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Expert-verified Found in: Page 936 ### Fundamentals Of Physics

Book edition 10th Edition
Author(s) David Halliday
Pages 1328 pages
ISBN 9781118230718 # A variable capacitor with a range from 10 to 365pF is used with a coil to form a variable-frequency LC circuit 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 ratio of maximum to minimum frequency range is 6.0.
2. The capacitance to add is 36 pF.
3. The inductance of the coil is $2.2×{10}^{-4}\text{H}$.
See the step by step solution

## Step 1: The given data

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, ${f}_{min}=1.60\text{MHz}$.

## Step 2: Understanding the concept of frequency of LC circuit

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)

## Step 3: a) Calculation of maximum and minimum frequency

From equation (i), we conclude that the frequency is inversely proportional to the square 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.

## Step 4: b) Calculation of the capacitance to be added

An additional capacitor is added to the capacitor so that we take the ratio of the frequency as follows:

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

Now, by using equation (i) and the given 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.

## Step 5: c) Calculation of the inductance in the coil

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:

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

Hence, the value of the inductance is $2.2×{10}^{-4}\text{H}$. ### Want to see more solutions like these? 