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: Read off the two branches.
Looking at the network, current $I_C$ flows through $R_C$ in series with $C$, and current $I_L$ flows through $R_L$ in series with $L$, with both branches sitting across the same source $V$.
Step 2: Write each current's phase relative to V.
For the capacitive branch, the impedance is $R_C-\dfrac{j}{\omega C}$, a negative imaginary part, so $I_C$ sits ahead of $V$ by angle $\theta_1$, with
\[ \tan\theta_1=\frac{1/(\omega C)}{R_C} \]
For the inductive branch, the impedance is $R_L+j\omega L$, a positive imaginary part, so $I_L$ sits behind $V$ by angle $\theta_2$, with
\[ \tan\theta_2=\frac{\omega L}{R_L} \]
Step 3: Use the 90 degree fact as a product rule.
Two positive angles that add to $90^\circ$ are complementary, and for complementary angles $\tan\theta_1\tan\theta_2=1$. So
\[ \left(\frac{1}{\omega C R_C}\right)\left(\frac{\omega L}{R_L}\right)=1 \]
Step 4: Cancel omega and rearrange.
\[ \frac{L}{C\,R_C R_L}=1\quad\Rightarrow\quad R_C R_L=\frac{L}{C} \]
This is frequency independent, so the exact value $159.15$ kHz given in the problem is only there to assure us that such a frequency exists, not to be plugged into the formula.
Step 5: Put in L and C.
\[ R_C R_L=\frac{1\ \mu\text{H}}{1\ \mu\text{F}}=\frac{1\times10^{-6}}{1\times10^{-6}}=1 \]
\[ \boxed{1} \]
