Question E5D09:
What happens to reactive power in ideal inductors and capacitors?  

Choices:  
A: It is dissipated as heat in the circuit  
B: Energy is stored in magnetic or electric fields, but power is not dissipated  
C: It is canceled by Coulomb forces in the capacitor and inductor  
D: It is dissipated in the formation of inductive and capacitive fields    

Correct Answer is  B: Energy is stored in magnetic or electric fields, but power is not dissipated  

Correct answer (B) explained:
- Ideal inductors and capacitors are lossless. They store energy in fields and then return it to the source each cycle, so the average real power dissipated is zero.
- Inductor stores magnetic field energy: $$E_L = \tfrac{1}{2} L i^2$$
- Capacitor stores electric field energy: $$E_C = \tfrac{1}{2} C v^2$$
- For a purely reactive load with sinusoidal signals, the instantaneous power oscillates positive and negative and averages to zero:  
  If $v(t)=V_m\cos\omega t$ and $i(t)=I_m\sin\omega t$ (90° phase shift), then  
  $$p(t)=v(t)i(t)=\tfrac{V_m I_m}{2}\sin(2\omega t)$$  
  The average over a cycle is 0, so no net dissipation.

Why the other choices are incorrect:
- A: “Dissipated as heat” describes resistive loss. Ideal L and C have no resistance; heat requires real power in a resistive element.
- C: “Canceled by Coulomb forces” is not how AC reactive energy works; the energy is stored in fields and exchanged with the source, not “canceled.”
- D: “Dissipated in the formation of fields” is wrong. Field formation stores energy; dissipation would require loss (resistance), which ideal reactances do not have.

Useful formulas and rules of thumb:
- Reactance: $$X_L=\omega L,\quad X_C=\frac{1}{\omega C}$$
- Power relationships with phase angle $\phi$:  
  $$P=VI\cos\phi \quad(\text{real power, W})$$
  $$Q=VI\sin\phi \quad(\text{reactive power, var})$$
  For a pure inductor or capacitor, $\phi=\pm 90^\circ \Rightarrow \cos\phi=0 \Rightarrow P=0$ and $|Q|=VI$.
- Phase rule: “ELI the ICE man”  
  - Inductor: E (voltage) leads I (current) by 90°  
  - Capacitor: I (current) leads E (voltage) by 90°
- Real components have ESR/ESL causing some loss, but the question assumes ideal parts.