GATE 2023 IN – Question 46
For the circuit shown, $\omega=100\pi$ rad/s, $R_1=R_2=2.2$ Ω, and $L=7$ mH. The capacitance making input admittance $Y_{in}$ purely real is ______ mF (two decimal places).
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Correct answer: 1.40 to 1.50
Explanation
Write $Y_{in}=1/(R+jX_L)+1/(R-jX_C)$, with $X_L=\omega L$, $X_C=1/(\omega C)$. Zero imaginary part requires $X_C/(R^2+X_C^2)=X_L/(R^2+X_L^2)$. Factoring gives $X_C=X_L$ or $X_C=R^2/X_L$. Here $X_L=100\pi(0.007)\approx2.199$ Ω and R = 2.2 Ω. Both capacitance solutions round to **1.45 mF**.