In the given circuit, \(L=1\ \mu H\) and \(C=1\ \mu F\). The phasor diagram for \(I_C\) and \(I_L\) is also shown. Assume that the phase \((\theta_1+\theta_2)\) is \(90^{\circ}\) at a frequency of \(159.15\) kHz.
Among the following options, what is the nearest integer value of \(R_C\times R_L\)?
Show Hint
Write tan(theta1) and tan(theta2) for the two branch currents, use theta1+theta2=90 degrees, and simplify: the result is RC times RL = L/C.
Step 1: Identify the two parallel branches from the figure.
The source \(V\) drives two parallel branches. One branch has a resistor \(R_C\) in series with the capacitor \(C\), carrying current \(I_C\). The other branch has a resistor \(R_L\) in series with the inductor \(L\), carrying current \(I_L\).
Step 2: Write the branch impedances.
For the RC branch,
\[
Z_C=R_C+\frac{1}{j\omega C}=R_C-\frac{j}{\omega C}
\]
For the RL branch,
\[
Z_L=R_L+j\omega L
\]
Step 3: Express the phase angles of the two branch currents.
Since \(I_C=\frac{V}{Z_C}\) and the phase of \(V\) is taken as the reference (\(0^{\circ}\)), the phase of \(I_C\) is the negative of the phase of \(Z_C\). Because \(Z_C\) has a negative imaginary part, \(I_C\) leads \(V\) by an angle \(\theta_1\) where
\[
\tan\theta_1=\frac{1}{\omega C R_C}
\]
Similarly, since \(Z_L\) has a positive imaginary part, \(I_L\) lags \(V\) by an angle \(\theta_2\) where
\[
\tan\theta_2=\frac{\omega L}{R_L}
\]
This matches the figure, where \(I_C\) is drawn above \(V\) at angle \(\theta_1\) and \(I_L\) is drawn below \(V\) at angle \(\theta_2\).
Step 4: Use the given condition on the angles.
We are given \(\theta_1+\theta_2=90^{\circ}\), so \(\theta_2=90^{\circ}-\theta_1\), which means
\[
\tan\theta_2=\tan\left(90^{\circ}-\theta_1\right)=\cot\theta_1=\frac{1}{\tan\theta_1}
\]
Step 5: Substitute the two tangent expressions.
\[
\frac{\omega L}{R_L}=\frac{1}{\frac{1}{\omega C R_C}}=\omega C R_C
\]
Step 6: Simplify to find the product R_C R_L.
Cancel \(\omega\) from both sides:
\[
\frac{L}{R_L}=C R_C\quad\Rightarrow\quad L=C R_C R_L\quad\Rightarrow\quad R_C R_L=\frac{L}{C}
\]
Notice that this relation does not depend on the operating frequency at all, so the value of \(159.15\) kHz given in the question is not needed for this particular calculation; it is only there to confirm that the \(90^{\circ}\) condition genuinely occurs at some real frequency.
Step 7: Substitute the given values of L and C.
\[
R_C R_L=\frac{L}{C}=\frac{1\times10^{-6}}{1\times10^{-6}}=1
\]
Step 8: Analyze the options.
(A) 0: Would mean either \(R_C\) or \(R_L\) is zero, which is not implied by the given condition. Incorrect.
(B) 1: Matches the value computed directly from \(R_CR_L=L/C\). Correct.
(C) 2: Does not match the computed ratio \(L/C=1\). Incorrect.
(D) 10: Far larger than the computed value of \(1\). Incorrect.
Step 9: Final conclusion.
The nearest integer value of \(R_C\times R_L\) is
\[
\boxed{1}
\]
