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 angular frequency is $\omega = 2\pi f = 2\pi 	imes 1

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

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(B) Given: Impedance $Z = 10 \Omega$,Frequency $f = 1000 \ Hz$,Phase angle $\phi = 45^\circ$.The angular frequency is $\omega = 2\pi f = 2\pi 	imes 1000 = 2000\pi \ rad/s$.In an $RL$ circuit,the phase angle $\phi$ is given by $	an \phi = \frac{\omega L}{R}$.Since $\phi = 45^\circ$,$	an 45^\circ = 1$,so $\frac{\omega L}{R} = 1$,which implies $R = \omega L$.The impedance $Z$ is given by $Z = \sqrt{R^2 + (\omega L)^2}$.Substituting $R = \omega L$,we get $Z = \sqrt{(\omega L)^2 + (\omega L)^2} = \sqrt{2(\omega L)^2} = \omega L \sqrt{2}$.Given $Z = 10 \Omega$,we have $10 = \omega L \sqrt{2}$.Therefore,$\omega L = \frac{10}{\sqrt{2}} = 5\sqrt{2}$.Since $\omega L = 2000\pi L$,we have $2000\pi L = 5\sqrt{2}$.$L = \frac{5\sqrt{2}}{2000\pi} = \frac{\sqrt{2}}{400\pi} = \frac{1}{\sqrt{2} 	imes 200\pi} \ H$.

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

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