In the circuit shown, \(\omega=100\pi\ \text{rad/s}\), \(R_1=R_2=2.2~\Omega\), and \(L=7~\text{mH}\). The capacitance \(C\) for which \(Y_{\text{in}}\) is purely real is _____ mF (rounded to two decimal places).
Show Hint
Make \(Y_{\text{in}}\) real by cancelling branch susceptances: for a series branch \(R\pm jX\), the susceptance is \(\mp X/(R^2+X^2)\).
Two parallel branches: \(R_1\!+\!j\omega L\) in series, and \(R_2\!-\!jX_C\) in series (with \(X_C=\frac{1}{\omega C}\)).
Admittance of a series branch \(R\pm jX\): \(Y=\dfrac{1}{R\pm jX}=\dfrac{R\mp jX}{R^2+X^2}\).
Hence susceptances (imaginary parts) are
\[
b_1=-\frac{\omega L}{R_1^2+(\omega L)^2},\qquad
b_2=+\frac{X_C}{R_2^2+X_C^2}.
\]
For \(Y_{\text{in}}\) to be real: \(b_1+b_2=0\Rightarrow
\frac{X_C}{R_2^2+X_C^2}=\frac{\omega L}{R_1^2+(\omega L)^2}.
\]
Compute \(\omega L=100\pi\times0.007=2.199~\Omega\).
With \(R_1=R_2=2.2~\Omega\),
\[
\frac{\omega L}{R_1^2+(\omega L)^2}=
\frac{2.199}{2.2^2+2.199^2}=\frac{2.199}{9.676}=0.2273.
\]
Solve \(\dfrac{X_C}{4.84+X_C^2}=0.2273\) \(\Rightarrow\) \(X_C=2.2~\Omega\).
Therefore
\[
C=\frac{1}{\omega X_C}=\frac{1}{(100\pi)(2.2)}=1.446\times10^{-3}\ \text{F}
=1.45\ \text{mF (to two decimals)}.
\]
Final Answer: 1.45 mF
