Question

The inductor in the RLC tuning circuit of an AM radio has a value of 9 mH. What should be the value of the variable capacitor, in picofarads, in the circuit to tune the radio to 66 kHz

Answer 1

Given:

The inductance is,

`\begin{gathered} L=9\text{ mH} \\ =9*10^(-3)\text{ H} \end{gathered}`

The radio frequency is,

`\begin{gathered} f=66\text{ kHz} \\ =66*10^3\text{ Hz} \end{gathered}`

To find:

value of the variable capacitor, in picofarads

Step-by-step explanation:

The frequency of the AM is,

`\begin{gathered} f=(1)/(2\pi√(LC)) \\ \end{gathered}`

Substituting the values we get,

`\begin{gathered} 66*10^3=\frac{1}{2\pi\sqrt{9*10^(-3)* C}} \\ \sqrt{9*10^(-3)* C}=(1)/(2\pi*66*10^3) \\ \sqrt{9*10^(-3)* C}=2.41*10^(-6) \\ 9*10^(-3)* C=5.81*10^(-12) \\ C=6.45*10^(-10) \\ C=645*10^(-12)\text{ F} \\ C=645\text{ pF} \end{gathered}`

Hence, the capacitance is 645 pF.