JEE MAIN - Physics (2022 - 27th July Evening Shift - No. 12)
A series LCR circuit has $$\mathrm{L}=0.01\, \mathrm{H}, \mathrm{R}=10\, \Omega$$ and $$\mathrm{C}=1 \mu \mathrm{F}$$ and it is connected to ac voltage of amplitude $$\left(\mathrm{V}_{\mathrm{m}}\right) 50 \mathrm{~V}$$. At frequency $$60 \%$$ lower than resonant frequency, the amplitude of current will be approximately :
466 mA
312 mA
238 mA
196 mA
Explanation
$$\omega = 0.4{\omega _0}$$ ...... (i)
$$ \Rightarrow I = {V \over Z} = {{50} \over {\sqrt {{R^2} + {{\left( {\omega L - {1 \over {\omega C}}} \right)}^2}} }}$$ ..... (ii)
$$ \Rightarrow I = 238$$ mA
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