JEE MAIN - Physics (2022 - 28th July Morning Shift - No. 20)
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
250
Explanation
$${i \over {{i_{\max }}}} = {1 \over {\sqrt 2 }}$$
$$ = {{{{{V_0}} \over Z}} \over {{{{V_0}} \over R}}}$$
$$ \Rightarrow {R \over Z} = {1 \over {\sqrt 2 }}$$
and $${1 \over {212C}} - 212L = 232L - {1 \over {232C}}$$
so $$212L = {1 \over {232C}}$$
so $${R \over {\sqrt {{R^2} + {{\left( {232L + {1 \over {232C}}} \right)}^2}} }} = {1 \over {\sqrt 2 }}$$
$${{{R^2}} \over {{R^2} + {{(20L)}^2}}} = {1 \over 2}$$
$$400{L^2} = {R^2}$$
$$L = {5 \over {20}}$$
$$H = {5 \over {20}} \times 1000$$ mH
$$= 250$$ mH
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