91Ó°ÊÓ

We are given that A television channel is assigned the frequency range from $54MHZ$to $60MHZ$. A series RLC tuning circuit in a TV receiver resonates in the middle of this frequency range. The circuit uses a $16pF$capacitor. And we need to find that What is the value of the inductor?

In this q we are given that $C=16pF$ and resonance frequency is $$\begin{gathered} f=\frac{54+60}{2} \\ f=57mHZ \end{gathered}$$

and now applying the formula

localid="1650736156693" $$\begin{gathered} 2×\mathrm{π}×\mathrm{f}=\frac{1}{\sqrt{\mathrm{L}×\mathrm{C}}} \\ 2×3.14×57×{10}^6=\frac{1}{\sqrt{\mathrm{L}×16×{10}^{-12}}} \\ \mathrm{L}=0.49\mathrm{μ}\mathrm{H} \end{gathered}$$

so, the value of inductor islocalid="1650736174059" $0.49\mathrm{μ}H$

We are given that A television channel is assigned the frequency range from $54mHZ$to $60mHZ$. A series RLC tuning circuit in a TV receiver resonates in the middle of this frequency range. The circuit uses a $16pF$capacitor. In order to function properly, the current throughout the frequency range must be at least $50\%$ of the current at the resonance frequency. and we need to find that What is the minimum possible value of the circuit’s resistance?

Here current at final frequency equals to half of the current at resonant frequency so ,

$I_{resonant}=\frac{{E_0}^2}{R×\sqrt{R^2+{(X_L-X_C)}^2}}$

As we mentioned above current I at the final frequency equals half of the current at resonant frequency so we divide by two the

$I=\frac{E_0}{2R}$

And we know $I_{resonant}=\frac{E_0}{\sqrt{R^2+{(X_L-X_C)}^2}}$

so

SO the minimum possible value of current is$10.9\mathrm{Î\copyright }$