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# In a medium ware broadcast radio can be tuned in frequency range $300kHz$ to $1200kHz$ in circuit of this radio effective inductance is $300\mu H$, what range of variable capacitor?

Last updated date: 02nd Aug 2024
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Hint:When $L$ or $C$ is present in an ac circuit, energy is required to build up a magnetic field around $L$ or electric field in $C$. This energy comes from the source. However, power consumed in $L$ or $C$is zero because all the power received from the source in a quarter cycle is returned to the source in the next quarter cycle.

Formula used:
The frequency of $LC$ circuit is,
$f = \dfrac{1}{{2\pi \sqrt {LC} }}$
Where, $L$ is the inductance of an inductor $\left( H \right)$
$C$ is the capacitance of the capacitor $\left( F \right)$

For tuning the radio, the frequency of the $LC$ circuit must be equal to the frequency of the radio wave. The frequency of $LC$ circuit is,
$f = \dfrac{1}{{2\pi \sqrt {LC} }}$
……………….(1)
Where, $L$ is the inductance of an inductor $\left( H \right)$
$C$ is the capacitance of the capacitor$\left( F \right)$
Squaring on both the equation (1) we get
${f^2} = \dfrac{1}{{4{\pi ^2}LC}}$
The above equation can also be written as,
$C = \dfrac{1}{{4{\pi ^2}{f^2}L}}$ …………………….(2)
Given: Effective inductance,
$\Rightarrow L = 300\mu H$
We can convert the value of $\mu H$into $H$, we get
$\Rightarrow 300 \times {10^{ - 6}}H$
For frequency,
$\Rightarrow {f_1} = 300kHz$
We can convert the $kHz$ into $Hz$, we get
$\Rightarrow 300 \times {10^3}Hz$
Value of capacitance will be,
Consider equation (2) we get
$\Rightarrow {C_1} = \dfrac{1}{{4{\pi ^2}{f_1}^2L}}$
We can substitute the given values in the equations, we get,
$\Rightarrow {C_1} = \dfrac{1}{{4 \times {{3.14}^2} \times {{(300 \times {{10}^3})}^2} \times 300 \times {{10}^{ - 6}}}}$
Solving the above equation,
$\Rightarrow {C_1} = \dfrac{1}{{1.0648 \times {{10}^9}}}$
By using division we get,
$\Rightarrow {C_1} = 0.9147 \times {10^{ - 9}}$
$\Rightarrow 914 \times {10^{ - 12}}F$
We can convert the $F$into $pF$, we get
$\therefore {C_1} = 914pF$
Similarly, for frequency
$\Rightarrow {f_2} = 1200kHz$
We can convert the $kHz$ into $Hz$, we get

$\Rightarrow 1200 \times {10^3}Hz$
The value of capacitance will be,
Consider equation (2) we get
$\Rightarrow {C_2} = \dfrac{1}{{4{\pi ^2}{f_2}^2L}}$
We can substitute the values in the given equation, we get
$\Rightarrow {C_2} = \dfrac{1}{{4 \times {{3.14}^2} \times {{(1200 \times {{10}^3})}^2} \times 300 \times {{10}^{ - 6}}}}$
Solving the above equation,
$\Rightarrow {C_2} = \dfrac{1}{{1.7037 \times {{10}^{10}}}}$
We can use division to simplify the given equations we get,
$\Rightarrow {C_2} = 0.5869 \times {10^{ - 10}}$
$\Rightarrow 58.6 \times {10^{ - 12}}F$
We can convert the $F$into $pF$, we get
$\therefore {C_2} = 58.6pF$
Therefore, the variable capacitor should have a range of$58.6pF$to $914pF$.

When a charged capacitor is connected to an inductor, the charge oscillates from one plate of the capacitor to the other through the inductor. This results in the production of electrical oscillations called electromagnetic oscillations. The physical reason is that energy moves back and forth between the magnetic field of the inductor and the electric field of the capacitor. Therefore, when a charged capacitor $C$ is allowed to discharge through an inductor $L$, electrical oscillations are produced. These oscillations are called $LC$ oscillations.