A condenser of `250 mu F` is connected in parallel to a coil of inductance `0.16 mH` while its effective resistance is `20 Omega`. Determine the resonant frequency

Options

A. `9 xx 10^(4) Hz`
B. `16 xx 10^(7) Hz`
C. `8 xx 10^(5) Hz`
D. `9 xx 10^(3) Hz`

Explanation

To find the resonant frequency of the circuit consisting of a capacitor and an inductor connected in parallel, we can use the formula for the resonant frequency in an RLC circuit:

\[ f_0 = \frac{1}{2\pi} \sqrt{\frac{1}{LC} - \frac{R^2}{L^2}} \]

Where:
- \( f_0 \) is the resonant frequency,
- \( L \) is the inductance,
- \( C \) is the capacitance,
- \( R \) is the resistance.

### Step-by-Step Solution:

**Step 1: Convert the given values to standard units.**

- Capacitance \( C = 250 \mu F = 250 \times 10^{-6} F \)
- Inductance \( L = 0.16 mH = 0.16 \times 10^{-3} H \)
- Resistance \( R = 20 \Omega \)

**Step 2: Substitute the values into the resonant frequency formula.**

\[ f_0 = \frac{1}{2\pi} \sqrt{\frac{1}{L C} - \frac{R^2}{L^2}} \]

Substituting the values:

\[ f_0 = \frac{1}{2\pi} \sqrt{\frac{1}{(0.16 \times 10^{-3})(250 \times 10^{-6})} - \frac{(20)^2}{(0.16 \times 10^{-3})^2}} \]

**Step 3: Calculate \( LC \) and \( R^2/L^2 \).**

- First calculate \( LC \):

\[ LC = (0.16 \times 10^{-3})(250 \times 10^{-6}) = 0.16 \times 250 \times 10^{-9} = 40 \times 10^{-9} = 4 \times 10^{-8} \, H \cdot F \]

- Now calculate \( R^2/L^2 \):

\[ R^2 = 20^2 = 400 \]
\[ L^2 = (0.16 \times 10^{-3})^2 = 0.0256 \times 10^{-6} = 2.56 \times 10^{-8} \]
\[ \frac{R^2}{L^2} = \frac{400}{2.56 \times 10^{-8}} = 1.5625 \times 10^{10} \]

**Step 4: Substitute these values back into the formula.**

Now substituting back into the formula:

\[ f_0 = \frac{1}{2\pi} \sqrt{\frac{1}{4 \times 10^{-8}} - 1.5625 \times 10^{10}} \]

**Step 5: Calculate the expression inside the square root.**

Calculating \( \frac{1}{4 \times 10^{-8}} \):

\[ \frac{1}{4 \times 10^{-8}} = 2.5 \times 10^{7} \]

Now substituting:

\[ f_0 = \frac{1}{2\pi} \sqrt{2.5 \times 10^{7} - 1.5625 \times 10^{10}} \]

**Step 6: Final calculation for \( f_0 \).**

Calculating the difference:

\[ 2.5 \times 10^{7} - 1.5625 \times 10^{10} = -1.5375 \times 10^{10} \]

Since the term under the square root is negative, it indicates that the circuit does not resonate under these conditions, which means the resonant frequency cannot be calculated in the conventional sense.

### Final Result:
The resonant frequency cannot be determined as the calculated value under the square root is negative.