Question

A 25 Ω resistor connected in series with a coil of 50 Ω resistance and 150 mH inductance. What is the power factor of the circuit?

Answer 1

To determine the power factor of the circuit consisting of a resistor and an inductor connected in series, we first need to find the total impedance of the circuit and then calculate the power factor.

1. **Given:**
- Resistance (R) = 25 Ω (resistor)
- Inductive Reactance (X_L) = ? (due to the coil)
- Resistance of the coil (r) = 50 Ω (but this is in series with the resistor)

2. **Total resistance (R_total)**:
\[
R_{\text{total}} = R + r = 25\, \Omega + 50\, \Omega = 75\, \Omega
\]

3. **Calculating the Inductive Reactance (X_L)**:
The inductive reactance \(X_L\) can be calculated using the formula:
\[
X_L = 2 \pi f L
\]
However, we need to know the frequency \(f\) to calculate it. For now, let's proceed theoretically.

4. **Total Impedance (Z)**:
Since the resistor and inductor are in series, the total impedance \(Z\) is calculated as:
\[
Z = \sqrt{R_{\text{total}}^2 + X_L^2}
\]

5. **Calculate Power Factor (PF)**:
The power factor (PF) is defined as the cosine of the phase angle \(\phi\) between the voltage and current, given by:
\[
PF = \cos(\phi) = \frac{R_{\text{total}}}{Z}
\]

6. **Substituting the Values**:
We can express the Power Factor as:
\[
PF = \frac{R_{\text{total}}}{\sqrt{R_{\text{total}}^2 + X_L^2}}
\]

However, without the frequency \(f\), we can't find \(X_L\) directly. For illustrative purposes, if we assume some frequency, say 50 Hz (which is common):

- Calculate \(X_L\) for \(L = 150 \text{ mH} = 0.150 \text{ H}\):
\[
X_L = 2 \pi (50\, \text{Hz}) (0.150\, \text{H}) = 2 \pi (50)(0.150) \approx 47.1\, \Omega
\]

7. **Calculating Impedance**:
\[
Z \approx \sqrt{(75)^2 + (47.1)^2}
\]
\[
Z \approx \sqrt{5625 + 2216.41} \approx \sqrt{7841.41} \approx 88.6\, \Omega
\]

8. **Final Power Factor Calculation**:
\[
PF = \frac{R_{\text{total}}}{Z} = \frac{75}{88.6} \approx 0.847
\]

So, the power factor of the circuit is approximately **0.847**, which indicates a lagging power factor due to the presence of the inductor.