Physics - EMI/AC Question with Solution | TestHub

Current through $60 \mathrm{\Omega }$ resistance $=\frac{15}{60}=\frac{1}{4}\mathrm{A}$

thus capacitor current $=\frac{1}{4} \mathrm{A}$

$\because V_C=I X_C$

$10=\frac{1}{4}\times \frac{1}{\omega C}$

$\therefore C=\frac{1}{40\omega }=\frac{1}{4000}=250 \mu \mathrm{F}$

Now,

current through $40 \mathrm{\Omega }$ resistance $=\frac{20}{40}=\frac{1}{2} \mathrm{A}$

thus current through inductor $=\frac{1}{2}-\frac{1}{4}=\frac{1}{4} \mathrm{A}$

$V_L=I{X}_L=\frac{1}{4}\times \omega L$

$20=\frac{1}{4}\times 100\times L$

$\Rightarrow L=0.8 \mathrm{H}$