An inductor coil takes a current of $8 \, A$ when connected to a $100 \, V$ and $50 \, Hz$ $AC$ source. $A$ pure resistor under the same condition takes a current of $10 \, A$. If the inductor coil and resistor are connected in series to a $100 \, V$ and $40 \, Hz$ $AC$ supply, then the current in the series combination of the above resistor and inductor is:

  • A
    $\frac{10}{\sqrt{3}} \, A$
  • B
    $\frac{5}{\sqrt{2}} \, A$
  • C
    $10 \sqrt{2} \, A$
  • D
    $5 \sqrt{2} \, A$

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