Question 42: The net impedance of circuit (as shown in figure) will be :
(1) 5 \sqrt{5} \Omega
(2) 25 \Omega
(3) 10 \sqrt{2} \Omega
(4) 15 \Omega
Answer: Option (1)
Explanation:
The circuit is a series RLC circuit with resistance R=10 \Omega,
inductance L=\frac{50}{\pi} \mathrm{~mH},
and capacitance C=\frac{10^{3}}{\pi} \mu \mathrm{F}.
The supply frequency is f=50 \mathrm{~Hz}.
Hence, angular frequency is \omega=2\pi f = 100\pi \mathrm{~rad\,s^{-1}}.
Inductive reactance is given by X_L=\omega L. Substituting values,
X_L = 100\pi \times \frac{50}{\pi} \times 10^{-3} = 5 \Omega.
Capacitive reactance is given by X_C=\frac{1}{\omega C}.
Substituting values,
X_C = \frac{1}{100\pi \times \frac{10^{3}}{\pi} \times 10^{-6}} = 10 \Omega.
The net reactance of the circuit is
X = X_L - X_C = 5 - 10 = -5 \Omega.
The magnitude of impedance of a series RLC circuit is
Z = \sqrt{R^{2} + X^{2}}.
Substituting values,
Z = \sqrt{10^{2} + 5^{2}} = \sqrt{125}.
Thus,
Z = 5\sqrt{5} \Omega.
Hence, the correct answer is Option (1).