Physics - EMI/AC Question with Solution | TestHub

For ${\omega }_1=1000rad/s$: $X_{L_1}=1000\times 0.1=100\Omega$, $X_{C_1}=1/(1000\times 10^{-5})=100\Omega$.

The impedance $Z_1=\sqrt{R^2+(X_{L_1}-X_{C_1}{)}^2}=\sqrt{10^2+0^2}=10\Omega$.

The peak current $I_{01}=V_{01}/Z_1=100/10=10A$.

RMS current $I_{rms1}=I_{01}/\sqrt{2}=10/\sqrt{2}=5\sqrt{2}A$. So (P)-(2).

Power factor $\cos {\varphi }_1=R/Z_1=10/10=1$. So (S)-(1).

For ${\omega }_2=2000rad/s$: $X_{L_2}=2000\times 0.1=200\Omega$, $X_{C_2}=1/(2000\times 10^{-5})=50\Omega$.

The impedance $Z_2=\sqrt{R^2+(X_{L_2}-X_{C_2}{)}^2}=\sqrt{10^2+(200-50{)}^2}=\sqrt{100+150^2}=\sqrt{100+22500}=\sqrt{22600}\Omega$. The peak current $I_{02}=V_{02}/Z_2=50/\sqrt{22600}A$.

RMS current $I_{rms2}=I_{02}/\sqrt{2}=50/\sqrt{45200}=5/\sqrt{452}A$. So (Q)-(4).

Average power dissipated is the sum of powers for each frequency component: $P_{avg}=I_{rms1}^{2}R+I_{rms2}^{2}R=(5\sqrt{2}{)}^2\times 10+(5/\sqrt{452}{)}^2\times 10=(50\times 10)+(25/452\times 10)=500+250/452=500+125/226W$. So (R)-(3).