An oscillator circuit consists of an inductance of `0.5mH` and a capacitor of `20muF`. The resonant frequency of the circuit is nearly

Options

A. `15.92`Hz
B. `159.2` Hz
C. 1592 Hz
D. 15910 Hz

Answer

To find the resonant frequency of the oscillator circuit consisting of an inductance of \(0.5 \, \text{mH}\) and a capacitor of \(20 \, \mu\text{F}\), we can use the formula for the resonant frequency \(f\) of an LC circuit:

\[ f = \frac{1}{2\pi\sqrt{LC}} \]

### Step-by-Step Solution:

**Step 1: Convert the units of inductance and capacitance.**
- Inductance \(L = 0.5 \, \text{mH} = 0.5 \times 10^{-3} \, \text{H}\)
- Capacitance \(C = 20 \, \mu\text{F} = 20 \times 10^{-6} \, \text{F}\)

**Step 2: Substitute the values into the formula.**
\[ f = \frac{1}{2\pi\sqrt{(0.5 \times 10^{-3})(20 \times 10^{-6})}} \]

**Step 3: Calculate the product \(LC\).**
\[ LC = (0.5 \times 10^{-3})(20 \times 10^{-6}) = 10 \times 10^{-9} = 10^{-8} \, \text{H}\cdot\text{F} \]

**Step 4: Calculate the square root of \(LC\).**
\[ \sqrt{LC} = \sqrt{10^{-8}} = 10^{-4} \, \text{s} \]

**Step 5: Substitute back into the frequency formula.**
\[ f = \frac{1}{2\pi(10^{-4})} \]

**Step 6: Calculate the frequency.**
\[ f = \frac{1}{2\pi \times 10^{-4}} \approx \frac{1}{6.2832 \times 10^{-4}} \approx 1591.55 \, \text{Hz} \]

**Step 7: Round to the nearest whole number.**
\[ f \approx 1592 \, \text{Hz} \]

### Final Answer:
The resonant frequency of the circuit is approximately \(1592 \, \text{Hz}\).

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