Question

Determine the power factor of an RLC series circuit with given values.

Answer 1

When a wire is stretched to double its original length, its resistivity \( R \) (which is a material property and does not change) remains the same, but its resistance changes. Here’s how you can determine the new resistance:

### 1. Understanding the Relationship

Resistance \( R \) of a wire is given by:
\[ R = \rho \frac{L}{A} \]
where:
- \( \rho \) is the resistivity of the material.
- \( L \) is the length of the wire.
- \( A \) is the cross-sectional area of the wire.

### 2. Stretching the Wire

When the wire is stretched to double its length, the new length \( L' \) is:
\[ L' = 2L \]

The volume of the wire remains constant during stretching. So, the initial volume \( V \) is:
\[ V = L \cdot A \]

After stretching, the volume \( V' \) remains:
\[ V' = L' \cdot A' \]
where \( A' \) is the new cross-sectional area.

Since \( V = V' \), we have:
\[ L \cdot A = 2L \cdot A' \]
\[ A' = \frac{A}{2} \]

### 3. New Resistance Calculation

The new resistance \( R' \) after stretching is:
\[ R' = \rho \frac{L'}{A'} \]
Substitute \( L' = 2L \) and \( A' = \frac{A}{2} \):
\[ R' = \rho \frac{2L}{\frac{A}{2}} \]
\[ R' = \rho \frac{2L \cdot 2}{A} \]
\[ R' = 4 \rho \frac{L}{A} \]
\[ R' = 4R \]

### Conclusion

When a wire is stretched to double its length, its resistance increases by a factor of 4.

Answer 2

To determine the power factor of an RLC series circuit, you need to know the values of the resistor (R), inductor (L), and capacitor (C), as well as the frequency (f) of the AC source. Here's a step-by-step method to calculate the power factor:

### Step 1: Calculate Reactances

1. **Inductive Reactance (X_L)**:
   \[
   X_L = 2 \pi f L
   \]
   where \( f \) is the frequency in hertz (Hz) and \( L \) is the inductance in henrys (H).

2. **Capacitive Reactance (X_C)**:
   \[
   X_C = \frac{1}{2 \pi f C}
   \]
   where \( C \) is the capacitance in farads (F).

### Step 2: Calculate Impedance (Z)

The impedance \( Z \) of the series RLC circuit is given by:
   \[
   Z = \sqrt{R^2 + (X_L - X_C)^2}
   \]

Here, \( X_L - X_C \) is the net reactance of the circuit.

### Step 3: Calculate the Phase Angle (θ)

The phase angle \( \theta \) between the voltage and the current is:
   \[
   \theta = \arctan\left(\frac{X_L - X_C}{R}\right)
   \]

### Step 4: Calculate the Power Factor (PF)

The power factor is the cosine of the phase angle \( \theta \):
   \[
   \text{PF} = \cos(\theta)
   \]

### Example Calculation

Let's go through an example with specific values:

- Resistor \( R = 10 \, \Omega \)
- Inductor \( L = 0.1 \, \text{H} \)
- Capacitor \( C = 100 \, \mu\text{F} = 100 \times 10^{-6} \, \text{F} \)
- Frequency \( f = 50 \, \text{Hz} \)

1. **Calculate Reactances**:
   - \( X_L = 2 \pi \times 50 \times 0.1 = 31.4 \, \Omega \)
   - \( X_C = \frac{1}{2 \pi \times 50 \times 100 \times 10^{-6}} = 31.4 \, \Omega \)

2. **Calculate Impedance**:
   \[
   Z = \sqrt{10^2 + (31.4 - 31.4)^2} = \sqrt{10^2} = 10 \, \Omega
   \]

3. **Calculate Phase Angle**:
   \[
   \theta = \arctan\left(\frac{31.4 - 31.4}{10}\right) = \arctan(0) = 0^\circ
   \]

4. **Calculate Power Factor**:
   \[
   \text{PF} = \cos(0^\circ) = 1
   \]

### Conclusion

In this example, the power factor is 1, indicating that the circuit is perfectly in phase with the AC source and there is no reactive power. The actual power factor can vary based on the values of \( R \), \( L \), \( C \), and \( f \). If \( X_L \neq X_C \), the phase angle \( \theta \) will not be zero, and the power factor will be less than 1.