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A charged capacitor with \(C = 590 \, \mu \mathrm{F}\) is connected in series to an inductor that has \(L = 0.330\) H and negligible resistance. At an instant when the current in the inductor is \(i = 2.50\) A, the current is increasing at a rate of \(di/dt = 73.0\) A/s. During the current oscillations, what is the maximum voltage across the capacitor?

Short Answer

Expert verified
The maximum voltage across the capacitor is approximately 123.7 V.

Step by step solution

01

Determine the Angular Frequency

First, we need to find the angular frequency of the oscillation. The formula for the angular frequency \( \omega \) in an LC circuit is given by:\[\omega = \frac{1}{\sqrt{LC}}\]Substitute \( L = 0.330 \, \text{H} \) and \( C = 590 \, \mu \text{F} = 590 \times 10^{-6} \, \text{F} \):\[\omega = \frac{1}{\sqrt{0.330 \times 590 \times 10^{-6}}}\]
02

Calculate Angular Frequency

Compute \( \omega \) using the values from Step 1:\[\omega = \frac{1}{\sqrt{0.330 \times 590 \times 10^{-6}}} \approx 74.19 \, \text{rad/s}\]
03

Find Maximum Voltage Across Capacitor

The maximum voltage \( V_C \) across the capacitor in an LC circuit is given by the relationship between the rate of change of current \( \frac{di}{dt} \) and \( \omega \):\[|V_C| = \frac{1}{C} \cdot \left| \frac{di}{dt} \right|\]Substitute \( \frac{di}{dt} = 73.0 \, \text{A/s} \) and \( C = 590 \times 10^{-6} \, \text{F} \):\[|V_C| = \frac{1}{590 \times 10^{-6}} \times 73.0 \]
04

Complete the Calculation

Finally, compute the maximum voltage across the capacitor using the expression from Step 3:\[|V_C| = \frac{73.0}{590 \times 10^{-6}} \approx 123.7 \, \text{V}\]This is the maximum voltage across the capacitor during the oscillations.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Angular Frequency
In an LC circuit, angular frequency is a crucial concept that determines the speed of oscillations. This is particularly significant because it describes how fast the circuit energy alternates between the capacitor and the inductor. To determine the angular frequency, we use the formula:
  • \( \omega = \frac{1}{\sqrt{LC}} \)
This formula shows that angular frequency \( \omega \) is inversely related to the square root of the product of the inductance \( L \) and capacitance \( C \). For our exercise, substituting \( L = 0.330 \) H and \( C = 590 \mu \text{F} = 590 \times 10^{-6} \) F gives:
  • \( \omega = \frac{1}{\sqrt{0.330 \times 590 \times 10^{-6}}} \)
This results in an angular frequency of approximately \( 74.19 \) rad/s. Understanding angular frequency helps in analyzing how quickly these electrical oscillations occur within the circuit.
Maximum Voltage
In the context of an LC circuit, finding the maximum voltage across the capacitor at any point during the oscillation is fundamental. This voltage represents the peak energy stored in the capacitor at a certain instant. To determine this maximum voltage \( |V_C| \), we apply the formula related to the rate of change of current:
  • \( |V_C| = \frac{1}{C} \times \left| \frac{di}{dt} \right| \)
For the given circuit, we substitute the rate of change of the current \( \frac{di}{dt} = 73.0 \) A/s and the capacitance \( C = 590 \mu \text{F} \) as follows:
  • \( |V_C| = \frac{1}{590 \times 10^{-6}} \times 73.0 \)
The calculation yields a maximum voltage across the capacitor of approximately \( 123.7 \) V. Hence, this voltage peak is a key parameter in determining the maximum potential energy stored during oscillations.
Oscillations
In LC circuits, oscillations describe the periodic exchange of energy between the capacitor and inductor. This energy transfer governs the behavior of the circuit, as the capacitor stores energy in the electric field while the inductor does so in the magnetic field. When these components are connected, energy oscillates back and forth:
  • The capacitor releases its energy, creating a current through the inductor.
  • The inductor's magnetic field stores energy as the current decreases, and then releases it back into the circuit.
This process repeats, leading to oscillations characterized by the angular frequency \( \omega \), calculated initially. During each cycle, the voltage and current reach their maximum and minimum values, governed by the LC parameters. Understanding these oscillations provides insights into how LC circuits can be used in applications like radio receivers and signal processing.

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Most popular questions from this chapter

An air-filled toroidal solenoid has 300 turns of wire, a mean radius of 12.0 cm, and a cross-sectional area of 4.00 cm\(^2\). If the current is 5.00 A, calculate: (a) the magnetic field in the solenoid; (b) the self inductance of the solenoid; (c) the energy stored in the magnetic field; (d) the energy density in the magnetic field. (e) Check your answer for part (d) by dividing your answer to part (c) by the volume of the solenoid.

An \(L\)-\(C\) circuit containing an 80.0-mH inductor and a 1.25-nF capacitor oscillates with a maximum current of 0.750 A. Calculate: (a) the maximum charge on the capacitor and (b) the oscillation frequency of the circuit. (c) Assuming the capacitor had its maximum charge at time \(t = 0\), calculate the energy stored in the inductor after 2.50 ms of oscillation.

A toroidal solenoid has 500 turns, cross-sectional area 6.25 cm\(^2\), and mean radius 4.00 cm. (a) Calculate the coil's selfinductance. (b) If the current decreases uniformly from 5.00 A to 2.00 A in 3.00 ms, calculate the self- induced emf in the coil. (c) The current is directed from terminal \(a\) of the coil to terminal \(b\). Is the direction of the induced emf from \(a\) to \(b\) or from \(b\) to \(a\)?

A solenoidal coil with 25 turns of wire is wound tightly around another coil with 300 turns (see Example 30.1). The inner solenoid is 25.0 cm long and has a diameter of 2.00 cm. At a certain time, the current in the inner solenoid is 0.120 A and is increasing at a rate of \(1.75 \times 10^3\) A/s. For this time, calculate: (a) the average magnetic flux through each turn of the inner solenoid; (b) the mutual inductance of the two solenoids; (c) the emf induced in the outer solenoid by the changing current in the inner solenoid.

The minimum capacitance of a variable capacitor in a radio is 4.18 pF. (a) What is the inductance of a coil connected to this capacitor if the oscillation frequency of the \(L\)-\(C\) circuit is \(1600 \times 10^3\) Hz, corresponding to one end of the \(\textbf{AM}\) radio broadcast band, when the capacitor is set to its minimum capacitance? (b) The frequency at the other end of the broadcast band is \(540 \times 10^3\) Hz. What is the maximum capacitance of the capacitor if the oscillation frequency is adjustable over the range of the broadcast band?

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