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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
$1.1 \times 10^{-2}\; H$
$1.1 \times 10^{-1} \;H$
$5.5 \times 10^{-5} \;H$
$6.7 \times 10^{-7}\; H$
Solution
Reactance of inductance coil
$=\sqrt{{ R ^{2}+ x _{ L }^{2}}}=100$ $…(i)$
$f =1000 Hz$ of applied $AC$ signal
Voltage leads current bly $45^{\circ}$
$\tan 45^{\circ}=\frac{ i X _{ L }}{ iR }=\frac{\omega L }{ R }$
ie $R = X _{ L }=\omega L$
Putting in eqn $(i):$ $\sqrt{ X _{ L }^{2}+ X _{ L }^{2}}=100$
$\sqrt{2} X _{ L }=100 \Rightarrow X _{ L }=50 \sqrt{2}$
ie $\omega L =50 \sqrt{2}$
$L =\frac{50 \sqrt{2}}{\omega}=\frac{50 \sqrt{2}}{2 \pi f }=\frac{25 \sqrt{2}}{\pi \times 1000} H$
$=1.125 \times 10^{-2} H$
