Question 3.36: An inductor, a variable capacitor, and a resistor are connec......
An inductor, a variable capacitor, and a resistor are connected in series across a constant voltage, 100 Hz power supply. When the capacitor value is fixed at 100 μF, the current reaches its maximum value. Current gets reduced to half its maximum value when the capacitor value is 200 μF. Calculate the values of circuit parameters and the Q-factor of the circuit.
Let resonant frequency be f_0.
\begin{aligned}\text{At resonance, }\quad X_L & =X_C \\ 2 \pi f_0 L & =\frac{1}{2 \pi f_0 C} \end{aligned}or, \mathrm{LC}=\frac{1}{\left(2 \pi f_0\right)^2}
or, \mathrm{L}=\frac{1}{\left(2 \pi \mathrm{f}_0\right)^2 \mathrm{C}}
Substituting values,
\begin{aligned} \mathrm{L} & =\frac{1}{(6.28 \times 100)^2 \times 100 \times 10^{-6}} \\ & =25.3 \times 10^{-3} \mathrm{H} \end{aligned}Since X_L = X_C, the value of impedance at = R.
Maximum value of current, \mathrm{I}_{\mathrm{m}}=\mathrm{I}_0=\frac{\mathrm{V}}{\mathrm{R}} \mathrm{A}
At a frequency of 100 Hz, \mathrm{C}=200 \times 10^{-6} \mathrm{~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+\left(X_L-X_C\right)^2}
Current, I=\frac{V}{Z}=\frac{V}{\sqrt{R^2+\left(X_L-X_C\right)^2}}
According to the problem
I=\frac{I_m}{2}or, \frac{\mathrm{V}}{\sqrt{R^2+\left(X_L-X_C\right)^2}}=\frac{I_m}{2}=\frac{V}{2 R}
or, \sqrt{\mathrm{R}^2+\left(\mathrm{X}_{\mathrm{L}}-\mathrm{X}_{\mathrm{C}}\right)^2}=2 \mathrm{R}
or, \left(X_L-X_C\right)^2=3 R^2.
\begin{gathered}or, \quad \sqrt{3} \mathrm{R}=\mathrm{X}_{\mathrm{L}}-\mathrm{X}_{\mathrm{C}} \\ \mathrm{X}_{\mathrm{L}}=2 \pi \mathrm{f}_0 \mathrm{~L}=628 \times 25.3 \times 10^{-3}=15.88 \Omega \\ \mathrm{X}_{\mathrm{C}}=\frac{1}{2 \pi \mathrm{f}_0 \mathrm{C}}=\frac{1}{6.28 \times 200 \times 10^{-6}}=7.96 \Omega \\ \mathrm{R}=\frac{\mathrm{X}_{\mathrm{L}}-\mathrm{X}_{\mathrm{C}}}{\sqrt{3}}=\frac{15.88-7.96}{1.732}=4.57 \Omega \end{gathered}