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Expert-verified Found in: Page 936 ### Fundamentals Of Physics

Book edition 10th Edition
Author(s) David Halliday
Pages 1328 pages
ISBN 9781118230718 # In a certain oscillating LC circuit, the total energy is converted from electrical energy in the capacitor to magnetic energy in the inductor in ${\mathbf{1}}{\mathbf{.}}{\mathbf{50}}{\mathbf{}}{\mathbit{\mu }}{\mathbit{s}}$. (a) What is the period of oscillation? (b) What is the frequency of oscillation? (c) How long after the magnetic energy will be a maximum again?

1. The period of oscillation is $6.00\mu s$.
2. The frequency of the oscillation is $1.67×{10}^{5}\text{Hz}$.
3. The magnetic energy will be a maximum again at $3.00\mu s$ .
See the step by step solution

## Step 1: The given data

The total energy in conserved from electric energy in the capacitor to magnetic energy in the inductor in $t=1.50\mu s$.

## Step 2: Understanding the concept of oscillation in LC circuit

From the given condition, we can find the period of oscillation. Using the formula for the frequency of the oscillation, we can find the frequency of the oscillation, and using the relation between the energy stored in the magnetic field of the inductor at any time and the current through it, we can find the time after which the magnetic energy is a maximum again.

Formulae:

The frequency of an oscillation in an LC circuit, ${\mathbit{f}}{\mathbf{=}}\frac{\mathbf{1}}{\mathbf{T}}$ (i)

The energy stored in the magnetic field of the inductor at any time, ${{\mathbit{U}}}_{{\mathbf{B}}}{\mathbf{=}}\frac{\mathbf{L}{\mathbf{i}}^{\mathbf{2}}}{\mathbf{2}}$ (ii)

## Step 3: a) Calculation of the period of oscillation

The total energy is conserved from electric energy in the capacitor to magnetic energy in the inductor $1.50\mu s$.

Thus, the period of oscillation is given by:

$T=4\left(1.50\mu s\right)\phantom{\rule{0ex}{0ex}}=6.00\mu s$

Hence, the value of the period of oscillation is localid="1662757013743" $6.00\mu s$.

## Step 4: b) Calculation of the frequency of oscillation

The frequency of the oscillation is given using equation (i) as follows:

$f=\frac{1}{\left(6.00×{10}^{-6}\text{s}\right)}\phantom{\rule{0ex}{0ex}}=1.67×{10}^{5}\text{Hz}\phantom{\rule{0ex}{0ex}}$

Hence, the value of the frequency is $1.67×{10}^{5}\text{Hz}$.

## Step 5: c) Calculation of the time at which the magnetic field is maximum

From equation (ii), we can get that ${U}_{B}\propto {i}^{2}$.

Thus, the magnetic energy does not depend on the direction of the current, so this will occur after one-half of a period. Therefore, the magnetic energy is a maximum again after the time:

$3.00\mu s$

Hence, the value of the time is $3.00\mu s$. ### Want to see more solutions like these? 