In the given circuit, $L=1\mathrm{\mu H}$ and $C=1\mathrm{\mu F}$. The phasor diagram for $I_C$ and $I_L$ is also shown. Assume that the phase $({\theta }_1+{\theta }_2)$ is $90°$ at a frequency of $159.15\mathrm{kHz}$. Among the following options, what is the nearest integer value of $R_C\times R_L$ ?

The correct option is 1.

Analysis of the Circuit and Phasor Diagram:
The given circuit diagram shows an AC voltage source $V$ connected in parallel with two branches:
1. A capacitive branch consisting of a resistor $R_C$ in series with a capacitor $C$. The current through this branch is denoted as $I_C$.
2. An inductive branch consisting of a resistor $R_L$ in series with an inductor $L$. The current through this branch is denoted as $I_L$.

From the phasor diagram, the reference phasor is the voltage $V$.
The capacitive current $I_C$ leads the voltage $V$ by a phase angle ${\theta }_1$.
The inductive current $I_L$ lags the voltage $V$ by a phase angle ${\theta }_2$.

Formulating the Phase Angle Relations:
For the capacitive branch, the phase angle ${\theta }_1$ is given by:
$\tan {\theta }_1=\frac{X_C}{R_C}=\frac{1}{\omega C{R}_C}$

For the inductive branch, the phase angle ${\theta }_2$ is given by:
$\tan {\theta }_2=\frac{X_L}{R_L}=\frac{\omega L}{R_L}$

Applying the Phase Condition:
We are given that the sum of the angles is:
${\theta }_1+{\theta }_2=90°$

Taking the tangent on both sides:
$\tan ({\theta }_1+{\theta }_2)=\tan (90°)=\infty$

Using the trigonometric identity for the tangent of a sum:
$\frac{\tan {\theta }_1+\tan {\theta }_2}{1-\tan {\theta }_1\tan {\theta }_2}=\infty$

For the fraction to become infinite, the denominator must equal zero:
$1-\tan {\theta }_1\tan {\theta }_2=0\Rightarrow \tan {\theta }_1\tan {\theta }_2=1$

Substituting the values of the tangents:
$\left(\frac{1}{\omega C{R}_C}\right)\times \left(\frac{\omega L}{R_L}\right)=1$

Simplifying the expression by canceling $\omega$ from the numerator and denominator:
$\frac{L}{C{R}_C{R}_L}=1\Rightarrow R_C\times R_L=\frac{L}{C}$

Calculating the Product:
Given component values:
$L=1\mathrm{\mu H}=1\times {10}^{-6}H$
$C=1\mathrm{\mu F}=1\times {10}^{-6}F$

Substituting these values:
$R_C\times R_L=\frac{1\times {10}^{-6}}{1\times {10}^{-6}}=1$

Thus, the nearest integer value of $R_C\times R_L$ is 1.