An inductance L , a capacitor of `20mu`F and a resistor of `10Omega` are connected in series with an AC source of frequency 50 Hz . If the current is in phase with the voltage, then the inductance of the inductor is

Options

A. 2.00 H
B. 0.51 H
C. 1.5 H
D. 0.99 H

Explanation

To solve the problem step by step, we will analyze the given information and apply the relevant formulas for an LCR circuit where the current is in phase with the voltage.

### Step 1: Identify the given values
- Capacitance, \( C = 20 \, \mu F = 20 \times 10^{-6} \, F \)
- Resistance, \( R = 10 \, \Omega \)
- Frequency, \( f = 50 \, Hz \)

### Step 2: Calculate the angular frequency \( \omega \)
The angular frequency \( \omega \) is given by the formula:
\[ \omega = 2 \pi f \]
Substituting the value of \( f \):
\[ \omega = 2 \pi \times 50 = 100 \pi \, \text{rad/s} \approx 314 \, \text{rad/s} \]

### Step 3: Use the condition for current and voltage being in phase
In an LCR circuit, the current is in phase with the voltage when:
\[ \omega L = \frac{1}{\omega C} \]
This means that the inductive reactance \( X_L \) equals the capacitive reactance \( X_C \).

### Step 4: Rearranging the equation to find \( L \)
From the equation \( \omega L = \frac{1}{\omega C} \), we can rearrange it to find \( L \):
\[ L = \frac{1}{\omega^2 C} \]

### Step 5: Substitute the values of \( \omega \) and \( C \)
Now, substituting the values of \( \omega \) and \( C \):
\[ L = \frac{1}{(314)^2 \times (20 \times 10^{-6})} \]

### Step 6: Calculate \( L \)
Calculating \( (314)^2 \):
\[ (314)^2 = 98596 \]
Now substituting this value into the equation for \( L \):
\[ L = \frac{1}{98596 \times 20 \times 10^{-6}} = \frac{1}{1.9712} \approx 0.507 \, H \]

### Step 7: Round the value
Rounding \( 0.507 \, H \) gives approximately \( 0.51 \, H \).

### Conclusion
Thus, the inductance \( L \) of the inductor is approximately \( 0.51 \, H \).

### Final Answer
The inductance \( L \) is \( 0.51 \, H \).

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