Question

To construct an oscillating LCsystem, you can choose from a 10mH inductor, a 5.0 µF capacitor, and a 2.0 µF capacitor. (a)What is the smallest largest oscillation frequency that can be set up by these elements in various combinations? (b)What is the second smallest largest oscillation frequency that can be set up by these elements in various combinations? (c)What is the second largest oscillation frequency that can be set up by these elements in various combinations? (d)What is the largest oscillation frequency that can be set up by these elements in various combinations?

Short answer

1. The smallest oscillation frequency is 600 Hz.
2. The second smallest oscillation frequency is 710 Hz.
3. The second largest oscillation frequency is 1100 Hz.
4. The largest oscillation frequency is 1300 Hz.

Step-by-step solution

The given data

The given LC oscillating system has

1. The inductance of the inductor,$L=10\text{mH or}0.01\text{H}$.
2. The capacitance of the first capacitor,$C_1=5\mathrm{\mu }\text{For}5\times {10}^{-6}\text{F}$.
3. The capacitance of the second capacitor,$C_2=2\mathrm{\mu }\text{F or}2\times {10}^{-6}\text{F}$.

 Step 2: Understanding the concept of the oscillation frequency of LC system

Frequency is inversely proportional to the square root of the capacitor, so for a small frequency, the capacitor must be in a parallel combination. From that, we can calculate the small, second small frequency, and second-largest oscillation frequency.

Formulae:

The resonance frequency of the LC circuit, $f=\frac{1}{2\pi \sqrt{LC}}$ (i)

The equivalent capacitance of a parallel combination, $C_{eq}=\sum _{i=1}^n{C}_i$ (ii)

The equivalent capacitance of a series combination, $C_{eq}=\sum _{i=1}^n\frac{1}{C_i}$ (iii)

a) Calculation of the smallest oscillation frequency.

Capacitors 1and2can be used in four different ways as

1)1, 2in parallel combination,

2) Only1

3) only2and

4)1and2in parallel combination.

For a small frequency, two capacitors must be in parallel combination so that the resultant capacitor is more as compared to any other combination. Thus, the equivalent capacitance can be given using equation (ii) as follows:

$\begin{array}{rcl}C& =& 5+2\\ & =& 7\times {10}^{-6}\text{F}\\ & & \end{array}$

So the smallest frequency is given using the above capacitance in equation (i) as follows:

$\begin{array}{rcl}f& =& \frac{1}{2\pi \sqrt{0.01\times 7\times {10}^{-6}}}\\ & =& \frac{1}{2\pi \sqrt{7\times {10}^{-8}}}\\ & =& 601\text{Hz}\\ & \approx & 600\text{Hz}\\ & & \end{array}$

Hence, the smallest frequency is 600 Hz.

b) Calculation of the second smallest oscillation frequency

The second smallest oscillation frequency is when$C=C_1$.

Thus, the second smallest frequency using the capacitance value in equation (i) as follows:

$\begin{array}{rcl}f& =& \frac{1}{2\pi \sqrt{0.01\times 5\times {10}^{-6}}}\\ & =& 711\text{Hz}\\ & \approx & 710\text{Hz}\\ & & \end{array}$

Hence, the value of the frequency is 710 Hz.

c) Calculation of the second largest oscillation frequency.

The second largest frequency is when$C=C_2$.

Thus, the second largest frequency using the capacitance value in equation (i) as follows:

$\begin{array}{rcl}f& =& \frac{1}{2\pi \sqrt{0.01\times 2\times {10}^{-6}}}\\ & =& 1125\text{Hz}\\ & \approx & 1100\text{Hz}\\ & & \end{array}$

Hence, the value of frequency is 1100 Hz.

d) Calculation of the largest oscillation frequency

The largest frequency is when the capacitor is in series combination so the resultant capacitor is given using equation (iii) as follows:

$\begin{array}{rcl}C& =& \frac{C_1{C}_2}{C_1+C_2}\\ & =& \frac{5\times 2}{5+2}\\ & =& \frac{10}{7}\\ & =& 1.42\mathrm{\mu }\text{F}\\ & =& 1.42\times {10}^{-6}\text{F}\\ & & \end{array}$

Thus, the largest frequency using the capacitance value in equation (i) as follows:

$\begin{array}{rcl}f& =& \frac{1}{2\pi \sqrt{0.01\times 1.42\times {10}^{-6}}}\\ & =& 1335\text{Hz}\\ & \approx & 1300\text{Hz}\\ & & \end{array}$

Hence, the value of frequency is 1300 Hz.