Question E5D12:
What is reactive power?  

Choices:  
A: Power consumed in circuit Q  
B: Power consumed by an inductor’s wire resistance  
C: The power consumed in inductors and capacitors  
D: Wattless, nonproductive power    

Correct Answer is  D: Wattless, nonproductive power  

Reactive power is energy that sloshes back and forth between the source and reactive elements (inductors and capacitors) each cycle. It does not perform net work over time, so it is “wattless” or “nonproductive.”

Why D is correct (Wattless, nonproductive power):
- Reactive power represents energy storage and return, not energy conversion to heat or work.
- Over a full cycle in ideal L or C, the average power is zero.
- Measured in volt-ampere reactive (VAR), not watts.

Why the others are incorrect:
- A: “Power consumed in circuit Q”  
  The symbol $Q$ here is being confused with the quality factor (a dimensionless ratio). Circuit “Q” is not power; reactive power is a different concept with units of VAR.
- B: “Power consumed by an inductor’s wire resistance”  
  That is real power loss in resistance (in watts), not reactive power. Reactive power is associated with reactance, not resistance.
- C: “The power consumed in inductors and capacitors”  
  Ideal inductors and capacitors do not consume power; they store and release it. Any consumption in real components (ESR, core loss) is real power, not reactive.

Useful formulas and relationships:
- Real (active) power: $$P = V_{\text{rms}} I_{\text{rms}} \cos\varphi \quad \text{(watts)}$$
- Reactive power: $$Q = V_{\text{rms}} I_{\text{rms}} \sin\varphi \quad \text{(VAR)}$$
- Apparent power: $$S = V_{\text{rms}} I_{\text{rms}} \quad \text{(VA)}$$
- Power triangle: $$S^2 = P^2 + Q^2$$
- Power factor: $$\text{PF} = \cos\varphi = \frac{P}{S}$$

Rules of thumb:
- Purely resistive load: $\varphi = 0^\circ$, $P = VI$, $Q = 0$ (PF = 1).
- Purely reactive (ideal L or C): $\varphi = \pm 90^\circ$, $P = 0$, $|Q| = VI$ (PF = 0).
- Inductive loads have positive $Q$ (current lags); capacitive loads have negative $Q$ (current leads).