question_answer

A 200 V variable frequency AC source is connected to a series combination of L = 5 H, \[C=80\mu F\text{ }\]and \[R=40\,\Omega .\] Calculate

(i) angular frequency of source to get the maximum current in the circuit

(ii) current amplitude at resonance

(iii) power dissipation in the circuit

Answer:

Given,   \[{{E}_{rms}}=200\,V,\]            \[L=5\,H,\] \[C=80\mu F=80\times {{10}^{-6}}F\] and \[R=40\,\Omega \] (i) For the maximum current in the circuit, \[{{X}_{L}}={{X}_{C}}\] \[\Rightarrow \]   \[\omega L=\frac{1}{\omega C}\] \[\therefore \] Resonant frequency, \[{{\omega }_{r}}=\frac{1}{\sqrt{LC}}=\frac{1}{\sqrt{5\times 80\times {{10}^{-6}}}}=50\,rad/s\] (ii) Impedance,   \[Z=\sqrt{{{R}^{2}}+{{({{X}_{L}}-{{X}_{C}})}^{2}}}\]             \[Z=R=40\,\Omega \]                 \[[\because \,{{X}_{L}}={{X}_{C}}]\] Current, \[{{I}_{rms}}=\frac{{{E}_{rms}}}{Z}=\frac{200}{40}=5\,A\] Current amplitude at resonance,             \[{{I}_{0}}={{I}_{rms}}\sqrt{2}=5\times 1.414=7.07\,A\] (iii) Power dissipated in circuit, \[P={{E}_{rms}}{{I}_{rms}}\cos 0{}^\circ =200\times 5\times 1=1000\,W\]