JEE Mains · Physics · STD 12 -7. Alternating current
A series \(L C R\) circuit has \(L=0.01\,H, R=10\,\Omega\) and \(C =1\,\mu F\) and it is connected to ac voltage of amplitude \(\left( V _{ m }\right) 50\,V\). At frequency \(60 \%\) lower than resonant frequency, the amplitude of current will be approximately \(...............\,mA\)
- A \(466\)
- B \(312\)
- C \(238\)
- D \(196\)
Answer & Solution
Correct Answer
(C) \(238\)
Step-by-step Solution
Detailed explanation
Resonant frequency, \(\omega_{0}=\frac{1}{\sqrt{ LC }}=10^{4} rad / sec\) \(\omega^{\prime}=.4 \times 10^{4}=4000 rad / sec\) \(i_{0}=\frac{ V _{0}}{\sqrt{ R ^{2}+\left( X _{ C }^{\prime}- X _{ L }^{\prime}\right)^{2}}}=238\,mA\)
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