A variable capacitor with a range from 10 to 365pF is used with a coil to form a variable-frequency LCcircuit to tune the input to a radio. (a) What is the ratio of maximum frequency to minimum frequency that can be obtained with such a capacitor? If this circuit is to obtain frequencies from 0.54 MHz to 1.6 0MHz, the ratio computed in (a) is too large. By adding a capacitor in parallel to the variable capacitor, this range can be adjusted. To obtain the desired frequency range,(b) What capacitance should be added ? (c) What inductance should the coil have?

1. The capacitance ranges from $C_{min}=10\text{pF}$to $C_{max}=365\text{pF}$.
2. Maximum frequency, $f_{max}=0.54\text{MHz}$.
3. Minimum frequency, $f_{min}=1.60\text{MHz}$.

We can use the concept of frequency of the oscillator, which is inversely proportional to the capacitance. So from the minimum value of the capacitor, we get the maximum frequency. Now, we can take the ratio of the maximum frequency to the minimum frequency. Adding the additional capacitor, which is parallel to the original capacitor, and by using the condition that the frequency range must be the same, we can first take the ratio of frequency and find the capacitance of the additional capacitor. Finally, we can find the inductance of the coil.

Formula:

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

From equation (i), we conclude that the frequency is inversely proportional to thesquare root of the capacitor. So, the ratio of the maximum frequency to the minimum frequency range using equation (i) can be written as follows:

$\begin{array}{rcl}\frac{f_{max}}{f_{min}}& =& \frac{\sqrt{C_{min}}}{\sqrt{C_{max}}}\\ & =& \frac{\sqrt{365}}{\sqrt{10}}\\ & =& 6.0\\ & & \end{array}$

Hence, the value of the ratio is 6.0.

An additional capacitor is added to the capacitor so that we taketheratio ofthefrequency as follows:

$\begin{array}{rcl}\frac{f_{max}}{f_{min}}& =& \frac{1.60}{0.54}\\ & =& 2.96\\ & & \end{array}$

Now, by using equation (i) andthegiven condition, we can get the added capacitor as follows:

$\begin{array}{rcl}\frac{f_{max}}{f_{min}}& =& \frac{\sqrt{C+365}}{\sqrt{C+10}}\\ \frac{\sqrt{C+365}}{\sqrt{C+10}}& =& 2.96\\ \frac{C+365}{C+10}& =& 8.76\\ C+365& =& \left(C+10\right)8.76\\ 365-87.6& =& \left(8.76C-C\right)\\ 7.76C& =& 277.4\\ C& =& 35.74\text{pF}\\ & \approx & 36\text{pF}\\ & & \end{array}$

Hence, the value of the added capacitance is 36 pF.

Now we have to solve the frequency equation to find out the inductance. For the minimum frequency, the new capacitance can be given as:

$\begin{array}{rcl}C& =& 365+36\\ & =& 401\text{pF}\\ & & \end{array}$

And localid="1663085663189" $f=0.54\text{MHz}$

Now, using the given data in equation (i), we can get the inductance value as follows:

$\begin{array}{rcl}L& =& \frac{1}{{\left(2\pi \right)}^2{f}^{2}C}\\ & =& \frac{1}{{\left(2\pi \right)}^2{\left(0.54\times {10}^6\right)}^2\times \left(401\times {10}^{-12}\right)}\\ & =& \frac{1}{4{\pi }^2\times 0.29\times {10}^{12}\times \left(401\times {10}^{-12}\right)}\\ & =& 2.2\times {10}^{-4}\text{H}\\ & & \end{array}$

Hence, the value of the inductance is $2.2\times {10}^{-4}\text{H}$.