An inductance coil has a reactance of 100, Om | vedclass

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(A) The impedance $Z$ of an $RL$ circuit is given by $Z = \sqrt{R^2 + X_L^2} = 100\, \Omega$. Given that the voltage leads the current by $\phi = 45^{

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$1.1 	imes 10^{-2}\; H$

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(A) The impedance $Z$ of an $RL$ circuit is given by $Z = \sqrt{R^2 + X_L^2} = 100\, \Omega$. Given that the voltage leads the current by $\phi = 45^{\circ}$,we have $	an \phi = \frac{X_L}{R}$. Since $	an 45^{\circ} = 1$,it follows that $X_L = R$. Substituting $X_L = R$ into the impedance equation: $\sqrt{X_L^2 + X_L^2} = 100 \Rightarrow \sqrt{2} X_L = 100$. Thus,$X_L = \frac{100}{\sqrt{2}} = 50\sqrt{2}\, \Omega$. Since $X_L = 2\pi f L$,we have $L = \frac{X_L}{2\pi f} = \frac{50\sqrt{2}}{2 	imes \pi 	imes 1000}$. $L = \frac{25\sqrt{2}}{1000\pi} \approx \frac{25 	imes 1.414}{3141.59} \approx 0.01125\, H = 1.125 	imes 10^{-2}\, H$.

An inductance coil has a reactance of $100\, \Omega$. When an $AC$ signal of frequency $1000\, Hz$ is applied to the coil,the applied voltage leads the current by $45^{\circ}$. The self-inductance of the coil is

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