A resonant `AC` circuit contains a capacitor of capacitance `10^(-6)F` and an inductor of `10^(-4)H`. The frequency of electrical oscillation will be

Options

A. (a)`10^(5)`Hz
B. (b)10 Hz
C. (c)`(10^(5))/(2pi)`Hz
D. (d)`10/(2pi)Hz`

Answer

To find the frequency of electrical oscillation in a resonant AC circuit containing a capacitor and an inductor, we can follow these steps:

### Step 1: Write the formula for the resonant frequency
The resonant frequency \( f \) of an LC circuit is given by the formula:
\[ f = \frac{1}{2\pi\sqrt{LC}} \]
where \( L \) is the inductance and \( C \) is the capacitance.

### Step 2: Identify the values of \( L \) and \( C \)
From the problem, we have:
- Capacitance \( C = 10^{-6} \, \text{F} \)
- Inductance \( L = 10^{-4} \, \text{H} \)

### Step 3: Substitute the values into the formula
Now we substitute the values of \( L \) and \( C \) into the resonant frequency formula:
\[ f = \frac{1}{2\pi\sqrt{(10^{-4})(10^{-6})}} \]

### Step 4: Simplify the expression under the square root
Calculating the product under the square root:
\[ LC = 10^{-4} \times 10^{-6} = 10^{-10} \]
Thus,
\[ \sqrt{LC} = \sqrt{10^{-10}} = 10^{-5} \]

### Step 5: Substitute back into the frequency formula
Now substitute \( \sqrt{LC} \) back into the frequency formula:
\[ f = \frac{1}{2\pi(10^{-5})} \]

### Step 6: Simplify the frequency expression
This can be simplified to:
\[ f = \frac{10^5}{2\pi} \]

### Step 7: Calculate the numerical value
Using \( \pi \approx 3.14 \):
\[ f \approx \frac{10^5}{6.28} \approx 15915.5 \, \text{Hz} \]

### Conclusion
Thus, the frequency of electrical oscillation in the circuit is approximately:
\[ f \approx 15915.5 \, \text{Hz} \]