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An inductance coil has a resistance of $100 \Omega$. When are a.c. signal of frequency $100 \mathrm{Hzis}$ applied to the coil, the voltage leads the current by $45^{\circ}$. The inductance of the coil in henry is $\left[\sin 45^{\circ}=\cos 45^{\circ}=\frac{1}{\sqrt{2}}\right]$
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The correct answer is:
$\frac{1}{2 \pi}$
The phase angle in an LR circuit is given by, since $\phi=45^{\circ}$,
$\tan \phi=\frac{X_L}{X_R}=1$
Resistance of inductive coil $X_L=X_R=100 \Omega$
Thus, inductance of coil $L=\frac{X_L}{2 \pi f}$
$\therefore L=\frac{100}{2 \pi \times 100 \mathrm{~s}^{-1}}=\frac{1}{2 \pi} \mathrm{H}$
$\tan \phi=\frac{X_L}{X_R}=1$
Resistance of inductive coil $X_L=X_R=100 \Omega$
Thus, inductance of coil $L=\frac{X_L}{2 \pi f}$
$\therefore L=\frac{100}{2 \pi \times 100 \mathrm{~s}^{-1}}=\frac{1}{2 \pi} \mathrm{H}$
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