An a.c. source is connected across an LCR series circuit with `L = 100 mH, C = 0.1muF` and `R = 50 Omega`. The frequency of ac to make the power factor of the circuit, unity is

Options

A. `(10^(4))/(2pi)Hz`
B. `(10^(3))/(2pi)Hz`
C. `(10^(-4))/(2pi)Hz`
D. `(10^(-3))/(2pi)Hz`

Answer

To find the frequency of the AC source that makes the power factor of the LCR circuit unity, we can follow these steps:

### Step 1: Understand the Condition for Unity Power Factor
The power factor (PF) is defined as the cosine of the phase angle (Φ) between the voltage and current in the circuit. For the power factor to be unity (PF = 1), the circuit must be in resonance, which means the inductive reactance (XL) equals the capacitive reactance (XC).

### Step 2: Write the Formulas for Inductive and Capacitive Reactance
The inductive reactance (XL) is given by:
\[ X_L = 2\pi f L \]
The capacitive reactance (XC) is given by:
\[ X_C = \frac{1}{2\pi f C} \]

### Step 3: Set the Inductive and Capacitive Reactance Equal
At resonance:
\[ X_L = X_C \]
Substituting the formulas:
\[ 2\pi f L = \frac{1}{2\pi f C} \]

### Step 4: Rearrange the Equation
Cross-multiplying gives:
\[ (2\pi f)^2 = \frac{1}{LC} \]
Thus:
\[ f^2 = \frac{1}{4\pi^2 LC} \]

### Step 5: Solve for Frequency (f)
Taking the square root:
\[ f = \frac{1}{2\pi \sqrt{LC}} \]

### Step 6: Substitute the Given Values
Given:
- \( L = 100 \, \text{mH} = 100 \times 10^{-3} \, \text{H} = 0.1 \, \text{H} \)
- \( C = 0.1 \, \mu\text{F} = 0.1 \times 10^{-6} \, \text{F} = 0.1 \times 10^{-6} \, \text{F} \)

Now substituting these values into the frequency formula:
\[ f = \frac{1}{2\pi \sqrt{(0.1 \times 10^{-3})(0.1 \times 10^{-6})}} \]

### Step 7: Calculate the Value
Calculating the product:
\[ LC = (0.1 \times 10^{-3})(0.1 \times 10^{-6}) = 0.1 \times 10^{-9} = 10^{-10} \]
Then:
\[ \sqrt{LC} = \sqrt{10^{-10}} = 10^{-5} \]
Now substituting back into the frequency formula:
\[ f = \frac{1}{2\pi (10^{-5})} = \frac{10^5}{2\pi} \]

### Step 8: Final Calculation
Calculating the numerical value:
\[ f \approx \frac{10^5}{6.2832} \approx 15915.5 \, \text{Hz} \]

Thus, the frequency of the AC source to make the power factor of the circuit unity is approximately:
\[ f \approx 15915.5 \, \text{Hz} \]