Physics - EMI/AC Question with Solution | TestHub

Given:

Resistance $R=10^3\Omega$

Capacitance $C=10^{-6}\mathrm{F}$

Inductance $L=2\mathrm{H}$

Voltage $V=200\sin (1000\mathrm{t})$ volt

From the voltage equation, the peak voltage $V_0=200\mathrm{V}$ and angular frequency $\omega =1000rad/s$.

1. Calculate inductive reactance $X_L$:

$X_L=\omega L=1000\times 2=2000\Omega$

2. Calculate capacitive reactance $X_C$:

$X_C=\frac{1}{\omega C}=\frac{1}{1000\times 10^{-6}}=\frac{1}{10^{-3}}=1000\Omega$

3. Calculate impedance $Z$:

$Z=\sqrt{R^2+(X_L-X_C{)}^2}$

$Z=\sqrt{(10^3{)}^2+(2000-1000{)}^2}$

$Z=\sqrt{(1000{)}^2+(1000{)}^2}=\sqrt{10^6+10^6}=\sqrt{2\times 10^6}=1000\sqrt{2}\Omega$

$Z\approx 1000\times 1.414=1414\Omega$

So, option B is correct.

4. Calculate the peak current $I_0$:

$I_0=\frac{V_0}{Z}=\frac{200}{1000\sqrt{2}}=\frac{0.2}{\sqrt{2}}=\frac{0.2\sqrt{2}}{2}=0.1\sqrt{2}\mathrm{A}$

$I_0\approx 0.1\times 1.414=0.1414\mathrm{A}$

The RMS current $I_{rms}=\frac{I_0}{\sqrt{2}}=\frac{0.1\sqrt{2}}{\sqrt{2}}=0.1\mathrm{A}$.

So, option A is correct.

5. Calculate the phase angle $\varphi$:

$\tan \varphi =\frac{X_L-X_C}{R}=\frac{2000-1000}{1000}=\frac{1000}{1000}=1$

$\varphi =\arctan (1)=45^{\circ }$

Since $X_L>X_C$, the circuit is inductive, and the voltage leads the current.

So, option C is correct.

6. Calculate the power factor $\cos \varphi$:

$\cos \varphi =\cos (45^{\circ })=\frac{1}{\sqrt{2}}\approx 0.707$

So, option D is correct.

All options A, B, C, D are correct.