A `100Omega` resistance, a `0.1 muF` capacitor and an inductor are connected in series across a 250 V supply at variable frequency. Calculate the value of inductance of inductor at which resonance will occur. Given the resonant frequency is 60 Hz.

Options

A. `0.70` H
B. `70.3` mH
C. `7.03xx10^(-5)` H
D. `70.9` H

Answer

To find the value of inductance \( L \) at which resonance occurs in a series RLC circuit, we can use the relationship between the resonant frequency \( f \), the capacitance \( C \), and the inductance \( L \). The resonant frequency is given by the formula:

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

Given:
- \( C = 0.1 \, \mu F = 0.1 \times 10^{-6} \, F \)
- \( f = 60 \, Hz \)

We can rearrange the formula to find \( L \):

\[ L = \frac{1}{(2\pi f)^2 C} \]

Now, substituting the values into the equation:

1. Calculate \( 2\pi f \):
\[ 2\pi f = 2 \times 3.14 \times 60 \approx 376.99 \, rad/s \]

2. Now square this value:
\[ (2\pi f)^2 \approx (376.99)^2 \approx 142,000 \, rad^2/s^2 \]

3. Now substitute this value and \( C \) into the formula for \( L \):
\[ L = \frac{1}{142,000 \times 0.1 \times 10^{-6}} = \frac{1}{0.0000142} \approx 70.4 \, H \]

Thus, the value of inductance \( L \) at which resonance occurs is approximately \( 70.4 \, H \).

### Summary of Steps:
1. Use the resonant frequency formula to express \( L \).
2. Substitute the given values of \( f \) and \( C \).
3. Calculate \( 2\pi f \) and its square.
4. Substitute back to find \( L \).