91Ó°ÊÓ

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,$C_1=5\mathrm{μ}\text{For}5×{10}^{-6}\text{F}$.
3. The capacitance of the second capacitor,$C_2=2\mathrm{μ}\text{F or}2×{10}^{-6}\text{F}$.

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, $f=\frac{1}{2\mathrm{Ï€}\sqrt{LC}}$ (i)

The equivalent capacitance of a parallel combination, $C_{eq}=\underset{i=1}{\overset{n}{∑}}{C}_i$ (ii)

The equivalent capacitance of a series combination, $C_{eq}=\underset{i=1}{\overset{n}{∑}}\frac{1}{C_i}$ (iii)

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:

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

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

$\begin{array}{rcl}f& =& \frac{1}{2\mathrm{π}\sqrt{0.01×7×{10}^{-6}}}\\ & =& \frac{1}{2\mathrm{π}\sqrt{7×{10}^{-8}}}\\ & =& 601\text{Hz}\\ & ≈& 600\text{Hz}\\ & & \end{array}$

Hence, the smallest frequency is 600 Hz.

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

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

$\begin{array}{rcl}f& =& \frac{1}{2\mathrm{π}\sqrt{0.01×5×{10}^{-6}}}\\ & =& 711\text{Hz}\\ & ≈& 710\text{Hz}\\ & & \end{array}$

Hence, the value of the frequency is 710 Hz.

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

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

$\begin{array}{rcl}f& =& \frac{1}{2\mathrm{π}\sqrt{0.01×2×{10}^{-6}}}\\ & =& 1125\text{Hz}\\ & ≈& 1100\text{Hz}\\ & & \end{array}$

Hence, the value of frequency is 1100 Hz.

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

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

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

$\begin{array}{rcl}f& =& \frac{1}{2\mathrm{π}\sqrt{0.01×1.42×{10}^{-6}}}\\ & =& 1335\text{Hz}\\ & ≈& 1300\text{Hz}\\ & & \end{array}$

Hence, the value of frequency is 1300 Hz.