Question

An inductor is connected across a capacitor whose capacitance can be varied by turning a knob. We wish to make the frequency of oscillation of this LC circuit vary linearly with the angle of rotation of the knob, going from$\mathbf{2}\mathbf{\times }{\mathbf{10}}^{\mathbf{5}}\mathbf{}\mathbf{Hzto}\mathbf{}\mathbf{}\mathbf{4}\mathbf{\times }{\mathbf{10}}^{\mathbf{5}}\mathbf{}\mathbf{Hz}$as the knob turns through 180°. If $\mathit{L}\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{0}\mathbf{}\mathbf{mH}$, plot the required capacitance C as a function of the angle of rotation of the knob.

Short answer

The graph for the required capacitance as a function of the angle of rotation of the knob is plotted.

Step-by-step solution

The given data

1. Value of the inductance of the inductor,$L=1.0\mathrm{mH}\mathrm{or}1.0\times {10}^{-3}\mathrm{H}$
2. Frequency of the oscillation linearly varies from $f_0=2\times {10}^5\mathrm{Hz}$to as the knob turn through $\theta ={180}^0$.

Understanding the concept of capacitance and frequency as a function of an angle

By using the linear relationship between the angle and frequency, we can find the capacitance in Pico-farads for various angles and plot the capacitance as a function of angle.

Formulae:

The frequency of an oscillation due to the rotation of the knob, $f=f_0\left(1+\frac{\theta }{180°}\right)\left(\mathrm{i}\right)$

The frequency of an oscillation, $f=\frac{\omega }{2\pi }\left(\mathrm{ii}\right)$

The angular frequency of an LC oscillation,$\omega =1/\sqrt{LC}\left(\mathrm{iii}\right)$

Calculation for plotting the graph for the required capacitance as a function of an angle

Substituting value of angular frequency from equation (iii) in equation (ii), we can get the frequency of the oscillations as follows:

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

Rearrange this formula for the capacitance and substituting frequency value from equation (i), we get

$\begin{array}{c}{f}^2=\frac{1}{4{\pi }^{2}LC}\\ C=\frac{1}{4{\pi }^{2}L{f}^2}\\ =\frac{1}{4{\pi }^{2}L{f}_0^2{\left(1+\frac{\theta }{180°}\right)}^2}\end{array}$

By putting the value of the inductance and the inductance and the given value of lower frequency$f_0$, we get the capacitance as

$\begin{array}{c}C=\frac{{180}^2}{4{\pi }^2\left(1.0\times {10}^{-3}\right)\left(2\times {10}^5\right){\left(180+\theta \right)}^2}\\ =\frac{{180}^2}{8{\pi }^2{\left(180+\theta \right)}^2}\times {10}^{10}\mathrm{pF}\end{array}$

Now vary the θ and measure the capacitance in the Pico farad shown in the following table:

Obs. no Angle$\theta$ Capacitance in F Capacitance in pF 1 0 6.339E-10 633.8999554 2 10 5.6893E-10 568.9295998 3 20 5.13459E-10 513.4589639 4 30 4.65722E-10 465.7224162 5 40 4.24346E-10 424.3462511 6 50 3.88249E-10 388.2487439 7 60 3.56569E-10 356.5687249 8 70 3.28614E-10 328.6137369 9 80 3.03822E-10 303.8218721 10 90 2.81733E-10 281.7333135 11 100 2.61969E-10 261.9688591 12 110 2.44214E-10 244.2135381 13 120 2.28204E-10 228.2039839 14 130 2.13719E-10 213.7186114 15 140 2.0057E-10 200.5699078 16 150 1.88598E-10 188.5983338 17 160 1.77667E-10 177.6674615 18 170 1.6766E-10 167.6600698 19 180 1.58475E-10 158.4749888

Now plot the value $C\left(\mathrm{pF}\right)\mathrm{versus}\theta$:

By using capacitance as a function of angle, we plotted on the graph.