An inductance coil has a reactance


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

  1. (1) $1.1 \times 10^{-2} \mathrm{H}$

  2. (2) $1.1 \times 10^{-1} \mathrm{H}$

  3. (3) $5.5 \times 10^{-5} \mathrm{H}$

  4. (4) $6.7 \times 10^{-7} \mathrm{H}$

Correct Option: 1


(1) Given,

Reactance of inductance coil, $Z=100 \Omega$

Frequency of $\mathrm{AC}$ signal, $v=1000 \mathrm{~Hz}$

Phase angle, $\phi=45^{\circ}$

$\tan \phi=\frac{X_{L}}{R}=\tan 45^{\circ}=1$

$\Rightarrow X_{L}=R$

Reactance, $Z=100=\sqrt{X_{L}^{2}+R^{2}}$

$\Rightarrow 100=\sqrt{R^{2}+R^{2}}$

$\Rightarrow \sqrt{2} R=100 \Rightarrow R=50 \sqrt{2}$

$\therefore X_{L}=50 \sqrt{2}$

$\Rightarrow L \omega=50 \sqrt{2}$                             $\left(\because X_{L}=\omega L\right)$

$\Rightarrow L=\frac{50 \sqrt{2}}{2 \pi \times 1000}$    $(\because \omega=2 \pi v)$

$=\frac{25 \sqrt{2}}{\pi} \mathrm{mH}=1.1 \times 10^{-2} \mathrm{H}$


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