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 :
Solution
<p>$\omega = 0.4{\omega _0}$ ...... (i)</p>
<p>$$ \Rightarrow I = {V \over Z} = {{50} \over {\sqrt {{R^2} + {{\left( {\omega L - {1 \over {\omega C}}} \right)}^2}} }}$$ ..... (ii)</p>
<p>$\Rightarrow I = 238$ mA</p>
About this question
Subject: Physics · Chapter: Alternating Current · Topic: AC Circuits: R, L, C
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