Correct Option is (C)

$\begin{array}{rl}& {\mathrm{X}}_{\mathrm{C}}=\frac{1}{\omega \mathrm{C}}=\frac{1}{(2\pi \mathrm{f})\mathrm{C}}\\ \\ & \therefore {\mathrm{X}}_{\mathrm{C}}\propto \frac{1}{\mathrm{f}}\\ \\ & \therefore \text{ Curve }\mathrm{A}\\ \\ & {\mathrm{X}}_{\mathrm{L}}=\omega \mathrm{L}=(2\pi \mathrm{f})\mathrm{L}\\ \\ & \therefore {\mathrm{X}}_{\mathrm{L}}\propto \mathrm{f}\\ \\ & \therefore \text{ Curve }\mathrm{B}\end{array}$

Correct Option is (D)

For a capacitor connected to an AC source, the maximum current $I_{\text{max}}$ can be calculated using the formula:

$I_{\text{max}}=E_{\text{max}}\cdot \omega C$

where $E_{\text{max}}$ is the maximum voltage, $\omega$ is the angular frequency, and $C$ is the capacitance.

Given the emf equation: $E=36\sin (120\pi t)\text{V}$, we can determine that $E_{\text{max}}=36\text{V}$ and $\omega =120\pi \text{rad/s}$.

The capacitance is given as $150.0\mu \text{F}=150.0\times {10}^{-6}\text{F}$.

Now, we can calculate the maximum current:

$I_{\text{max}}=36\cdot (120\pi )\cdot (150.0\times {10}^{-6})$

$I_{\text{max}}\approx 2\text{A}$

Thus, the correct answer is $2\text{A}$.

Correct Option is (B)

$I_{rms}=\frac{V_{rms}}{z}=\frac{200\sqrt{2}}{\sqrt{{100}^2+{(200-100)}^2}}$

$=\frac{200\sqrt{2}}{100\sqrt{2}}$

= 2 A

Correct Option is (A)

$X_L=R$

$\Rightarrow P_1=\frac{R}{\sqrt{X_L^2+R^2}}=\frac{1}{\sqrt{2}}$

Now, $X_L=X_C=R$

$\Rightarrow P_2=\frac{R}{\sqrt{R^2+{(X_L-X_C)}^2}}=1$

$\Rightarrow \frac{P_1}{P_2}=\frac{1}{\sqrt{2}}$