A resistor of resistance 100 `Omega` is connected in series with an inductor of self-inductance `sqrt(3)`. H. this combination is connected to an AC source rated as 220 V-50 /`pi` Hz. Power factor of the circuit is found to be 1/n. What is the value of n ?

Explanation

To solve the problem, we need to find the value of \( n \) given the power factor of the circuit. Let's go through the steps systematically.

### Step 1: Identify the given values
- Resistance \( R = 100 \, \Omega \)
- Inductance \( L = \sqrt{3} \, \text{H} \)
- AC source voltage \( V = 220 \, \text{V} \)
- Frequency \( f = \frac{50}{\pi} \, \text{Hz} \)

### Step 2: Calculate the inductive reactance \( X_L \)
The inductive reactance \( X_L \) is given by the formula:
\[ X_L = 2\pi f L \]
Substituting the values:
\[ X_L = 2\pi \left(\frac{50}{\pi}\right) \sqrt{3} = 100 \sqrt{3} \, \Omega \]

### Step 3: Calculate the impedance \( Z \)
The total impedance \( Z \) in a series circuit with resistance \( R \) and inductive reactance \( X_L \) is given by:
\[ Z = \sqrt{R^2 + X_L^2} \]
Substituting the values:
\[ Z = \sqrt{(100)^2 + (100\sqrt{3})^2} = \sqrt{10000 + 30000} = \sqrt{40000} = 200 \, \Omega \]

### Step 4: Calculate the angle \( \phi \)
The power factor \( \cos \phi \) is defined as:
\[ \cos \phi = \frac{R}{Z} \]
Substituting the values:
\[ \cos \phi = \frac{100}{200} = \frac{1}{2} \]

### Step 5: Relate the power factor to \( n \)
According to the problem, the power factor is given as \( \frac{1}{n} \). We have found that:
\[ \cos \phi = \frac{1}{2} \]
Thus, we can set up the equation:
\[ \frac{1}{n} = \frac{1}{2} \]

### Step 6: Solve for \( n \)
From the equation above, we can find \( n \):
\[ n = 2 \]

### Final Answer
The value of \( n \) is \( 2 \).

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