Question E5A01:
What can cause the voltage across reactances in a series RLC circuit to be higher than the voltage applied to the entire circuit?  

Choices:  
A: Resonance  
B: Capacitance  
C: Low quality factor (Q)  
D: Resistance    

Correct Answer is  A: Resonance  

At resonance in a series RLC, the inductor and capacitor voltages can be much larger than the source due to voltage magnification.

Why A (Resonance) is correct:
- In a series RLC at resonance, $X_L = X_C$ and the net reactance cancels, so the total impedance is minimum: $Z = R$.
- The current is maximized: $I = \dfrac{V}{R}$.
- The individual reactive voltages are:
  - $V_L = I X_L$, 
  - $V_C = I X_C$,
  and since $X_L = X_C$, each can be much larger than $V$.
- The magnification equals the quality factor:
  $$Q = \frac{X_L}{R} = \frac{X_C}{R} \quad \Rightarrow \quad V_L = QV, \; V_C = QV$$
- Although $V_L$ and $V_C$ are large, they are 180° out of phase and cancel in the sum, leaving the source voltage across $R$.

Why the other choices are incorrect:
- B: Capacitance — Having a capacitor present does not by itself create high reactive voltages; the magnification occurs specifically at resonance with sufficient Q.
- C: Low quality factor (Q) — Low Q means higher losses (larger $R$), reducing current and thus reducing $V_L$ and $V_C$. High Q increases magnification.
- D: Resistance — Increasing $R$ raises damping, decreases current, and reduces voltages across the reactances; it does not cause them to exceed the source.

Useful formulas and rules:
- Reactances: $X_L = 2\pi fL$, $X_C = \dfrac{1}{2\pi fC}$
- Impedance: $Z = R + j(X_L - X_C)$
- Current: $I = \dfrac{V}{|Z|}$
- At resonance: $X_L = X_C$, so $Z = R$, $I = \dfrac{V}{R}$
- Voltage magnification in series resonance: $V_L/V = V_C/V = Q$
- Rule of thumb: Higher Q ⇒ higher circulating current and higher inductor/capacitor voltages; design parts for the elevated voltage stress.