Chapter 21, Problem 71P

To determine

(a) To Determine:

The inductance from the resonance frequency

Answer to Problem 71P

Solution:

The inductance from the resonance frequency will be $0.032\text{H}$

Explanation of Solution

Given:

$\begin{array}{l}C=2200\text{pF}\\ V=120\text{V}\\ f=19\text{kHz}\end{array}$

Formula used:

$L=\frac{1}{4{\pi }^2{f}_0^{2}C}$

Where,

$C=$ Capacitor

$f=$ Frequency

Calculation:

$\begin{array}{l}L=\frac{1}{4{\pi }^2{f}_0^{2}C}\\ =\frac{1}{4{\pi }^2\left(19\times {10}^{3}Hz\right)\left(2.2\times {10}^{-9}F\right)}\\ =0.03187H\\ =0.032H\end{array}$

To determine

(b)

The peak value of the current

Answer to Problem 71P

Solution:

The peak value of the current will be $0.032A$

Explanation of Solution

The peak value of the current

$\frac{1}{2}C{V}_0^2=\frac{1}{2}L{I}_{MAX}{}^2\to I_{\max }=\sqrt{\frac{C{V}_0^2}{L}}=\sqrt{\frac{\left(2.2\times {10}^{-9}\text{F}\right){\left(120\text{V}\right)}^2}{\left(0.03189\text{H}\right)}}=0.032\text{A}$

To determine

(c)

The maximum energy stored in the magnetic field of the inductor

Answer to Problem 71P

Solution:

The maximum energy stored in the magnetic field of the inductor will be $16\text{μJ}$

Explanation of Solution

The maximum energy stored in the magnetic field of the inductor

$U_{L,\max }=\frac{1}{2}C{V}_0^2=\frac{1}{2}\left(2.2\times {10}^{-9}\text{F}\right){\left(120\text{V}\right)}^2=16\text{μJ}$