*Question G5A01:
What happens when inductive and capacitive reactance are equal in a series LC circuit?*  

*Correct Answer is  C: Resonance causes impedance to be very low*  

### AI Response: (may contain errors)  

In a series LC circuit, when inductive reactance ($X_L$) and capacitive reactance ($X_C$) are equal, the circuit reaches a condition known as resonance. 

### Correct Answer: C - Resonance causes impedance to be very low

- At resonance, the inductive reactance ($X_L$) and capacitive reactance ($X_C$) cancel each other out:
  
  $$X_L = X_C$$

- The total impedance ($Z$) of the circuit is given by:

  $$Z = R + j(X_L - X_C)$$

  Since $X_L = X_C$, the imaginary part becomes zero, and the impedance is purely resistive. If the resistance ($R$) is low, the overall impedance can be very low, allowing maximum current to flow through the circuit.

### Why the Other Choices are Incorrect:

- **A: Resonance causes impedance to be very high**
  - This is incorrect because at resonance, the impedance is minimized, not maximized. The circuit behaves like a short circuit (ideally) at resonance.

- **B: Impedance is equal to the geometric mean of the inductance and capacitance**
  - This is incorrect because impedance at resonance is not calculated using the geometric mean of inductance and capacitance. Instead, it is determined by the resistance in the circuit.

- **D: Impedance is equal to the arithmetic mean of the inductance and capacitance**
  - This is also incorrect. The impedance at resonance does not relate to the arithmetic mean of inductance and capacitance. The impedance is determined by the resistance when reactances cancel out.

### Useful Formulas:

- Inductive reactance: 
  $$X_L = 2\pi f L$$
  
- Capacitive reactance: 
  $$X_C = \frac{1}{2\pi f C}$$

- At resonance:
  $$X_L = X_C$$

This understanding of resonance is crucial for analyzing LC circuits and their behavior in radio frequency applications.