Question

A$\mathbf{45}\mathbf{}\mathbf{mH}$inductor has a reactance ofrole="math" localid="1662986007307" $\mathbf{1}\mathbf{.}\mathbf{30}\mathbf{}\mathbf{k\Omega }$ . (a) What is its operating frequency? (b) What is the capacitance of a capacitor with the same reactance at that frequency? If the frequency is doubled, what is the new reactance of (c) the inductor and (d) the capacitor?

Short answer

1. Operating frequency is $4.6\mathrm{kHz}$.
2. Capacitance is $26.62\mathrm{nF}$.
3. If the frequency is doubled, then the reactance of the inductor is$2.6\mathrm{k\Omega }$ .
4. If the frequency is doubled, then the reactance of capacitor is $0.65\mathrm{k\Omega }$.

Step-by-step solution

Listing the given quantities

Inductor is $L=45\mathrm{mH}$.

Reactance is$R=1.30\text{kΩ}$ .

Understanding the concepts of the reactance of capacitor and inductor

Reactance withstands currents without loss of power, unlike resistors. Inductive reactance increases with frequency and inductance. Capacitive reactance decreases with frequency and capacitance. Impedance represents the total opposition provided by reactance and resistance.

Formula:

The inductive reactance is,

$X_L=2\pi fL$

The capacitive reactance is,

$X_c=\frac{1}{2\pi fC}$

(a) Calculations of the Operating frequency

Write inductive capacitance as

$\begin{array}{rcl}{X}_L& =& 2\pi fL\\ 1300& =& 2\times 3.14\times f\times 45\times {10}^{-3}\\ f& =& 4.6\mathrm{kHz}\end{array}$

Hence, the operating frequency is$4.6\mathrm{kHz}$ .

(b) Calculations of the Capacitance

As given that $X_L=X_c$

Therefore,

$\begin{array}{rcl}2\pi fL& =& \frac{1}{2\pi fC}\\ f^2& =& \frac{1}{4{\pi }^{2}LC}\end{array}$

$\begin{array}{rcl}C& =& \frac{L}{\frac{1}{4{\pi }^2{L}^2{f}^2}}\\ & =& \frac{L}{X_l^2}\end{array}$

$\begin{array}{rcl}C& =& \frac{45\times {10}^{-3}}{{\left(1300\right)}^2}\\ & =& 26.62\times {10}^{-9}\mathrm{F}\\ & =& 26.62\mathrm{nF}\end{array}$

Hence, the capacitance is$26.62\mathrm{nF}$.

(c) Calculations of the reactance of inductor, if frequency is doubled

When frequency is double

The inductive capacitance is,

$\begin{array}{rcl}{X}_L& =& 2\times 3.14\times 9\times {10}^3\times 45\times {10}^{-3}\\ & =& 2.6\mathrm{k\Omega }\end{array}$

If frequency is doubled, then reactance of inductor is $2.6\mathrm{k\Omega }$.

(d) Calculations of the reactance of capacitor

If frequency is doubled,the reactance of capacitor:

$\begin{array}{rcl}{X}_c& =& \frac{1}{2\pi fC}\\ & =& \frac{1}{2\times 3.14\times 9\times {10}^3\times 26.62\times {10}^{-9}}\\ & =& 0.65\mathrm{k\Omega }\end{array}$

Hence, if frequency is doubled, then reactance of capacitor is $0.65\mathrm{k\Omega }$.