Question

When 10V, DC is applies across a coil current through it is 2.5 A, if 10V, 50Hz A. C. is applied current reduces to 2 A. Calculate reluctance of the coil.

Answer 1

$so,I=2.5$

$\frac{V}{R}=I$

$R=\frac{10}{2.5}=4\Omega$

$I=\frac{V}{Z}$

$Z=(R+j\omega L)$

$=\sqrt{R^2+{\omega }^2{L}^2}$

$2=\frac{10}{\sqrt{16+{\omega }^2{L}^2}}$

$16+{\omega }^2{L}^2=5^2$

${\omega }^2{L}^2=9$

$L^2=\frac{9}{{\omega }^2}$

$L=\frac{3}{\omega }$

$=\frac{3}{2050}$ $\therefore \omega =2\pi f=2\pi 50=100\pi$

$=9.5mH$