/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 75 A \(3.0-\mu \mathrm{F}\) capacit... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 22: Problem 75

A \(3.0-\mu \mathrm{F}\) capacitor has a voltage of \(35 \mathrm{V}\) between its plates. What must be the current in a \(5.0-\mathrm{mH}\) inductor so that the energy stored in the inductor equals the energy stored in the capacitor?

Short Answer

Expert verified
The current in the inductor must be approximately 2.71 A.

Step by step solution

01

Calculate energy stored in the capacitor

First, we calculate the energy stored in the capacitor using the formula for the energy stored in a capacitor: \[ E_C = \frac{1}{2} C V^2 \]where \(C = 3.0\, \mu F = 3.0 \times 10^{-6} F\) and \(V = 35\, V\).Substitute these values into the formula:\[ E_C = \frac{1}{2} \times 3.0 \times 10^{-6} \times (35)^2 \]\[ E_C = \frac{1}{2} \times 3.0 \times 10^{-6} \times 1225 \]\[ E_C = 1.8375 \times 10^{-2} \text{ Joules} \].
02

Setup the equation for energy in the inductor

Next, we set up the equation for energy stored in the inductor using the formula:\[ E_L = \frac{1}{2} L I^2 \]where \(L = 5.0\, mH = 5.0 \times 10^{-3} H\) and \(I\) is the current in the inductor.We want the energy in the inductor to equal the energy in the capacitor, so:\[ \frac{1}{2} L I^2 = E_C \]
03

Solve for current in the inductor

Solve the energy equation for the current \(I\):\[ \frac{1}{2} \times 5.0 \times 10^{-3} \times I^2 = 1.8375 \times 10^{-2} \]\[ 2.5 \times 10^{-3} \times I^2 = 1.8375 \times 10^{-2} \]Divide both sides by \(2.5 \times 10^{-3}\):\[ I^2 = \frac{1.8375 \times 10^{-2}}{2.5 \times 10^{-3}} \]\[ I^2 = 7.35\]Take the square root of both sides:\[ I = \sqrt{7.35} \approx 2.71 \text{ A} \].

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 91影视!

Key Concepts

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

Capacitor Energy Formula
To find the energy stored in a capacitor, we use a specific formula. The energy stored, denoted as \(E_C\), can be calculated with:
  • \(E_C = \frac{1}{2} C V^2\)
  • \(C\) represents the capacitance in farads (F).
  • \(V\) stands for voltage in volts (V).
In our example, a capacitor of \(3.0\, \mu F\) and a voltage of \(35\, V\) results in an energy of \(1.8375 \times 10^{-2}\) Joules.
The formula shows that the energy is directly proportional to both the square of the voltage and the capacitance.
This means that doubling the voltage will increase the energy stored by a factor of four.
Inductor Energy Formula
Inductors, like capacitors, store energy, but they store it in magnetic fields. The energy stored in an inductor, \(E_L\), is given by:
  • \(E_L = \frac{1}{2} L I^2\)
  • \(L\) is the inductance in henrys (H).
  • \(I\) is the current flowing through the inductor in amperes (A).
In the problem, with \(L = 5.0\, mH\), finding \(I\) involves ensuring \(E_L = E_C\).
This highlights the proportionality between energy storage and the square of the current, similar to how energy in capacitors relies on voltage squared.
This analogy underscores a balanced relationship between electricity in a capacitor and magnetism in an inductor.
Electrical Current Calculation
Calculating electrical current involves applying the equations for energy stored. To find \(I\) in an inductor so that it matches energy stored in a capacitor:
  • Start with \(\frac{1}{2} L I^2 = E_C\).
  • Re-arrange to isolate \(I^2\): \(I^2 = \frac{2 E_C}{L}\).
  • Using the given values, \(I^2 = \frac{2 \times 1.8375 \times 10^{-2}}{5.0 \times 10^{-3}}\).
  • Solve for \(I\) by taking the square root: \(I = \sqrt{7.35}\).
  • The result, \(I \approx 2.71\, \text{A}\), provides the requisite current.
This calculation shows how identifying energy equivalences aids in determining unknown current values in electrical components.
Ensuring energy balance requires precise calculations, emphasizing the harmony in electrical principles.

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 circular loop of wire rests on a table. A long, straight wire lies on this loop, directly over its center, as the drawing illustrates. The current \(I\) in the straight wire is decreasing. In what direction is the induced current, if any, in the loop? Give your reasoning.

Two coils of wire are placed close together. Initially, a current of \(2.5 \mathrm{A}\) exists in one of the coils, but there is no current in the other. The current is then switched off in a time of \(3.7 \times 10^{-2} \mathrm{s}\). During this time, the average emf induced in the other coil is \(1.7 \mathrm{V}\). What is the mutual inductance of the two-coil system?

A uniform magnetic field is perpendicular to the plane of a single-turn circular coil. The magnitude of the field is changing, so that an emf of \(0.80 \mathrm{V}\) and a current of \(3.2 \mathrm{A}\) are induced in the coil. The wire is then re-formed into a single-turn square coil, which is used in the same magnetic field (again perpendicular to the plane of the coil and with a magnitude changing at the same rate). What emf and current are induced in the square coil?

In each of two coils the rate of change of the magnetic flux in a single loop is the same. The emf induced in coil \(1,\) which has 184 loops, is \(2.82 \mathrm{V}\). The emf induced in coil 2 is 4.23 V. How many loops does coil 2 have?

In a television set the power needed to operate the picture tube comes from the secondary of a transformer. The primary of the transformer is connected to a \(120-\mathrm{V}\) receptacle on a wall. The picture tube of the television set uses \(91 \mathrm{W}\), and there is \(5.5 \mathrm{mA}\) of current in the secondary coil of the transformer to which the tube is connected. Find the turns ratio \(N_{\mathrm{s}} / N_{\mathrm{p}}\) of the transformer.
See all solutions

Recommended explanations on Physics Textbooks

Force

Read Explanation

Wave Optics

Read Explanation

Radiation

Read Explanation

Atoms and Radioactivity

Read Explanation

Solid State Physics

Read Explanation

Further Mechanics and Thermal Physics

Read Explanation
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.