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BCECE Engineering BCECE Engineering Solved Paper-2011

  • question_answer
    A series resonant circuit contains \[L=\frac{5}{\pi }mH,\]  \[C=\frac{200}{\pi }\mu F\] and R = 100 \[\Omega \] If a source of emf It e = 200 sin 1000 \[1000\,\pi t\] is applied, then the rms current is

    A)  2 A                        

    B)         200\[\sqrt{2}\] A

    C)  100\[\sqrt{2}\] A            

    D)         1.41 A

    Correct Answer: D

    Solution :

    \[e=220\sin 100\,\pi t,\,\omega =1000\,\pi \] \[\Rightarrow \]               \[f=\frac{\omega }{2\pi }=500\,Hz\] \[{{e}_{0}}=200\,V\] At resonance Z = R So,          \[{{i}_{0}}=\frac{200}{R}=\frac{200}{100}=2A\] \[{{i}_{rms}}=\frac{{{i}_{0}}}{\sqrt{2}}=\frac{2}{\sqrt{2}}=\sqrt{2}=1.41\,A\]


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