question_answer

A series L-C-R circuit is connected to a 220V variable frequency AC supply. If L = 20 mH,

\[C=\left( \frac{800}{{{\pi }^{2}}} \right)\mu F\] and \[R=110\,\Omega .\]

(i) Find the frequency of the source for which average power absorbed by the circuit is maximum.

(ii) Calculate the value of maximum current amplitude.

Answer:

(i) 125 Hz          (ii) 2.83A Given, \[L=20\,\,mH,C=\frac{800}{{{\pi }^{2}}}\mu F,R=110\Omega \] (i) When average power absorbed by the circuit is maximum, then             \[{{X}_{L}}={{X}_{C}}\Rightarrow \omega L=\frac{1}{\omega C}\]             \[v=\frac{1}{2\pi \sqrt{LC}}\]                 \[(\because \omega \le 2\pi v)\] or         \[=\frac{1}{2\pi \sqrt{20\times {{10}^{-3}}\times \frac{800}{{{\pi }^{2}}}\times {{10}^{-6}}}}\]             \[\text{v}\,\,\text{=}\,\,\text{125}\,\,\text{Hz}\] (ii) Here,\[{{X}_{L}}=\omega \,\,L=2\pi \,\,vL\] \[=2\times 314\times 125\times 20\times {{10}^{-3}}H=15.7\Omega \] \[{{X}_{C}}=\frac{1}{\omega C}=\frac{1}{2\times 3.14\times 125\times \frac{800}{{{\pi }^{2}}}\times {{10}^{-6}}}=15.7\Omega \] \[R=110\Omega ,{{E}_{0}}=\sqrt{2}\,\,{{E}_{rms}}=\sqrt{2}\times 220=311\,V\]             \[\therefore \]      \[Z=\sqrt{{{({{X}_{L}}-{{X}_{C}})}^{2}}+{{R}^{2}}}\]                         \[=\sqrt{{{(15.7-15.7)}^{2}}+{{(110)}^{2}}}\]                         \[Z=110\,\,\Omega \] Also,     \[Z=\frac{{{E}_{0}}}{{{I}_{0}}}\] \[\Rightarrow \therefore \] \[{{I}_{0}}=\frac{{{E}_{0}}}{Z}=\frac{311}{110}=2.83\,\,A\]