Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

A 7.00-\(\mu\)F capacitor is initially charged to a potential of 16.0 V. It is then connected in series with a 3.75-mH inductor. (a) What is the total energy stored in this circuit? (b) What is the maximum current in the inductor? What is the charge on the capacitor plates at the instant the current in the inductor is maximal?

Short Answer

Expert verified
(a) 0.000896 J; (b) Maximum current is 0.691 A; Charge is 0 C.

Step by step solution

01

Calculate Initial Energy Stored in Capacitor

The energy stored in a capacitor is given by the formula \[ U = \frac{1}{2} C V^2 \] Substitute the given values: \[ C = 7.00\ \mu F = 7.00 \times 10^{-6}\ F \text{, } V = 16.0\ V \] \[ U = \frac{1}{2} \times 7.00 \times 10^{-6} \times (16.0)^2 \] \[ U = \frac{1}{2} \times 7.00 \times 10^{-6} \times 256 \] \[ U = 0.000896\ J \] Thus, the total energy stored in this circuit is 0.000896 J.
02

Calculate Maximum Current in Inductor

When the capacitor is fully discharged, the energy stored in the inductor is maximum. The energy stored in the inductor is given by \[ U = \frac{1}{2} L I^2 \] Since the energy is conserved, it remains the same, so set the energy stored in capacitor equal to that stored in the inductor:\[ 0.000896 = \frac{1}{2} \times 3.75 \times 10^{-3} \times I^2 \] Solve for \( I \):\[ I^2 = \frac{2 \times 0.000896}{3.75 \times 10^{-3}} \] \[ I^2 = 0.477867 \] \[ I = 0.691 \ A \] Thus, the maximum current in the inductor is 0.691 A.
03

Charge on Capacitor Plates When Current is Maximal

When the current in the inductor is at its maximum, the charge on the capacitor is zero. This is because all the energy initially stored in the capacitor is now in the inductor based on the energy conservation in an LC oscillator.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

Key Concepts

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

LC Circuit
An LC circuit is a fundamental component in understanding electromagnetic oscillations. It consists of an inductor (L) and a capacitor (C) connected together. This circuit type is crucial because it can naturally oscillate electrical energy back and forth between the capacitor and the inductor.

In simple terms, when you have a charged capacitor in the LC circuit, it will start by transferring its energy to the inductor, turning the electric energy into magnetic energy. The inductor, with its property to resist changes in the current, will cause the energy to progressively transfer back to the capacitor, and this process will keep oscillating.

Key aspects of LC circuits:
  • The oscillations are sinusoidal, meaning they follow a smooth, wave-like pattern.
  • The frequency of oscillation is determined by both the inductance and capacitance, specifically by the formula: \[ f = \frac{1}{2\pi\sqrt{LC}} \]
  • The circuit is often used in radio transmitters, tuners, and filters due to its ability to select particular frequencies.

This kind of circuit demonstrates fundamental principles of physics, such as resonance and energy transfer in systems, providing a playground for understanding deeper electrical engineering concepts.
Energy Conservation in Circuits
Energy conservation is a pivotal concept in the study of LC circuits. This principle states that the total energy within a closed system remains constant over time. In the context of an LC circuit, this means that the total energy will oscillate between the capacitor and the inductor, but the sum will not change.

Initially, when the capacitor is charged, it holds all the potential energy defined by the formula: \[ U = \frac{1}{2} C V^2 \] where \(U\) is energy, \(C\) is capacitance, and \(V\) is voltage.

As the energy transfers to the inductor, it converts into magnetic energy, described by the formula: \[ U = \frac{1}{2} L I^2 \] where \(L\) is inductance and \(I\) is the current.

This constant back and forth shows that even though the forms of energy are changing, the total amount does not. This conversion is the essence of electromagnetic oscillations in LC circuits:
  • At the peak of the current flow, all the energy is in the inductor, and the capacitor is uncharged.
  • At the peak of voltage across the capacitor, the current momentarily drops to zero.

Understanding this flow of energy is crucial, as it forms the basis for designing devices that rely on alternating currents and frequencies.
Maximum Current Calculation
The calculation of maximum current in an LC circuit is a direct application of energy conservation. When the capacitor is initially fully charged, it holds the maximum energy. As this energy fully transfers to the inductor, the current reaches its peak.

The energy relationship is determined by equating the energy in the capacitor to the energy in the inductor:
\[ \frac{1}{2} C V^2 = \frac{1}{2} L I^2 \]

By rearranging the terms to solve for \(I\), the maximum current, we get:
\[ I = \sqrt{\frac{C}{L}} \times V \]

In the specific example of the exercise, with given values for \(C\) and \(L\), the maximum current calculation ensures the understanding that current and voltage in such oscillating circuits follow an interdependent relationship.

This calculation is significant for the design of circuits where the safe maximum current is a critical parameter. It helps prevent overloading and ensures the circuit works within its intended capacity, upholding performance and safety standards.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

A capacitor with capacitance \(6.00 \times 10^{-5}\) F is charged by connecting it to a 12.0-V battery. The capacitor is disconnected from the battery and connected across an inductor with \(L = 1.50\) H. (a) What are the angular frequency \(\omega\) of the electrical oscillations and the period of these oscillations (the time for one oscillation)? (b) What is the initial charge on the capacitor? (c) How much energy is initially stored in the capacitor? (d) What is the charge on the capacitor 0.0230 s after the connection to the inductor is made? Interpret the sign of your answer. (e) At the time given in part (d), what is the current in the inductor? Interpret the sign of your answer. (f) At the time given in part (d), how much electrical energy is stored in the capacitor and how much is stored in the inductor?

A 10.0-cm-long solenoid of diameter 0.400 cm is wound uniformly with 800 turns. A second coil with 50 turns is wound around the solenoid at its center. What is the mutual inductance of the combination of the two coils?

A 15.0-\(\mu\)F capacitor is charged by a 150.0-V power supply, then disconnected from the power and connected in series with a 0.280-mH inductor. Calculate: (a) the oscillation frequency of the circuit; (b) the energy stored in the capacitor at time \(t = 0\) ms (the moment of connection with the inductor); (c) the energy stored in the inductor at \(t = 1.30\) ms.

(a) A long, straight solenoid has \(N\) turns, uniform cross sectional area \(A\), and length \(l\). Show that the inductance of this solenoid is given by the equation \(L = \mu_0 AN^2/l\). Assume that the magnetic field is uniform inside the solenoid and zero outside. (Your answer is approximate because \(B\) is actually smaller at the ends than at the center. For this reason, your answer is actually an upper limit on the inductance.) (b) A metallic laboratory spring is typically 5.00 cm long and 0.150 cm in diameter and has 50 coils. If you connect such a spring in an electric circuit, how much self-inductance must you include for it if you model it as an ideal solenoid?

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?

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free