Obtain the resonant frequency `(omega_(r))` of a series LCR circuit withL = 2.0 H, C = 32 `muF` and R = 10 ohm. What is the Q value of this circuit ?

Options

A. 15
B. 20
C. 25
D. 30

Explanation

To solve the problem, we need to find the resonant frequency \((\omega_r)\) and the Q factor of a series LCR circuit given the values of inductance \(L\), capacitance \(C\), and resistance \(R\).

### Step 1: Calculate the Resonant Frequency \((\omega_r)\)

The formula for the resonant frequency in a series LCR circuit is given by:

\[ \omega_r = \frac{1}{\sqrt{LC}} \]

Where:
- \(L = 2.0 \, \text{H}\) (inductance)
- \(C = 32 \, \mu\text{F} = 32 \times 10^{-6} \, \text{F}\) (capacitance)

Now, substituting the values into the formula:

\[ \omega_r = \frac{1}{\sqrt{2.0 \times 32 \times 10^{-6}}} \]

Calculating the product inside the square root:

\[ LC = 2.0 \times 32 \times 10^{-6} = 64 \times 10^{-6} = 6.4 \times 10^{-5} \]

Now, taking the square root:

\[ \sqrt{LC} = \sqrt{6.4 \times 10^{-5}} \approx 0.008 \]

Now, calculating \(\omega_r\):

\[ \omega_r = \frac{1}{0.008} = 125 \, \text{rad/s} \]

### Step 2: Calculate the Q Factor

The formula for the Q factor of a series LCR circuit is given by:

\[ Q = \frac{\omega_r L}{R} \]

Where:
- \(R = 10 \, \Omega\) (resistance)

Substituting the values into the formula:

\[ Q = \frac{125 \times 2.0}{10} \]

Calculating the numerator:

\[ 125 \times 2.0 = 250 \]

Now, dividing by \(R\):

\[ Q = \frac{250}{10} = 25 \]

### Final Results

- The resonant frequency \((\omega_r)\) is \(125 \, \text{rad/s}\).
- The Q factor of the circuit is \(25\).