Q 12-07-008NEETNEET 2023Top questionMedium
The net impedance of circuit (as shown in figure) will be :
Answer: (C) $5\sqrt{5}\ \Omega$
$\omega = 2\pi \times 50 = 100\pi$ rad/s.
$$X_L = \omega L = 100\pi \times \frac{50}{\pi} \times 10^{-3} = 5\ \Omega$$
$$X_C = \frac{1}{\omega C} = \frac{1}{100\pi \times \frac{10^3}{\pi} \times 10^{-6}} = \frac{1}{0.1} = 10\ \Omega$$
$$Z = \sqrt{R^2 + (X_C - X_L)^2} = \sqrt{10^2 + 5^2} = \sqrt{125} = 5\sqrt{5}\ \Omega$$
Solution by Sreeraj P, M.Sc Physics