Problem 58 - University Physics with Modern Physics

Question

An \(L-C\) circuit consists of a \(60.0-\mathrm{mH}\) inductor and a \(250-\mu \mathrm{F}\) capacitor. The initial charge on the capacitor is 6.00\(\mu \mathrm{C}\) , and the initial current in the inductor is zero. (a) What is the maximum voltage across the capacitor? (b) What is the maximum current in the inductor? (c) What is the maximum energy stored in the inductor? (d) When the current in the inductor has half its maximum value, what is the charge on the capacitor and what is the energy stored in the inductor?

Step-by-Step Solution

Answer

(a) 24 V; (b) 1.548 mA; (c) 7.19 x 10^-8 J; (d) Q = 5.832 μC, U = 1.799 x 10^-8 J.

Step 1: Determine the Angular Frequency of the Circuit

The angular frequency \( \omega \) for an \( L-C \) circuit is given by the formula \( \omega = \frac{1}{\sqrt{LC}} \), where \( L \) is the inductance and \( C \) is the capacitance. Substituting the given values, \( L = 60.0\times10^{-3} \) H and \( C = 250\times10^{-6} \) F, you get \[ \omega = \frac{1}{\sqrt{(60.0\times10^{-3})(250\times10^{-6})}} = \frac{1}{\sqrt{15\times10^{-6}}} = \frac{1}{\sqrt{15}\times10^{-3}} \approx \frac{1}{3.87\times10^{-3}} \approx 258 \text{ rad/s}. \]

Step 2: Calculate Maximum Voltage Across the Capacitor

The maximum voltage \( V_{max} \) across the capacitor is given by \( V = \frac{Q}{C} \), where \( Q \) is the initial charge. \[ V_{max} = \frac{6.00\times10^{-6}}{250\times10^{-6}} = 24 \text{ V}. \]

Step 3: Calculate Maximum Current in the Inductor

The maximum current \( I_{max} \) in the inductor is related to the maximum charge \( Q \) and angular frequency \( \omega \) by \( I_{max} = Q \cdot \omega \). \[ I_{max} = 6.00\times10^{-6} \times 258 \approx 1.548\times10^{-3} = 1.548 \text{ mA}. \]

Step 4: Calculate Maximum Energy Stored in the Inductor

The maximum energy \( U_{max} \) stored in the inductor at any point is given by \( U_{max} = \frac{1}{2}LI_{max}^2 \).\[ U_{max} = \frac{1}{2} \times 60.0\times10^{-3} \times (1.548\times10^{-3})^2 \approx \frac{1}{2} \times 60.0\times10^{-3} \times 2.397\times10^{-6} \approx 7.19\times10^{-8} \text{ J}. \]

Step 5: Determine Charge on Capacitor When Current is Half Maximum Value

When the current is half its maximum value, \( I = \frac{I_{max}}{2} = 0.774 \text{ mA}. \) Using the energy conservation in an oscillating \( L-C \) system, \( Q^2 + (LI^2) = Q_{max}^2 \), where \( Q_{max} = 6.00 \mu \text{C} \). Using \( I = 0.774 \text{ mA} = 0.774\times10^{-3} \text{ A}, \)\[\left(\frac{Q}{C}\right)^2 + \frac{1}{2}LI^2 = \frac{Q_{max}^2}{C^2}\]Substitute:\[\left(\frac{Q}{250 \times 10^{-6}}\right)^2 + \frac{1}{2}(60.0\times10^{-3})(0.774\times10^{-3})^2 = \left(\frac{6.00 \times10^{-6}}{250 \times 10^{-6}}\right)^2\] Solving gives \( Q \approx 5.832 \mu \text{C}. \)

Step 6: Calculate Energy in Inductor at Half Maximum Current

Using the previously determined value of current \( I = 0.774 \text{ mA} \), the energy in the inductor can be calculated using \( U = \frac{1}{2}LI^2 \) at half the maximum current,\[ U \approx \frac{1}{2} \times 60.0\times10^{-3} \times (0.774\times10^{-3})^2 \approx 1.799 \times 10^{-8} \text{ J}. \]

Key Concepts

Inductor

An inductor in an L-C circuit is a crucial component. It stores energy in the form of a magnetic field when current flows through it. Inductance, measured in henries (H), is a property that indicates how effectively the inductor can store this magnetic energy. In our exercise, we have a 60.0-mH inductor, which is equivalent to 0.06 H. This parameter plays a pivotal role in determining the dynamics of the circuit, such as the angular frequency and the maximum energy stored.

Capacitor

A capacitor in an L-C circuit stores energy as an electric field by separating charges between its plates. The capacitance is measured in farads (F), and it tells you how much charge the capacitor can hold per volt of potential. In the given problem, the capacitor has a capacitance of 250-µF, or 0.00025 F. Initially, it is charged to 6.00 µC. The capacitance value affects the angular frequency of the circuit and helps determine maximum voltage across the capacitor.

Angular Frequency

The angular frequency (\( \omega \)) of an L-C circuit is a measure of how rapidly the current and voltage oscillate in the circuit. It is given by the formula \( \omega = \frac{1}{\sqrt{LC}} \), where \( L \) is the inductance and \( C \) is the capacitance. For our circuit, substituting the given values results in an angular frequency of approximately 258 rad/s. This frequency is crucial for calculating other parameters like maximum current and maximum voltage.

Maximum Voltage

The maximum voltage (\( V_{max} \)) across the capacitor is the highest potential difference the capacitor can achieve during the oscillation. It is calculated using \( V = \frac{Q}{C} \), where \( Q \) is the initial charge and \( C \) is the capacitance. In the exercise, with an initial charge of 6.00 µC and a capacitance of 250 µF, the maximum voltage comes out to be 24 V. This value indicates the peak energy conversion from electric to magnetic form in the circuit.

Maximum Current

The maximum current (\( I_{max} \)) in the inductor is the highest current that flows through the circuit. It can be calculated using the relationship \( I_{max} = Q \cdot \omega \), where \( Q \) is the initial charge and \( \omega \) is the angular frequency. Given the parameters, the maximum current is approximately 1.548 mA. This peak current shows the highest point of magnetic energy storage in the inductor before it starts converting back to electric energy.

Energy Stored

The energy stored in an L-C circuit is the sum of electric energy in the capacitor and magnetic energy in the inductor. At maximum energy storage in the inductor, it is calculated using \( U_{max} = \frac{1}{2}LI_{max}^2 \). In this case, the inductor stores about 7.19 x 10^{-8} J. Even when the current in the inductor drops to half, the total energy is conserved and redistributed between capacitor and inductor, showcasing the oscillatory nature of the L-C circuit.