Question 3.33: An R–L–C series circuit has R = 10 Ω, L = 0.1 H, and C = 8 μ......
An R–L–C series circuit has R = 10 Ω, L = 0.1 H, and C = 8 μf. Calculate the following:
(a) resonant frequency;
(b) Q-factor of the circuit at resonance;
(c) half-power frequencies and bandwidth.
(a) We know that the condition for the series resonance is
X_L=X_Ci.e., 2 \pi \mathrm{f}_0 \mathrm{~L}=\frac{1}{2 \pi \mathrm{f}_0 \mathrm{C}}
or, f_0=\frac{1}{2 \pi \sqrt{\mathrm{LC}}}
Substituting values
\begin{aligned} \mathrm{f}_0 & =\frac{1}{2 \pi \sqrt{0.1 \times 8 \times 10^{-6}}}=178 \mathrm{~Hz} \\(b)\quad \mathrm{Q} \text {-factor } & =\frac{\text { voltage across } \mathrm{L}}{\text { supply voltage }}=\frac{\mathrm{I}_0 \mathrm{X}_{\mathrm{L}}}{\mathrm{I}_0 \mathrm{R}}=\frac{2 \pi \mathrm{f}_0 \mathrm{~L}}{\mathrm{R}} \\ & =\frac{2 \pi \mathrm{L}}{\mathrm{R}} \cdot \frac{1}{2 \pi \sqrt{\mathrm{LC}}}=\frac{1}{\mathrm{R}} \sqrt{\frac{\mathrm{L}}{\mathrm{C}}} \end{aligned}Substituting values
\mathrm{Q} \text {-factor }=\frac{1}{10} \sqrt{\frac{0.1}{8 \times 10^{-6}}}=11.2(c) Half-power frequencies are the frequencies corresponding to current which is 0.707 of the resonant current. They are f_1\ and\ f_2 , i.e., the lower, and upper frequencies forming the bandwidth.
f_1=f_0-\frac{R}{4 \pi L}and \mathrm{f}_2=\mathrm{f}_0+\frac{\mathrm{R}}{4 \pi \mathrm{L}}.
\begin{aligned}\text{Substituting}\quad & \mathrm{f}_1=178-\frac{10}{12.56 \times 0.1}=169 \mathrm{~Hz} \\ & \mathrm{f}_2=178+\frac{10}{12.56 \times 0.1}=187 \mathrm{~Hz} \end{aligned}Bandwidth =\mathrm{f}_2-\mathrm{f}_1=18 \mathrm{~Hz}