Q68 / 120 · NEET 2011 (Mains) · Physics

A coil has resistance $30 \Omega$ and inductive reactance $20 \Omega$ at $50 \mathrm{~Hz}$ frequency. If an $\mathrm{AC}$ source of $200 \mathrm{~V}, 100 \mathrm{~Hz}$, is connected across the coil, the current in the coil will be

  • Correct answer
    A
    $4.0 \mathrm{~A}$
  • B
    $8.0 \mathrm{~A}$
  • C
    $\frac{20}{\sqrt{13}} \mathrm{~A}$
  • D
    $2.0 \mathrm{~A}$

Solution

If $\omega=50 \times 2 \pi$ then $\omega L=20 \Omega$ Similarly $\omega^{\prime}=100 \times 2 \pi$ then $\omega^{\prime} L=40 \Omega$ $$ \begin{aligned} i & =\frac{200}{Z}=\frac{200}{\sqrt{R^2+\left(\omega^{\prime} L\right)^2}} \\ & =\frac{200}{\sqrt{(30)^2+(40)^2}}=4 \mathrm{~A} \end{aligned} $$

Hence, the correct answer is option: 'A'.

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