Problem — Capacitor of an AM radio receiver

Problem

An LC circuit is used as a resonator in the receiver of an AM radio broadcasting at frequency $f=1{,}0\ \text{MHz}$. If the coil’s inductance is $L=0{,}9\ \text{mH}$, what is the capacitance of the variable capacitor it must be tuned to?

Solution

Tuning occurs when the natural frequency of the LC circuit coincides with that of the broadcast: $f=\frac{1}{2\pi\sqrt{LC}}$

Solving for the capacitance: $C=\frac{1}{4\pi^2 f^2 L}$

Substituting $f=1{,}0\cdot 10^{6}\ \text{Hz}$ (so $f^2=10^{12}\ \text{Hz}^2$), $4\pi^2\approx 39{,}48$ and $L=0{,}9\cdot 10^{-3}\ \text{H}$: $C=\frac{1}{39{,}48\cdot 10^{12}\cdot 0{,}9\cdot 10^{-3}}=\frac{1}{39{,}48\cdot 0{,}9\cdot 10^{9}}=\frac{1}{3{,}55\cdot 10^{10}}\approx 2{,}8\cdot 10^{-11}\ \text{F}$

that is, about $28\ \text{pF}$.

$\ev{C=\frac{1}{4\pi^2 f^2 L}\approx 2{,}8\cdot 10^{-11}\ \text{F}\approx 28\ \text{pF}}$

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Topics: Electromagnetic induction Concepts: LC and RLC oscillations · Resonance