FAQ
TL;DR: To tell if a load is inductive or capacitive, check Q: your −11.75 var means capacitive; “reactive power for the capacitive load is negative.” [Elektroda, Anonymous, post #16934211]
Why it matters: Knowing Q’s sign lets you size compensation and avoid higher currents or penalties. This FAQ is for electricians and techs asking how to identify load type and pick capacitors from basic P–Q–S data.
Quick Facts
- Q < 0 → capacitive; Q > 0 → inductive. Field-ready rule for fast diagnostics. [Elektroda, Anonymous, post #16934211]
- Power triangle: S² = P² + Q². Use it to verify apparent, active, and reactive power. [Elektroda, Anonymous, post #16946655]
- Basic multimeters won’t resolve P, Q, S or Q’s sign; use a power analyzer. [Elektroda, Anonymous, post #16946655]
- Typical target after compensation: cos φ ≈ 0.95. Avoid overcorrection beyond this. [Elektroda, bodzio507, post #16951918]
- Example sizing: at 230 VAC, 11.75 var needs about 700 nF to reach unity. [Elektroda, Anonymous, post #16947983]
How do I quickly tell if my AC load is inductive or capacitive?
Read the sign of reactive power Q. Negative Q means a capacitive load, positive Q means an inductive load. Your measured −11.75 var indicates a capacitive receiver. “Reactive power for the capacitive load is negative.” Confirm with a capable power meter. [Elektroda, Anonymous, post #16934211]
Can I use a standard multimeter to calculate Q and cos φ?
No. A typical multimeter measures RMS V and I, not phase angle or Q. Use a dedicated power meter that reports P, Q, S, cos φ, and Q’s sign. These instruments simplify classification and compensation work. [Elektroda, Anonymous, post #16946655]
What formula links apparent, active, and reactive power?
Use the power triangle: S² = P² + Q². Verify any two quantities and compute the third. This helps cross-check questionable readings before you decide on compensation. “S^2 = P^2 + Q^2.” [Elektroda, Anonymous, post #16946655]
My data shows P = −32.499 W and Q = −11.75 var. What does that mean?
The negative Q flags a capacitive nature. The negative P suggests your sign convention or wiring may be reversed. Focus first on Q’s sign for type identification, then confirm meter orientation and CT polarity to resolve P’s sign. [Elektroda, Anonymous, post #16934211]
How do I size a capacitor to correct 11.75 var at 230 VAC, single‑phase?
One worked example shows 11.75 var needs about 700 nF at 230 VAC to reach unity power factor. That corresponds to roughly 4.5 kΩ capacitive reactance at 50 Hz. Use this as a sizing check against your analyzer’s readings. [Elektroda, Anonymous, post #16947983]
What power factor should I target when compensating?
Aim for cos φ around 0.95 rather than unity. This avoids overcompensation and keeps currents reasonable across load changes. “Usually the reactive power is compensated to cos (phi) 0.95.” [Elektroda, bodzio507, post #16951918]
Is there a simple field test to confirm inductive vs capacitive?
Yes. Temporarily connect a known capacitor and watch current or apparent power. If current increases, the system was already capacitive. If current decreases, the system was inductive. Remove the test part after observing the change. [Elektroda, bodzio507, post #16947956]
Can a wattmeter measure reactive power directly?
A wattmeter reads active power. In three-phase setups, specific connections allow inferring Q. For single-phase, use a power analyzer that directly reports Q and cos φ to avoid wiring tricks. [Elektroda, bodzio507, post #16948065]
Is this problem about a single‑phase or three‑phase system?
The original case involved a single‑phase system. That affects instrument choice and any special wattmeter methods. Use a single‑phase power analyzer for clarity. [Elektroda, Stanislaw elektryk12, post #16951876]
What’s a practical 3‑step method to tune compensation on site?
- Measure P, Q, S, and cos φ with a power analyzer.
- Add a trial capacitor; observe Q trending toward zero and current change.
- Adjust to reach about cos φ = 0.95 under typical load.
[Elektroda, bodzio507, post #16951918]
Will small loads benefit financially from compensation?
Savings may be negligible for very small P and Q. Hardware cost and labor can exceed any bill reduction. “You will spend more on a capacitor than you will save.” Prioritize larger offenders first. [Elektroda, bodzio507, post #16951918]
What if I have many receivers switching on and off?
Expect the power factor to change over time. Use automatic regulators or stepped capacitor banks to track the load. Fixed caps can mis-compensate when the mix shifts. [Elektroda, bodzio507, post #16947913]
How do constant-speed fans affect compensation planning?
Fans can run steadily, keeping power relatively constant. In such cases, a fixed compensation step may hold well. Verify stability with a data log before finalizing. [Elektroda, Anonymous, post #16947933]
What’s an edge case I should watch out for when testing with a capacitor?
If the load is already capacitive, adding capacitance will raise current instead of lowering it. That indicates overcompensation risk; remove or reduce capacitance immediately. [Elektroda, bodzio507, post #16947956]
What statistic from the thread confirms capacitive behavior?
The measured reactive power was −11.75 var with an apparent power of 34.56 VA and about −0.94 power factor. The negative Q establishes the capacitive character. [Elektroda, Stanislaw elektryk12, post #16933940]
