If a circuit load impedance is (25 - j25), find the power factor.
Impedance and Power Factor Calculation
The question asks us to determine the power factor of a circuit given its load impedance. Understanding the components of impedance and how they relate to the power factor is crucial in AC circuit analysis.
Understanding Circuit Load Impedance
The circuit load impedance, denoted as \(Z\), is a measure of the opposition a circuit presents to a current when a voltage is applied. In AC circuits, impedance is a complex quantity comprising two parts:
- Resistance (R): The real part, which dissipates energy.
- Reactance (X): The imaginary part, which stores and releases energy (either inductive or capacitive).
The given load impedance is \(Z = (25 - j25) \, \Omega\).
From this complex impedance, we can identify:
- Resistance, \(R = 25 \, \Omega\)
- Reactance, \(X = -25 \, \Omega\) (The negative sign indicates capacitive reactance)
Calculating the Power Factor
The power factor (PF) is a dimensionless quantity in AC circuits, ranging from 0 to 1, that represents the ratio of the real power flowing to the load to the apparent power in the circuit. It is a measure of how effectively electrical power is being converted into useful work.
The power factor can be calculated using the phase angle (\(\phi\)) of the impedance, or directly from the resistance and magnitude of the impedance.
Method 1: Using the Phase Angle (\(\phi\))
The phase angle \(\phi\) of the impedance \(Z = R + jX\) is given by:
\[ \phi = \arctan\left(\frac{X}{R}\right) \]Given \(R = 25 \, \Omega\) and \(X = -25 \, \Omega\):
\[ \phi = \arctan\left(\frac{-25}{25}\right) = \arctan(-1) \]The angle whose tangent is -1 is \(-45^\circ\).
\[ \phi = -45^\circ \]The power factor is then given by the cosine of this phase angle:
\[ \text{Power Factor (PF)} = \cos(\phi) = \cos(-45^\circ) \]Since \(\cos(-\theta) = \cos(\theta)\):
\[ \text{PF} = \cos(45^\circ) = \frac{1}{\sqrt{2}} \]As a decimal value:
\[ \text{PF} \approx 0.707 \]Method 2: Using Resistance and Impedance Magnitude
The power factor can also be calculated as the ratio of the resistance to the magnitude of the impedance:
\[ \text{Power Factor (PF)} = \frac{R}{|Z|} \]First, let's calculate the magnitude of the impedance \(|Z|\):
\[ |Z| = \sqrt{R^2 + X^2} \]Given \(R = 25 \, \Omega\) and \(X = -25 \, \Omega\):
\[ |Z| = \sqrt{(25)^2 + (-25)^2} = \sqrt{625 + 625} = \sqrt{1250} \]To simplify \(\sqrt{1250}\):
\[ \sqrt{1250} = \sqrt{625 \times 2} = \sqrt{625} \times \sqrt{2} = 25\sqrt{2} \, \Omega \]Now, substitute the values of \(R\) and \(|Z|\) into the power factor formula:
\[ \text{PF} = \frac{25}{25\sqrt{2}} = \frac{1}{\sqrt{2}} \]As a decimal value:
\[ \text{PF} \approx 0.707 \]Summary of Calculation
Both methods yield the same result. The power factor for the given circuit load impedance of \((25 - j25) \, \Omega\) is approximately 0.707.
| Parameter | Value |
|---|---|
| Resistance (R) | \(25 \, \Omega\) |
| Reactance (X) | \(-25 \, \Omega\) |
| Impedance Magnitude (\(|Z|\)) | \(25\sqrt{2} \, \Omega\) |
| Phase Angle (\(\phi\)) | \(-45^\circ\) |
| Power Factor (PF) | \(0.707\) |
This result indicates that the circuit has a lagging power factor since the reactance is capacitive (negative X), although for purely resistive-reactive circuits, power factor is often given as a positive value, with 'lagging' or 'leading' specified separately. Here, \(0.707\) is the magnitude of the power factor.
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