• # question_answer An induction coil has an impedance of $10\Omega .$When an AC signal of frequency 1000 Hz is applied to the coil, the voltage leads the current by $45{}^\circ$. The inductance of the coil is A)  $\frac{1}{20\pi }$    B)   $\frac{1}{\sqrt{2}\times 200\pi }$    C)    $\frac{1}{\sqrt{2}\times 20\pi }$     D)   $\frac{1}{200\pi }$

$\text{Tan}\,{{45}^{o}}=\frac{\omega L}{R}\Rightarrow L=R$ Also, $Z=\sqrt{{{R}^{2}}+{{(\omega L)}^{2}}}=10$ $\therefore$$L=\frac{5\sqrt{2}}{2\pi \times 1000}=\frac{1}{\sqrt{2}\times 200\pi }H$             Hence, the correction option is(b).