Question 439: AC fundamentals

Question

The net impedance of circuit (as shown in figure) will be :

Accepted Answer

To find the net impedance, Z, of the circuit, we need to calculate the inductive reactance (XL), the capacitive reactance (XC), and use the resistance (R) in the circuit. Since R is given as 10 $ \Omega $, we calculate the reactances as follows:
Inductive reactance, $ X_L = 2\pi fL $, where $ f $ is the frequency and $ L $ is the inductance. With $ f = 50 $ Hz and $ L $ given as $ \frac{50}{\pi} $ mH, or $ \frac{50 	imes 10^{-3}}{\pi} $ H, we find $ X_L = 2 \pi 	imes 50 	imes \frac{50 	imes 10^{-3}}{\pi} = 5 $ $ \Omega $.
Capacitive reactance, $ X_C = \frac{1}{2\pi fC} $, where $ C $ is the capacitance. Given $ C = \frac{10^3}{\pi} $ ÃÂ¼F, or $ \frac{10^{-3}}{\pi} $ F, results in $ X_C = \frac{1}{2 \pi 	imes 50 	imes \frac{10^{-3}}{\pi}} = 10 \Omega $.
With these values, the net impedance $ Z $ of the circuit can be calculated using the formula:
$ Z = \sqrt{R^2 + (X_L - X_C)^2} $
Substituting $ R = 10 \Omega $, $ X_L = 5 \Omega $, and $ X_C = 10 \Omega $, we find:
$ Z = \sqrt{10^2 + (5 - 10)^2} = \sqrt{100 + 25} = \sqrt{125} = 5\sqrt{5} \Omega $
Therefore, the correct option is $ 5\sqrt{5} \Omega $.

Answer

$$15 \Omega$$

Answer

$$5 \sqrt{5} \Omega$$

Answer

$$25 \Omega$$

Answer

$$10 \sqrt{2} \Omega$$

AC fundamentals

The net impedance of circuit (as shown in figure) will be :

Choose the correct answer: