Suggested languages for you:

Americas

Europe

Q20P

Expert-verifiedFound in: Page 936

Book edition
10th Edition

Author(s)
David Halliday

Pages
1328 pages

ISBN
9781118230718

**In an oscillating LC circuit in which ${\mathit{C}}{\mathbf{=}}{\mathbf{}}{\mathbf{4}}{\mathbf{.}}{\mathbf{00}}{\mathbf{}}{\mathit{\mu}}{\mathit{F}}$, the maximum potential difference across the capacitor during the oscillations is ** ${\mathbf{1}}{\mathbf{.}}{\mathbf{50}}{\mathbf{}}{\mathbf{V}}$

- The inductance is $3.60\mathrm{mH}$.
- The frequency of the oscillation is $1.33\times {10}^{3}\mathrm{Hz}$.
- Time required for the charge on the capacitor to rise from zero to its maximum value is $1.888\times {10}^{-4}\mathrm{s}$.

- Capacitance of the capacitor in LC circuit, $C=4.00\mathrm{\mu F}\mathrm{or}4.00\times {10}^{-6}\mathrm{F}$
- Maximum potential difference across the capacitor, $V=1.5\mathrm{V}$
- Maximum current through the inductor, $i=50.0\mathrm{mA}$

**In an oscillating LC circuit, energy is shuttled periodically between the electric field of the capacitor and the magnetic field of the inductor; instantaneous values of the two forms of energy are given by equations 31-1 and 31-2. By solving these equations and by substituting the values, we can calculate the inductance. From the relation between the linear frequency and the angular frequency, we can find the frequency. The period is reciprocal to frequency. From that, we can find the time required for the charge on the capacitor to rise from zero to its maximum value.**

Formulae:

The magnetic energy stored in the inductor, $U=\frac{L{i}^{2}}{2}$ (i)

The electric energy stored in the capacitor, $U=\frac{{Q}^{2}}{2C}$ (ii)

The frequency of the oscillation in the circuit, $f=\frac{\omega}{2\pi}$ (iii)

The charge across the capacitor, $Q=C{V}_{max}$ (iv)

The angular frequency of the LC oscillations, $\omega =\frac{1}{\sqrt{LC}}$ (v)

The period of an oscillation, $T=\frac{1}{f}$ (vi)

Equating equations (i) and (ii), we can get the value of the inductance as follows:

$\begin{array}{c}\frac{L{i}^{2}}{2}=\frac{{Q}^{2}}{2C}\\ L=\frac{1}{C}{\left(\frac{Q}{i}\right)}^{2}\\ =\frac{1}{C}{\left(\frac{C{V}_{max}}{i}\right)}^{2}(\mathrm{from}\mathrm{equation}\left(\mathrm{iv}\right))\\ =C{\left(\frac{{V}_{max}}{i}\right)}^{2}\phantom{\rule{0ex}{0ex}}=\left(4.00\times {10}^{-6}\right){\left(\frac{1.50}{50\times {10}^{-3}}\right)}^{2}\phantom{\rule{0ex}{0ex}}\begin{array}{l}=\left(4.00\times {10}^{-6}\right)\times 900\\ =3.60\times {10}^{-3}\mathrm{H}\mathrm{or}3.60\mathrm{mH}\end{array}\end{array}\phantom{\rule{0ex}{0ex}}$

Hence, the value of the inductance is $3.60\mathrm{mH}$.

Substituting value of equation (v) in equation (iii), we can get the frequency of the oscillations as follows:

$\begin{array}{c}f=\frac{1}{2\pi \sqrt{LC}}\\ =\frac{1}{2\pi \sqrt{\left(3.60\times {10}^{-3}\right)\left(4.00\times {10}^{-6}\mathrm{F}\right)}}\\ =1.33\times {10}^{3}\mathrm{Hz}\end{array}$

Hence, the value of the frequency is $1.33\times {10}^{3}\mathrm{Hz}$.

From figure 31-1, we see that the required time is one fourth of a period, so the period of oscillation can be calculated using equation (vi) as follows:

$\begin{array}{c}T=\frac{1}{1.33\times {10}^{3}}\\ =7.52\times {10}^{-4}\mathrm{s}\end{array}$

Now, the required time to charge the capacitor to the maximum value as follows:

role="math" localid="1663155617571" $\begin{array}{c}t=\frac{7.52\times {10}^{-4}}{4}\\ =1.888\times {10}^{-4}\mathrm{s}\end{array}$

Hence, the value of the required time is $1.888\times {10}^{-4}\mathrm{s}$.

94% of StudySmarter users get better grades.

Sign up for free