Problem 32 - University Physics with Modern Physics

Question

A \(20.0-\mu \mathrm{F}\) capacitor is charged by a \(150.0-\mathrm{V}\) power supply, then disconnected from the power and connected in series with a \(0.280-\mathrm{mH}\) inductor. Calculate: (a) the oscillation frequency of the circuit; (b) the energy stored in the capacitor at time \(t=0 \mathrm{ms}\) (the moment of connection with the inductor); (c) the energy stored in the inductor at \(t=1.30 \mathrm{ms}\) .

Step-by-Step Solution

Answer

(a) Oscillation frequency: \( 67.4 \text{ kHz} \). (b) Initial energy in capacitor: \( 0.225 \text{ J} \). (c) Energy in inductor at \( t = 1.30 \text{ ms} \): \( 0.225 \text{ J} \).

Step 1: Calculate the Oscillation Frequency

The oscillation frequency of an LC circuit can be determined using the formula \( f = \frac{1}{2\pi\sqrt{LC}} \), where \( L \) is the inductance and \( C \) is the capacitance.- Given, \( L = 0.280 \times 10^{-3} \text{ H} \) and \( C = 20.0 \times 10^{-6} \text{ F} \).We substitute these values into the formula:\[ f = \frac{1}{2\pi\sqrt{(0.280 \times 10^{-3}) \cdot (20.0 \times 10^{-6})}}\]Calculate the values:- \( \sqrt{LC} = \sqrt{(0.280 \times 10^{-3}) \cdot (20.0 \times 10^{-6})} \approx 2.36 \times 10^{-3} \text{ s} \).- \( f \approx \frac{1}{2\pi \times 2.36 \times 10^{-3}} \approx 67.4 \text{ kHz} \).Thus, the oscillation frequency is approximately \( 67.4 \text{ kHz} \).

Step 2: Calculate the Initial Energy in the Capacitor

The energy stored in a charged capacitor is given by \( U = \frac{1}{2}CV^2 \).- Given, \( C = 20.0 \times 10^{-6} \text{ F} \) and \( V = 150.0 \text{ V} \).Substituting these values into the formula:\[ U = \frac{1}{2} \cdot 20.0 \times 10^{-6} \cdot 150.0^2\]Calculate the values:- \( U = \frac{1}{2} \cdot 20.0 \times 10^{-6} \cdot 22500 = 0.225 \text{ J} \).Hence, the energy stored in the capacitor at \( t = 0 \text{ ms} \) is \( 0.225 \text{ J} \).

Step 3: Calculate the Energy in the Inductor at t = 1.30 ms

In a perfect LC circuit, the total energy is conserved and oscillates between the capacitor and inductor. At a quarter period (\( \frac{T}{4} \)), the energy in the inductor is maximized and equals the initial energy in the capacitor.First, calculate the period \( T \) of the oscillation using \( T = \frac{1}{f} \).- From the oscillation frequency \( f = 67.4 \text{ kHz} \), the period \( T \approx \frac{1}{67.4 \times 10^3} \approx 14.8 \times 10^{-6} \text{ s} \).Now determine the fraction of the period corresponding to \( t = 1.30 \text{ ms} \).- The time \( t = 1.30 \times 10^{-3} \text{ s} \) is approximately \( \frac{1.30 \times 10^{-3}}{14.8 \times 10^{-6}} \approx 87.8 \text{ cycles} \).Since energy oscillates completely every full cycle, and given the significant number of cycles, energy should have already transitioned fully to the inductor:- Hence, at maximum energy oscillation, the energy will be \( 0.225 \text{ J} \).Therefore, the energy in the inductor at \( t = 1.30 \text{ ms} \) is \( 0.225 \text{ J} \).

Key Concepts

Oscillation Frequency

In an LC circuit, the oscillation frequency is crucial as it dictates how rapidly the circuit oscillates between storing energy in the capacitor and the inductor. This frequency is determined by both the inductance \(L\) and the capacitance \(C\) of the circuit. The formula to calculate the oscillation frequency \(f\) is:

• \( f = \frac{1}{2\pi\sqrt{LC}} \)

To find the frequency, plug in the known values of inductance \(L = 0.280\, \text{mH}\) and capacitance \(C = 20.0\, \mu\text{F}\):\[ f = \frac{1}{2\pi\sqrt{(0.280 \times 10^{-3}) \cdot (20.0 \times 10^{-6})}} \approx 67.4 \text{ kHz} \]This frequency, approximately 67.4 kHz, means that the energy in the LC circuit switches back and forth quite rapidly. Understanding this can help in designing circuits for specific applications, ensuring optimal performance based on desired frequency outputs.

Energy in Capacitor

Capacitors are essential components in electronics known for storing energy in an electric field. When charged, a capacitor holds a certain amount of energy based on its capacitance and the voltage across it.The formula for the energy \(U\) stored in a capacitor is:

• \( U = \frac{1}{2}CV^2 \)

Given our capacitor's capacitance \(C = 20.0 \mu\text{F}\) and voltage \(V = 150.0 \text{V}\), the energy at \(t = 0\text{ ms}\) is:\[ U = \frac{1}{2} \cdot 20.0 \times 10^{-6} \cdot 150^2 = 0.225 \text{ J} \]This calculation shows that right at the moment of connection to the inductor, the capacitor is able to store 0.225 joules of energy. Comprehending how capacitors store and release energy is vital for managing power flow in various electronic devices, from small gadgets to large power systems.

Energy Conservation in Inductors

Inductors store energy in a magnetic field, and this energy fluctuates with currents through an oscillating LC circuit. A fundamental aspect of an LC circuit is that total energy is conserved and it oscillates back and forth between the capacitor and inductor.At the moment \(t = 1.30 \text{ ms}\), the energy concentration in the inductor reaches maximum based on its position in the oscillation cycle. Since the circuit in question begins with 0.225 joules in the capacitor, and energy oscillates between components over time equally:

• At a quarter of its cycle period, all available energy can be transferred to the inductor.
• The energy stored in the inductor at \(t = 1.30 \text{ ms}\), when full transfer occurs, equals the initial energy.

Thus, the magnetic energy at \(t = 1.30 \text{ ms}\) is the same as the initial capacitive energy, being approximately 0.225 J. This insight is crucial in ensuring the right timing and conditions for energy manipulation in electrical systems.