Problem 58 - University Physics with Modern Physics

Question

An \(L-C\) circuit consists of a \(60.0-\mathrm{mH}\) inductor and a \(250-\mu F\) 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) 0.024 V, (b) 4.90 mA, (c) 72 nJ, (d) 5.92 μC, 72 nJ in the inductor.

Step 1: Determine Natural Frequency

For an LC circuit, the natural angular frequency \( \omega \) is given by: \[ \omega = \frac{1}{\sqrt{LC}} \]Given:\[ L = 60.0 \text{ mH} = 60.0 \times 10^{-3} \text{ H} \]\[ C = 250 \mu \text{F} = 250 \times 10^{-6} \text{ F} \]Substitute the values:\[ \omega = \frac{1}{\sqrt{60.0 \times 10^{-3} \times 250 \times 10^{-6}}} = \frac{1}{\sqrt{15 \times 10^{-9}}} = \frac{1}{\sqrt{15} \times 10^{-9/2}} \approx 816.5 \text{ rad/s} \]

Step 2: Calculate Maximum Voltage across Capacitor

Maximum voltage across the capacitor is reached when the initial charge is fully across it:The maximum voltage \( V_{max} \) is:\[ V_{max} = \frac{Q}{C} \]where \( Q = 6.00 \mu \text{C} = 6.00 \times 10^{-6} \text{ C} \).Substitute the values:\[ V_{max} = \frac{6.00 \times 10^{-6}}{250 \times 10^{-6}} = 0.024 \text{ V} \]

Step 3: Calculate Maximum Current in Inductor

The maximum current occurs when the potential energy in the capacitor is completely transferred to the inductor. This is when all the charge is flowing through the circuit, and is given by:\[ I_{max} = Q \times \omega \]Substitute the values:\[ I_{max} = 6.00 \times 10^{-6} \times 816.5 = 4.90 \times 10^{-3} \text{ A} \]

Step 4: Calculate Maximum Energy in Inductor

The maximum energy stored in the inductor is equal to the total energy initially stored in the capacitor:Thus, the energy \( U \) is:\[ U = \frac{1}{2} C V_{max}^2 = \frac{1}{2} \times 250 \times 10^{-6} \times (0.024)^2 \]Calculate:\[ U = \frac{1}{2} \times 250 \times 10^{-6} \times 0.000576 = 0.072 \times 10^{-6} \text{ J} = 72 \text{ nJ} \]

Step 5: Calculate Charge on Capacitor and Energy in Inductor at Half Maximum Current

When the current is half of its maximum, use \( I = \frac{I_{max}}{2} \):\[ I = \frac{4.90 \times 10^{-3}}{2} = 2.45 \times 10^{-3} \text{ A} \]Use conservation of energy to find charge on the capacitor:\[ U = \frac{1}{2} L I^2 + \frac{1}{2} \frac{Q^2}{C} = 72 \text{ nJ} \]Substitute for energy in inductor:\[ \frac{1}{2} \times 60 \times 10^{-3} \times (2.45 \times 10^{-3})^2 + \frac{1}{2} \times \frac{Q^2}{250 \times 10^{-6}} = 72 \times 10^{-9} \]Calculate:\[ 0.5 \times 60 \times 10^{-3} \times 6.0025 \times 10^{-6} + \frac{Q^2}{5 \times 10^{-4}} \approx 72 \times 10^{-9} \]\[ 1.8 \times 10^{-7} + \frac{Q^2}{5 \times 10^{-4}} \approx 72 \times 10^{-9} \]Substitute and solve for \( Q \):\[ \frac{Q^2}{5 \times 10^{-4}} \approx 70.2 \times 10^{-9} \]\[ Q^2 \approx 3.51 \times 10^{-11} \]\[ Q \approx 5.92 \times 10^{-6} \text{ C} \]

Key Concepts

Inductor Energy

In an L-C circuit, the inductor stores energy in its magnetic field. This energy is at its maximum when the current through the inductor is at its peak. The energy (\[ U_L \]) stored in an inductor can be calculated using the formula:
\[ U_L = \frac{1}{2} L I^2 \], where \( L \) is the inductance and \( I \) is the current.

• In our example, when the current is at its maximum value of \(4.90 \times 10^{-3} \text{ A}\), the inductor holds a significant amount of energy.
• This energy is equal to the total initial energy stored in the capacitor since energy conservation holds in an ideal L-C circuit.

During oscillations in an L-C circuit, energy is continuously exchanged between the capacitor and the inductor. At some instances, particularly at the beginning, all energy resides in the capacitor. Later, it moves entirely to the inductor. This cycle continues repeatedly, evidencing the dynamic nature of L-C circuits.

Natural Frequency Calculation

In L-C circuits, the natural frequency determines how fast the energy oscillates between the inductor and capacitor. The natural angular frequency, \( \omega \), is given by:
\[ \omega = \frac{1}{\sqrt{LC}} \]. This formula highlights the dependence of frequency on both inductance and capacitance.

• Substituting the values from the example where \( L = 60.0 \times 10^{-3} \text{ H} \) and \( C = 250 \times 10^{-6} \text{ F} \), we find \( \omega \approx 816.5 \text{ rad/s} \).

The frequency describes how rapidly the energy exchange occurs. This is key in determining other quantities in the circuit — such as the maximum current in the inductor, calculated using the product of the initial charge and this frequency. Understanding the natural frequency helps predict the dynamic behavior of electrical circuits.

Capacitor Charge

The capacitor in an L-C circuit initially stores energy in an electric field created by storing charge. The amount of charge initially on the capacitor is crucial because it determines the maximum voltage across the capacitor, as well as the initial energy in the circuit. The maximum voltage, \( V_{max} \), is calculated by:
\[ V_{max} = \frac{Q}{C} \], where \( Q \) is the initial charge and \( C \) is the capacitance.

• For our example, with \( Q = 6.00 \times 10^{-6} \text{ C} \), the maximum voltage is found to be \( 0.024 \text{ V} \).

This voltage is the highest potential difference seen across the capacitor plates during oscillations. Furthermore, when the circuit oscillates, the charge on the capacitor shifts over time, representing the ongoing energy exchange with the inductor. The solution illustrates that even when the current is half of its maximum in an oscillation, calculations show how to determine the corresponding charge on the capacitor.