question_answer

How can you improve the quality factor of a parallel resonance circuit. Obtain the resonant frequency and quality factor of a series L-C-R circuit with L = 4 H, \[C=64\mu F\] and \[R=20\,\Omega .\]

Or

A resistor of \[200\,\Omega .\] and a capacitor of \[40\,\mu \,F\] are connected in series to 220 V AC source with angular frequency \[(\omega )=300\] Hz. Calculate the voltage (rms) across the resistor and the capacitor. Why is the algebraic sum of these voltages more than the source voltage? How do you resolve this paradox?

 

Answer:

In order to improve the quality factor of a parallel resonance circuit, ohmic resistance should be made as large as possible. Given, \[L=4H,\] \[C=64\mu F=64\times {{10}^{-6}}F,\] \[R=20\,\Omega .\] Resonant frequency, \[{{\omega }_{r}}=\frac{1}{\sqrt{LC}}\] (in rad/s) \[=\frac{1}{\sqrt{4\times 64\times {{10}^{-6}}}}=\frac{1}{2\times 8\times {{10}^{-3}}}=\frac{1}{0.016}=62.5\,\text{rad/s}\] Quality factor, \[Q=\frac{1}{R}\sqrt{\frac{L}{C}}\] \[=\frac{1}{20}\sqrt{\frac{4}{64\times {{10}^{-6}}}}=\frac{1}{20\times 4\times {{10}^{-3}}}=12.5\] Or \[R=200\,\Omega ,\]\[C=40\mu F,\]\[{{V}_{rms}}=220V,\]\[\omega =300\,Hz\] Inductive reactance             \[{{X}_{C}}=\frac{1}{\omega C}=\frac{1}{300\times 40\times {{10}^{-6}}}=83.33\,\Omega \] Impedance, \[Z=\sqrt{{{R}^{2}}+X_{C}^{2}}\]                         \[=\sqrt{{{(200)}^{2}}+{{(83.33)}^{2}}}=216.6\,\Omega \](approx) \[\therefore \]      \[{{I}_{rms}}=\frac{{{V}_{rms}}}{Z}=\frac{200}{216.6}=0.923\,A\] \[{{V}_{rms}}\,across\,\,R,\]                         \[{{V}_{R}}=IR=0.923\times 200=184.6\,V\] Across capacitor, \[{{V}_{C}}=I{{X}_{C}}=0.923\times 83.33\Rightarrow {{V}_{C}}=76.9\,V\] Thus,    \[V\ne {{V}_{R}}+{{V}_{C}}\]             Because \[{{V}_{C}}\] and \[{{V}_{R}}\] are not in same phase \[\therefore \] \[{{V}_{R}}\]leads \[{{V}_{C}}\] by phase \[\pi /2\]             \[\therefore \]      \[V=\sqrt{V_{R}^{2}+V_{C}^{2}}\]