A resistor of `500 Omega` and an inductance of 0.5 H are in series with an ac source which is given by `V=100sqrt(2)sin(1000t)`. The power factor of the combination is

Options

A. `(1)/(sqrt(2))`
B. `(1)/(sqrt(3))`
C. `0.5`
D. `0.6`

Answer

To find the power factor of the given circuit with a resistor and an inductor in series, we can follow these steps:

### Step 1: Identify Given Values
- Resistance, \( R = 500 \, \Omega \)
- Inductance, \( L = 0.5 \, H \)
- Voltage source, \( V = 100\sqrt{2} \sin(1000t) \)

### Step 2: Determine Angular Frequency
From the voltage equation, we can identify the angular frequency \( \omega \):
- \( \omega = 1000 \, \text{rad/s} \)

### Step 3: Calculate Inductive Reactance
The inductive reactance \( X_L \) is given by the formula:
\[ X_L = \omega L \]
Substituting the values:
\[ X_L = 1000 \times 0.5 = 500 \, \Omega \]

### Step 4: Calculate Impedance
The total impedance \( Z \) in a series circuit with resistance and inductive reactance is given by:
\[ Z = \sqrt{R^2 + X_L^2} \]
Substituting the values:
\[ Z = \sqrt{(500)^2 + (500)^2} = \sqrt{250000 + 250000} = \sqrt{500000} = 707.1 \, \Omega \]

### Step 5: Calculate Power Factor
The power factor \( \cos \phi \) is defined as:
\[ \cos \phi = \frac{R}{Z} \]
Substituting the values:
\[ \cos \phi = \frac{500}{707.1} \approx 0.707 \]

### Step 6: Conclusion
The power factor of the combination is:
\[ \cos \phi \approx 0.707 \quad \text{or} \quad \frac{1}{\sqrt{2}} \]