: Equation for Resonant Frequency

For an LC circuit, the resonant frequency \( f \) is given by the formula \( f = \frac{1}{2\pi\sqrt{LC}} \). Rearrange this equation to solve for the inductance \( L \), we get \( L = \frac{1}{(2\pi f)^2 C} \).

Step 2: Calculate Inductance

The frequency \( f \) is given as \(10.4 \mathrm{kHz} = 10,400 \mathrm{Hz}\) and the capacitance \( C \) is given as \(340 \mu \mathrm{F} = 340 \times 10^{-6} \mathrm{F}\). Substituting these into the equation:\[L = \frac{1}{(2\pi \cdot 10,400)^2 \cdot 340 \times 10^{-6}}\]Calculating this gives \( L \approx 1.40 \times 10^{-3} \mathrm{H} \) or \(1.40 \mathrm{mH}\).

Step 3: Energy in the LC Circuit

The energy in an LC circuit is given by \( E = \frac{1}{2}L I^2 \), where \( I \) is the maximum current. Substitute \( L = 1.40 \times 10^{-3} \mathrm{H} \) and \( I = 7.20 \times 10^{-3} \mathrm{A} \) into the formula:\[E = \frac{1}{2} \times 1.40 \times 10^{-3} \times (7.20 \times 10^{-3})^2\]Calculating this gives \( E \approx 3.63 \times 10^{-8} \mathrm{J} \).

Step 4: Maximum Charge on Capacitor

The maximum charge on the capacitor \( Q \) can be calculated using the formula \( Q = C \cdot V_{max} \), where \( V_{max} = I \cdot Z \) and the impedance \( Z \) is given by \( Z = \omega L = 2\pi f L \). Calculate \( \omega \),\[ \omega = 2\pi \times 10,400 \]Now find \( V_{max} = 7.20 \times 10^{-3} \cdot (2\pi \times 10,400 \times 1.40 \times 10^{-3}) \). Then \( Q = 340 \times 10^{-6} \times V_{max} \). Upon calculation, \( Q \approx 3.60 \times 10^{-3} \mathrm{C} \).

Key Concepts

Resonant Frequency

Understanding the resonant frequency of an LC circuit is crucial because it determines when the circuit naturally oscillates without any external input. This frequency can be calculated using the formula:

• For resonant frequency, \[ f = \frac{1}{2\pi\sqrt{LC}} \]

The symbol \( L \) represents the inductance, and \( C \) stands for capacitance in the circuit.
To find resonant frequency, rearrange this equation to solve for \( L \), resulting in:

• \[ L = \frac{1}{(2\pi f)^2 C} \]

Knowing the resonant frequency helps in achieving efficient energy transfer. It minimizes energy loss, which is particularly significant in wireless communication and power transfer circuits.
LC circuits are therefore pivotal in various technologies where resonance is key, making the concept of resonant frequency highly applicable across electronics.

Inductance

Inductance is a property of a coil or circuit that describes how well it can store energy in the form of a magnetic field. It is measured in henrys (H). In our LC circuit example, we calculated the inductance by using resonant frequency and capacitance.

• Given:
  • Resonant frequency \( f = 10.4 \, \text{kHz} = 10,400 \, \text{Hz} \)
  • Capacitance \( C = 340 \, \mu F = 340 \times 10^{-6} \, \text{F} \)

When these values are plugged into the relevant formula:

• \[ L = \frac{1}{(2\pi \cdot 10,400)^2 \cdot 340 \times 10^{-6}} \approx 1.40 \times 10^{-3} \, \text{H} \]

We find that the inductance \( L \) is approximately \( 1.40 \, \text{mH} \).
This measure helps define how the circuit reacts to changing currents and builds a more profound understanding of how LC circuits function in energy transfer applications.

Maximum Charge

The maximum charge that a capacitor can store directly influences the performance of an LC circuit. This is because the stored charge affects the maximum voltage, and thus the maximum current in the circuit.
We can calculate this using the formula for charge:

• \( Q = C \, \cdot \ V_{max} \)

To find \( V_{max} \), we use:

• \( V_{max} = I \cdot Z \)

where \( Z \) is the impedance calculated as \( Z = \omega L = 2\pi f L \).
By substituting \( \omega \), \( I \), and other values:

• Calculate \( \omega = 2\pi \times 10,400 \)
• Find \( V_{max} \) as \( 7.20 \times 10^{-3} \cdot (2\pi \times 10,400 \times 1.40 \times 10^{-3}) \)
• Finally, calculate \( Q = 340 \times 10^{-6} \times V_{max} \).

This results in approximately \( Q \approx 3.60 \times 10^{-3} \, \text{C} \).
Recognizing the maximum charge capacity of the capacitor is essential for optimizing circuit performance and ensuring safe operation in electrical designs.