If apparent power is found equal to active power, then the system power factor is:
1
In alternating current (AC) circuits, power is described using three terms: apparent power, active power, and reactive power. Understanding the relationship between these types of power is crucial for analyzing and designing AC systems.
These three power components are related by the power triangle, where the apparent power is the hypotenuse, active power is the adjacent side (usually on the horizontal axis), and reactive power is the opposite side (usually on the vertical axis). The relationship is given by the Pythagorean theorem:
$$S^2 = P^2 + Q^2$$
The power factor (PF) of a system is a dimensionless quantity between 0 and 1 (or 0% and 100%) that represents how effectively the apparent power is converted into active power. It is defined as the ratio of active power to apparent power:
$$PF = \frac{P}{S}$$
Power factor can also be expressed as the cosine of the phase angle ($\phi$) between the voltage and current waveforms:
$$PF = \cos(\phi)$$
A power factor close to 1 indicates efficient power usage, while a low power factor indicates inefficient usage due to a large amount of reactive power.
The question states that the apparent power ($S$) is equal to the active power ($P$).
Given: $$S = P$$
We use the definition of the power factor:
$$PF = \frac{P}{S}$$
Substitute $S = P$ into the power factor formula:
$$PF = \frac{P}{P}$$
Assuming the active power $P$ is not zero (as a non-zero apparent power equal to active power implies $P \neq 0$), the power factor is:
$$PF = 1$$
When the power factor is 1, it means that the active power is equal to the apparent power. This happens when the reactive power ($Q$) in the system is zero. Let's verify this using the power triangle equation:
$$S^2 = P^2 + Q^2$$
If $S = P$, then
$$P^2 = P^2 + Q^2$$
Subtracting $P^2$ from both sides gives:
$$0 = Q^2$$
This implies that $Q = 0$. A system with zero reactive power is purely resistive, meaning the current and voltage are perfectly in phase ($\phi = 0^\circ$). The power factor is $\cos(0^\circ) = 1$.
Therefore, if apparent power is found equal to active power, the system power factor is 1.
The final answer is 1.
| Concept | Symbol | Unit | Description |
|---|---|---|---|
| Apparent Power | $S$ | VA (Volt-Ampere) | Total power from source |
| Active Power | $P$ | W (Watt) | Power doing useful work |
| Reactive Power | $Q$ | VAR (Volt-Ampere Reactive) | Power exchanged by reactive components |
| Power Factor | PF | Dimensionless | Ratio of Active Power to Apparent Power ($P/S$) |
A power factor less than 1 occurs when there is significant reactive power in the circuit due to inductive loads (like motors, transformers) or capacitive loads (like capacitors, long transmission lines). Inductive loads cause the current to lag behind the voltage (lagging power factor), while capacitive loads cause the current to lead the voltage (leading power factor).
A low power factor can lead to several issues:
Power factor correction techniques, usually involving adding capacitors (for lagging PF) or inductors (for leading PF), are often employed to bring the power factor closer to unity (1).
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