A resistance of 40 `Omega` is connected in series with inductor of self-inductance 5 H and a capacitor of capacitance 80 `mu`F. This combination is connected to an AC source of rms voltage 220 V. frequency of AC source can changed continuously.
What should be the frequency of source which drives circuit to resonance ?

Options

A. `(100)/(pi)`
B. `(75)/(pi)`
C. `(50)/(pi)`
D. `(25)/(pi)`

Explanation

To find the frequency of the AC source that drives the circuit to resonance, we need to follow these steps:

### Step 1: Understand the Resonance Condition
In an RLC series circuit, resonance occurs when the inductive reactance (XL) equals the capacitive reactance (XC). This condition can be expressed mathematically as:
\[ X_L = X_C \]
Where:
- \( X_L = \omega L \) (Inductive reactance)
- \( X_C = \frac{1}{\omega C} \) (Capacitive reactance)

### Step 2: Set Up the Equation for Resonance
Setting the two reactances equal gives us:
\[ \omega L = \frac{1}{\omega C} \]
Rearranging this equation leads to:
\[ \omega^2 = \frac{1}{LC} \]
Thus, the resonant frequency \( \omega \) can be expressed as:
\[ \omega = \frac{1}{\sqrt{LC}} \]

### Step 3: Substitute the Given Values
We have:
- Inductance \( L = 5 \, H \)
- Capacitance \( C = 80 \, \mu F = 80 \times 10^{-6} \, F \)

Now substituting these values into the equation:
\[ \omega = \frac{1}{\sqrt{5 \times 80 \times 10^{-6}}} \]

### Step 4: Calculate the Resonant Frequency
Calculating the product inside the square root:
\[ LC = 5 \times 80 \times 10^{-6} = 400 \times 10^{-6} = 4 \times 10^{-4} \]
Now taking the square root:
\[ \sqrt{LC} = \sqrt{4 \times 10^{-4}} = 2 \times 10^{-2} \]
Thus:
\[ \omega = \frac{1}{2 \times 10^{-2}} = 50 \, \text{rad/s} \]

### Step 5: Convert Angular Frequency to Frequency in Hertz
To convert from angular frequency (rad/s) to frequency (Hz), we use:
\[ f = \frac{\omega}{2\pi} \]
Substituting the value of \( \omega \):
\[ f = \frac{50}{2\pi} \]
Calculating this gives:
\[ f \approx \frac{50}{6.2832} \approx 7.96 \, \text{Hz} \]

### Final Answer
The frequency of the source which drives the circuit to resonance is approximately:
\[ f \approx 7.96 \, \text{Hz} \]