BCECE Medical BCECE Medical Solved Papers-2010

  • question_answer
    An inductor of 1 H is connected across a 220 V. 50 Hz supply. The peak value of the current is approximately

    A)  0.5 A            

    B)  0.7 A

    C)  1A               

    D)  1.4 A

    Correct Answer: C

    Solution :

    Current \[({{i}_{0}})=\frac{{{E}_{0}}}{{{X}_{L}}}\] \[\therefore \] \[{{i}_{0}}=\frac{{{E}_{0}}}{L\omega }\] \[\therefore \] \[{{i}_{0}}=\frac{220\times \sqrt{2}}{1\times 2\pi \times 50}\] \[\Rightarrow \] \[{{i}_{0}}=\frac{220\times \sqrt{2}}{100\pi }=\frac{220\times \sqrt{2}}{100\times 3.14}\] \[\Rightarrow \] \[{{i}_{0}}=1\,A\]


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