Q 12-07-008NEETNEET 2023Top questionMedium
The net impedance of circuit (as shown in figure) will be :
Answer: (C) $5\sqrt{5}\ \Omega$
Concept. $Z = \sqrt{R^2 + (X_C - X_L)^2}$ with $\omega = 2\pi f$.
Step 1. Reactances. $\omega = 100\pi$:
$$X_L = 100\pi \times \frac{50}{\pi} \times 10^{-3} = 5\ \Omega,\qquad X_C = \frac{1}{100\pi \times \frac{10^3}{\pi} \times 10^{-6}} = 10\ \Omega$$
Step 2. Impedance. $Z = \sqrt{10^2 + 5^2} = 5\sqrt5\ \Omega$.
Answer. $5\sqrt5\ \Omega$ (Option C).
Solution by Sreeraj P, M.Sc Physics