Physics - EMI/AC Question with Solution | TestHub

Given, self-inductance,$L=1.8\times {10}^{-\mathrm{4 }}\mathrm{H}$

Resistance,$R=\mathrm{6 }\mathrm{\Omega }$

When self-inductance and resistance are broken up into identical coils.

Then, the self-inductance of each oil

$= \frac{1.8\times {10}^{-4}}{2} \mathrm{H}$

The resistance of each oil

$= \frac{6\mathrm{\Omega }}{2}=\mathrm{3 }\mathrm{\Omega }$

Coil are then connected in parallel
($L_{{}_{eq}}=\frac{L_1{L}_2}{L_1+L_2}$)

$\therefore L^{\prime }=\frac{\frac{1.8}{2}\times {10}^{-4}\times \frac{1.8}{2}\times {10}^{-4}}{\frac{1.8}{2}\times {10}^{-4}+\frac{1.8}{2}\times {10}^{-4}}$

$=0.45\times {10}^{-\mathrm{4 }}\mathrm{H}$

and$R^{\prime }=\frac{3\times 3}{3+3}=1.5\mathrm{\Omega }$
($R_{{}_{eq}}=\frac{R_1{R}_2}{R_1+R_2}$)

Time constant$=\frac{L^{\prime }}{R\prime }$

$= \frac{0.45\times {10}^{-4}}{1.5}=0.3\times {10}^{-\mathrm{4 }}\mathrm{s}$