AMU Medical AMU Solved Paper-2010

  • question_answer
    A pure inductor of 25 mH is connected to a source of 220 V. Given the frequency of the source as 50 Hz, the rms current in the circuit is

    A)  7 A                                        

    B)  14 A

    C)  28 A                                     

    D)  42 A

    Correct Answer: C

    Solution :

                     Given, \[L=25\,\,mH\]                 \[=25\times {{10}^{-3}}H\],                 \[f=50\,Hz\] \[{{V}_{rms}}=220\,V\] and \[{{f}_{2}}=50\,Hz\]                 \[{{X}_{L}}=2\pi fL\]                 \[=2\times \frac{22}{7}\times 50\times 25\times {{10}^{-3}}\Omega \] The rms current in the circuit is                 \[{{I}_{rms}}=\frac{{{V}_{rms}}}{{{X}_{L}}}\]                 \[=\frac{220}{2\times \frac{22}{7}\times 50\times 25\times {{10}^{-3}}}\]                 \[=\frac{7\times 1000}{2\times 5\times 25}\]                 \[{{I}_{rms}}=28\,A\]


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