Physics - EMI/AC Question with Solution | TestHub

Given that the lamp draws a current of $10\mathrm{A}$ upon a $80\mathrm{V}$ D.C. supply, the resistance is,

$R=\frac{80}{10}=8\Omega$

When connected to an AC supply with an inductor in series, as shown in the figure above, the current,

$I=10\mathrm{A}$

and the voltage drop over the lamp is $80\mathrm{V}$ and $V_L$ over the inductance.

Applying the rules of $L-R$ circuit,

$V_L^2+80^2=220^2$

$\Rightarrow V_L=\sqrt{48400-6400}=210$ V

The Inductive reactance of the given inductor of inductance $L$, for the frequency of supply $f$,

$X_L=2\pi fL...(1)$

The voltage drop across the inductance,

$V_L=I{X}_L=210$

$\Rightarrow X_L=2\pi fL=\frac{V_L}{I}=\frac{210}{10}=21\Omega$

Substituting values in equation $(1)$

$L=\frac{21}{2\pi f}=\frac{21}{2\pi \times 50}=\frac{21}{100\pi }=0.065\mathrm{H}$