A coil having an inductance of 50 mH and a resistance of `10 Omega` is connected in series with a `25 muF` capacitor across a 200 V supply. What is the Q factor of the circuit at resonance?

Options

A. 3.5
B. 4.47
C. 5.5
D. 7

Explanation

To solve the problem, we need to find the Q factor of the circuit at resonance. The Q factor (Quality factor) for a series RLC circuit can be calculated using the formula:

\[ Q = \frac{1}{R} \sqrt{\frac{L}{C}} \]

Where:
- \( Q \) is the quality factor,
- \( R \) is the resistance,
- \( L \) is the inductance,
- \( C \) is the capacitance.

### Step 1: Identify the given values
- Inductance \( L = 50 \, \text{mH} = 50 \times 10^{-3} \, \text{H} \)
- Resistance \( R = 10 \, \Omega \)
- Capacitance \( C = 25 \, \mu\text{F} = 25 \times 10^{-6} \, \text{F} \)

### Step 2: Substitute the values into the Q factor formula
Using the formula for Q factor:
\[ Q = \frac{1}{R} \sqrt{\frac{L}{C}} \]

Substituting the values:
\[ Q = \frac{1}{10} \sqrt{\frac{50 \times 10^{-3}}{25 \times 10^{-6}}} \]

### Step 3: Calculate \( \frac{L}{C} \)
First, we calculate \( \frac{L}{C} \):
\[ \frac{L}{C} = \frac{50 \times 10^{-3}}{25 \times 10^{-6}} = \frac{50}{25} \times 10^{3} = 2 \times 10^{3} = 2000 \]

### Step 4: Calculate the square root
Now, we take the square root:
\[ \sqrt{2000} = \sqrt{2000} = \sqrt{200 \times 10} = \sqrt{200} \times \sqrt{10} \]

Calculating \( \sqrt{200} \):
\[ \sqrt{200} = \sqrt{100 \times 2} = 10\sqrt{2} \approx 14.14 \]
And \( \sqrt{10} \approx 3.16 \).

Thus:
\[ \sqrt{2000} \approx 14.14 \times 3.16 \approx 44.72 \]

### Step 5: Substitute back to find Q
Now substituting back into the Q formula:
\[ Q = \frac{1}{10} \times 44.72 \]
\[ Q \approx 4.472 \]

### Final Answer
The Q factor of the circuit at resonance is approximately:
\[ Q \approx 4.47 \]