Chapter 21, Problem 68P

The variable capacitor in the tuner of an AM radio has a capacitance of 2800 pF when the radio is tuned to a station at 580 kHz. (a) What must be the capacitance for a station at 1600 kHz? (b) What is the inductance (assumed constant)?

To determine

(a)

The capacitance for a station at $1600KHz$

Answer to Problem 68P

Solution:

The capacitance for a station at $1600KHz$ will be $370\text{pF}$

Explanation of Solution

Given:

$\begin{array}{l}c=2800\text{pF}\\ f=580\text{kHz}\\ f_1=1600\text{kHz}\end{array}$

Formula used:

$f=\frac{1}{2\pi }\sqrt{\frac{1}{LC}}$

Where,

$\begin{array}{l}c=2800\text{pF}\\ f=580\text{kHz}\\ f_1=1600\text{kHz}\end{array}$

Calculation:

$\begin{array}{l}f=\frac{1}{2\pi }\sqrt{\frac{1}{LC}}\\ \frac{f_{580kHz}}{f_{1600kHZ}}=\frac{580}{1600}=\frac{\frac{1}{2\pi }\sqrt{\frac{1}{L{C}_{580kHz}}}}{\frac{1}{2\pi }\sqrt{\frac{1}{L{C}_{1600kHz}}}}\\ C_{1600kHz}={\left(\frac{580}{1600}\right)}^2{C}_{580kHz}={\left(\frac{580}{1600}\right)}^2\left(2800\text{pF}\right)=367.9\text{pF}\approx 370\text{pF}\\ \end{array}$

To determine

(b)

The inductance

Answer to Problem 68P

Solution:

The inductance will be $2.7\times {10}^{-5}\text{H}$

Explanation of Solution

Given info:

$\begin{array}{l}c=2800\text{pF}\\ f=580\text{kHz}\\ f_1=1600\text{kHz}\end{array}$

Formula Used:

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

Where,

$\begin{array}{l}c=2800\text{pF}\\ f=580\text{kHz}\end{array}$

Calculation:

$L=\frac{1}{4{\pi }^2{f}^{2}C}=\frac{1}{4{\pi }^2{\left(5.8\times {10}^5\text{Hz}\right)}^2\left(2800\times {10}^{-12}\text{F}\right)}=2.7\times {10}^{-5}\text{H}$