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  • A series circuit has <img alt="\mathrm{L=0.01 H, R=10 \Omega\: and \: C=1 \mu \mathrm{F}}

A series \mathrm{L C R} circuit has \mathrm{L=0.01 H, R=10 \Omega\: and \: C=1 \mu \mathrm{F}} and it is connected to ac voltage of amplitude \mathrm{\left(\mathrm{V}_{\mathrm{m}}\right) 50 \mathrm{~V}}. At frequency 60 % lower than resonant frequency, the amplitude of current will be approximately :
 

Option: 1

466 \mathrm{~mA}


Option: 2

312 \mathrm{~mA}


Option: 3

238 \mathrm{~mA}


Option: 4

196 \mathrm{~mA}


Answers (1)

best_answer

\mathrm{L=0.01H}

\mathrm{R=10\Omega }

\mathrm{C=10^{-6}F}

\mathrm{V_{m}=50v}

\mathrm{f=\left ( \frac{60}{100} \right )\times f_{resonant}}

    \mathrm{=\frac{3}{5}\times \frac{1}{2\pi \sqrt{LC}}}

\mathrm{f=\frac{3\times 10^{4}}{10\pi }\: \: or\: \: \omega =2\pi f}

\mathrm{\omega =\frac{3}{5}\times 10^{4}}

\mathrm{X_{L}=\omega L=\frac{300}{5}=60}

\mathrm{x_c=\frac{1}{\omega c}=\frac{}{\frac{3}{5} \times 10^4} \times 10^{-6}}

\mathrm{x_c=\frac{500}{3}}

\mathrm{z=\sqrt{R^2+\left(x_L-x_C\right)^2}}

\mathrm{=\sqrt{10^2+\left(\frac{320}{3}\right)^2}}

\mathrm{=10 \sqrt{1+\left(\frac{32}{3}\right)^2}}

\mathrm{z=\frac{10}{3} \times \sqrt{1033}}

\mathrm{I_m=\frac{V_m}{z}=\frac{50}{\frac{10}{3} \times \sqrt{1033}}}

\mathrm{=466 \mathrm{~mA}}

Hence 1 is correct option








 

Posted by

vishal kumar

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