An induction coil has an impedance of 10,Omeg | vedclass

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(B) Given: Impedance $Z = 10\,\Omega$,frequency $f = 1000\,Hz$,phase angle $\phi = 45^\circ$.The phase angle in an $RL$ circuit is given by $	an \phi

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$\frac{1}{\sqrt{2} 	imes 200\pi}\,H$

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(B) Given: Impedance $Z = 10\,\Omega$,frequency $f = 1000\,Hz$,phase angle $\phi = 45^\circ$.The phase angle in an $RL$ circuit is given by $	an \phi = \frac{\omega L}{R}$.Since $\phi = 45^\circ$,$	an 45^\circ = 1$,so $\omega L = R$.The impedance is $Z = \sqrt{R^2 + (\omega L)^2} = 10\,\Omega$.Substituting $R = \omega L$,we get $Z = \sqrt{(\omega L)^2 + (\omega L)^2} = \sqrt{2}(\omega L) = 10$.Thus,$\omega L = \frac{10}{\sqrt{2}} = 5\sqrt{2}\,\Omega$.We know $\omega = 2\pi f = 2\pi 	imes 1000 = 2000\pi\,rad/s$.Therefore,$L = \frac{5\sqrt{2}}{\omega} = \frac{5\sqrt{2}}{2000\pi} = \frac{\sqrt{2}}{400\pi} = \frac{1}{200\sqrt{2}\pi}\,H$.Wait,re-evaluating the calculation: $L = \frac{5\sqrt{2}}{2000\pi} = \frac{1}{400\pi/\sqrt{2}} = \frac{1}{200\sqrt{2}\pi}$.Checking the options provided,the correct expression is $L = \frac{1}{\sqrt{2} 	imes 200\pi}$.

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 $45^o$. The inductance of the coil is

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