Question

A coil has inductance of 1.3 mH and resonates at 600 kHz, and its Q value is 30. If the bandwidth is 50 kHz, then the required resistance divided by 100000 is (Until two digits after the decimal point).

Answer 1

Reactance of coil $=X_L=L\omega$
$\begin{array}{cc}=& 2\pi \times 600\times {10}^3\times 1.3\times {10}^3\\ =& 4900\Omega \\ \text{At resonance,}{X}_L& =X_C\\ X_C=4900\Omega & \end{array}$

Coil resistance $=Q{X}_C=30\times 4900=147000\Omega$

Required $Q$ for circuit is $\frac{f_0}{50}=\frac{600}{50}=12$

Equivalent input resistance required $=Q^{\prime }{X}_c$
$=12\times 4900=58800\Omega$

Let shunt resistance is $R^{\prime }$
Then, $58800=\frac{147000R^{\prime }}{147000+R^{\prime }}$

$R^{\prime }=98k$