Question 2.33: An inductor, a variable capacitor, and a resistor are connec...

Let resonant frequency be f_{0}.

At resonance,                                      X_{L} =X_{C}

2πf_{0}L=\frac{1}{2πf_{0}C}

or,                                                          LC =\frac{1}{(2πf_{0})^{2}}

or,                                                            L=\frac{1}{(2πf_{0})^{2}C}

Substituting values,

L=\frac{1}{(6.28 ×100)^{2} × 100  ×  10^{-6}}

= 25.3×10^{–3}  H

Since X_{L} = X_{C}, the value of impedance at = R.

Maximum value of current,            I_{m}=I_{0}=\frac{V}{R}  A

At a frequency of 100 Hz, C = 200 ×10^{–6} F, current is reduced to half, i.e., impedance becomes equal to twice its value at resonance, i.e., equals 2R.

Impedance,                       Z=\sqrt{R^{2}  +  (X_{L}  –  X_{C})^{2}}

Current,                               I=\frac{V}{Z}=\frac{V}{\sqrt{R^{2}+(X_{L}  –  X_{C})^{2}}}

According to the problem

I=\frac{I_{m}}{2}

or,                                           \frac{V}{\sqrt{R^{2}+(X_{L}  –  X_{C})^{2}}} = \frac{I_{m}}{2}=\frac{V}{2R}

or,                                          \sqrt{R^{2}+(X_{L}  –  X_{C})^{2}}=2R

or,                                           (X_{L}  –  X_{C})^{2}=3R^{2}

or,                                         \sqrt {3}  R = X_{L}  –  X_{C}

X_{L} = 2πf_{0}L = 628 × 25.3 × 10^{-3} = 15.88  Ω

X_{C}= \frac{1}{2πf_{0}C}= \frac{1}{6.28  ×  200  ×  10^{-6}}=7.96  Ω

R= \frac{X_{L}  –  X_{C}}{ \sqrt {3}}=\frac{15.88  –  7.96}{1.732}=4.57  Ω