Question

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

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Solution

## Step 1: Given dataThe lower tuning frequency of the radio is ${f}_{1}=800KHz=800×{10}^{3}Hz.$The upper tuning frequency of the radio is ${f}_{2}=1200KHz=1200×{10}^{3}Hz.$The effective inductance of the radio is $L=200\mu H=200×{10}^{-6}H.$Step 2: The capacitance of the LCR circuitWe know at the resonance of the LCR circuit the angular frequency is $\omega =\frac{1}{\sqrt{\mathrm{LC}}}$, where, L is the inductance, C is the capacitance, $\omega$ and is the angular frequency of the wave.Angular momentum, $\omega =2\mathrm{\pi f}$, where, f is the frequency of a wave.Step 3: Finding the range of the capacitorAs we know the angular frequency of an LCR circuit is $\omega =\frac{1}{\sqrt{\mathrm{LC}}}$.So, $C=\frac{1}{{\omega }^{2}L}.................\left(1\right)$ Now, the capacitance of the variable capacitor for frequency ${f}_{1}$ and angular frequency ${\mathrm{\omega }}_{1}$ is ${C}_{1}=\frac{1}{{{\omega }_{1}}^{2}L}=\frac{1}{{2{\mathrm{\pi f}}_{1}}^{2}×L}=\frac{1}{\left(2×3.14\right)×{\left(800×{10}^{3}\right)}^{2}×\left(200×{10}^{-6}\right)}\phantom{\rule{0ex}{0ex}}or{C}_{1}=1.98×{10}^{-10}F................\left(1\right)$And, the capacitance of the variable capacitor for frequency ${f}_{2}$ and angular frequency ${\mathrm{\omega }}_{2}$ is ${C}_{2}=\frac{1}{{{\omega }_{2}}^{2}L}=\frac{1}{{2{\mathrm{\pi f}}_{2}}^{2}×L}=\frac{1}{\left(2×3.14\right)×{\left(1200×{10}^{3}\right)}^{2}×\left(200×{10}^{-6}\right)}\phantom{\rule{0ex}{0ex}}or{C}_{2}=0.880×{10}^{-10}F..............\left(2\right)$From equations 1 and 2 we get, $C=\left[{C}_{1},{C}_{2}\right]=\left[\left(1.98×{10}^{-10}\right),\left(0.880×{10}^{-10}\right)\right]$Therefore, the range of its variable capacitor is $C=\left[\left(1.98×{10}^{-10}\right)F,\left(0.880×{10}^{-10}\right)F\right]$.

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