Question

A radio can tune over the frequency range of a portion of the MW broadcast band: ($800{\text{ }}kHz$ to $1200{\text{ }}kHz$). If its LC circuit has an effective inductance of $200\;\mu H$, what must be the range of its variable capacitor?

Answer 1

For tuning, the natural frequency i.eThe frequency of free oscillations of the LC circuit should be equal to the frequency of the radio wave. An LC circuit consists of circuit elements such as an inductor, and a capacitor connected. This circuit can also be called a resonant circuit. Because of no resistance in the circuit, the energy consumption of the LC circuit is less. But this is the ideal case. But in the practical circuit, it will always consume some energy because of the non-zero resistance and the components and connecting wires.

Complete step by step solution:
Given the range of frequency of the radio is from $800{\text{ }}kHz$ to $1200{\text{ }}kHz$ .
Let the lower frequency be ${v_1}$=$800{\text{ }}kHz$
Upper frequency is ${v_2}$=$1200{\text{ }}kHz$.
The effective inductance of the circuit is given as $L$=$200\;\mu H$
The angular frequency of which the energy is transferred in an oscillatory manner between the capacitor and inductor in an LC circuit is given as
$\omega = \dfrac{1}{{\sqrt {LC} }}$
We can rearrange this equation to find the variable capacitance as,
$C = \dfrac{1}{{{\omega ^2}L}}$……… (1)
Let the angular velocity of the capacitor ${C_1}$ be ${\omega _1}$
The angular frequency {\omega _1}$$$$ = 2\pi {v_1}
Substituting ${v_1}$=$800{\text{ }}kHz$
Therefore the angular velocity of the capacitor ${C_1}$,
${\omega _1} = 2\pi \times 800 \times {10^3}rad/s$
Substituting this in equation (1)
${C_1} = \dfrac{1}{{{{(2\pi \times 800 \times {{10}^3}rad/s)}^2} \times 200 \times {{10}^{ - 6}}}}$
${C_1} = 1.980 \times {10^{ - 10}}F$
${C_1} = 198pF$
Similarly, the capacitance for the variable capacitor
${C_2} = \dfrac{1}{{{\omega _2}^2L}}$…….. (2)
Let the angular velocity of the capacitor ${C_2}$ be ${\omega _2}$
The angular frequency ${\omega _2} = 2\pi {v_2}$
Substituting ${v_2} = 1200{\text{ }}kHz$
${\omega _2} = 2\pi \times 1200 \times {10^3}rad/s$
Substituting this in the above equation (2)
${C_2} = \dfrac{1}{{{{(2\pi \times 1200 \times {{10}^3}rad/s)}^2} \times 200 \times {{10}^{ - 6}}}}$
${C_2} = 0.8804F = 88.04pF$
The range of the variable capacitor is from $88.04pF$ to $198pF$.

Note:
The charges in the LC circuit move back and forth between the plates of the capacitor and an inductor. Therefore the energy oscillates between a capacitor and the inductor. These oscillations will eventually die out because of the internal resistance of the components of the connecting wires. This action mathematically is called a harmonic oscillator. This action is similar to water flowing back and forth in a tank. For this reason, the LC circuit is also called a tank circuit.