Question

Suppose you have a supply of inductors ranging from 1.00 nHto 10.0 H, and capacitors ranging from 0.100 pFto 0.100 F. What is the range of resonant frequencies that can be achieved from combinations of a single inductor and a single capacitor?

Short answer

The range of the frequency is from 0.159 Hz to 5.04 GHz .

Step-by-step solution

Definition of inductor and capacitor

Capacitor: A capacitor is an electric charge storage device made up of one or more pairs of conductors separated by an insulator.

Indicator: An inductor is a component with inductance in an electric or electronic circuit.

Given data

The minimum value of the inductor is $L_{min}=1.00nH\left(\frac{{10}^{-9}H}{1nH}\right)=1.00\times {10}^{-9}H$

The maximum value of the inductor is $L_{max}=10.0H$

The minimum value of the capacitor is $C_{min}=1.00pF\left(\frac{{10}^{-12}F}{1pF}\right)=1.00\times {10}^{-12}F$

The maximum value of the inductor is$C_{max}=0.100F$

Formula used

Using the formula for an LC circuit, the resonant frequency

$$\begin{gathered} f_0=\frac{1}{2\pi \sqrt{LC}} \\ C=\frac{1}{4{\pi }^2{L}^{2}f} \end{gathered}$$……………..(1)

Here, L is the self-inductance of the inductor and C is the capacitance.

Finding the range of the frequency

Substituting the given values in the above equation (1), we get

For maximum values,

$$\begin{gathered} f_0^{max}=\frac{1}{2\pi \sqrt{L_{min}{C}_{min}}} \\ =\frac{1}{2\times 3.14\sqrt{\left(1.00\times {10}^{-9}H\right)\times \left(1\times {10}^{-12}F\right)}} \\ =5.04\times {10}^{-9}Hz\left(\frac{1GHz}{{10}^{-9}Hz}\right) \\ =5.04GHz \end{gathered}$$

Therefore, the minimum range of frequency is$f_0^{max}=5.04GHz$

For minimum values,

$$\begin{gathered} f_0^{min}=\frac{1}{2\pi \sqrt{L_{max}{C}_{max}}} \\ =\frac{1}{2\times 3.14\sqrt{10.0H\times 0.100F}} \\ =0.159Hz \end{gathered}$$

Therefore, the minimum range of frequency is $f_0^{min}=0.159Hz$.