The frequencies at which the current amplitude in an LCR series circuit becomes $\frac{1}{\sqrt{2}}$ times its maximum value, are $212\,\mathrm{rad} \,\mathrm{s}^{-1}$ and $232 \,\mathrm{rad} \,\mathrm{s}^{-1}$. The value of resistance in the circuit is $R=5 \,\Omega$. The self inductance in the circuit is __________ $\mathrm{mH}$.
Answer (integer)
250
Solution
<p>${i \over {{i_{\max }}}} = {1 \over {\sqrt 2 }}$</p>
<p>$= {{{{{V_0}} \over Z}} \over {{{{V_0}} \over R}}}$</p>
<p>$\Rightarrow {R \over Z} = {1 \over {\sqrt 2 }}$</p>
<p>and ${1 \over {212C}} - 212L = 232L - {1 \over {232C}}$</p>
<p>so $212L = {1 \over {232C}}$</p>
<p>so $${R \over {\sqrt {{R^2} + {{\left( {232L + {1 \over {232C}}} \right)}^2}} }} = {1 \over {\sqrt 2 }}$$</p>
<p>${{{R^2}} \over {{R^2} + {{(20L)}^2}}} = {1 \over 2}$</p>
<p>$400{L^2} = {R^2}$</p>
<p>$L = {5 \over {20}}$</p>
<p>$H = {5 \over {20}} \times 1000$ mH</p>
<p>$= 250$ mH</p>
About this question
Subject: Physics · Chapter: Alternating Current · Topic: AC Circuits: R, L, C
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