Question

In the given circuit, the resonant frequency is

• A. 15.92 Hz
• B. 159.2 Hz
• C. 1592 Hz
• D. 15910 Hz

Answer 1

The Correct Option is C

The resonant frequency \( f_0 \) for an LC circuit is given by the formula: \[ f_0 = \frac{1}{2 \pi \sqrt{LC}} \] Where:
\( L \) is the inductance (0.5 mH = \( 0.5 \times 10^{-3} \, \text{H} \)),
\( C \) is the capacitance (20 µF = \( 20 \times 10^{-6} \, \text{F} \)).
Substituting the values into the formula: \[ f_0 = \frac{1}{2 \pi \sqrt{(0.5 \times 10^{-3}) \times (20 \times 10^{-6})}} = 1592 \, \text{Hz} \] Thus, the resonant frequency is 1592 Hz.

Answer 2

Resonant frequency of an circuit is given by
$f=\frac{1}{2 \pi \sqrt{L C}} $
$=\frac{1}{2 \pi \sqrt{0.5 \times 10^{-3} H \times 20 \times 10^{-6} F}} $
$=1592\, Hz$

Answer 3

The resonant frequency \( f_0 \) of an L-C circuit is given by the formula: \[ f_0 = \frac{1}{2 \pi \sqrt{LC}} \] where: - \( L \) is the inductance, - \( C \) is the capacitance. Given:
\( L = 0.5 \, \text{mH} = 0.5 \times 10^{-3} \, \text{H} \),
\( C = 20 \, \mu\text{F} = 20 \times 10^{-6} \, \text{F} \).

Now, substitute these values into the formula for the resonant frequency: \[ f_0 = \frac{1}{2 \pi \sqrt{(0.5 \times 10^{-3}) \times (20 \times 10^{-6})}} \] \[ f_0 = \frac{1}{2 \pi \sqrt{(10 \times 10^{-9})}} = \frac{1}{2 \pi \times (3.162 \times 10^{-5})} \] \[ f_0 \approx \frac{1}{1.986 \times 10^{-4}} = 1592 \, \text{Hz} \] Thus, the resonant frequency is approximately \( 1592 \, \text{Hz} \).

Therefore, the correct answer is (C).