Question E5A04:
What is the magnitude of the impedance of a parallel RLC circuit at resonance?  

Choices:  
A: Approximately equal to circuit resistance  
B: Approximately equal to inductive reactance  
C: Low compared to the circuit resistance  
D: High compared to the circuit resistance    

Correct Answer is  A: Approximately equal to circuit resistance  

At resonance in a parallel RLC, the inductor’s and capacitor’s susceptances cancel, leaving only the resistive part. Therefore the circuit appears purely resistive and its impedance equals the circuit’s resistance.

Why A is correct (Approximately equal to circuit resistance):
- The total admittance is
  $$Y = \frac{1}{R} + j\left(\omega C - \frac{1}{\omega L}\right).$$
  At resonance, $\omega_0 = \frac{1}{\sqrt{LC}}$, so the imaginary term is zero and
  $$Y(\omega_0) = \frac{1}{R} \;\Rightarrow\; Z(\omega_0) = \frac{1}{Y} = R.$$
- In a parallel tank, this is the maximum impedance, but its value is set by the resistive branch (the circuit’s equivalent parallel resistance).

Why the others are incorrect:
- B (Approximately equal to inductive reactance): At resonance the reactive parts cancel; the net reactance is zero. The impedance is not set by $X_L$.
- C (Low compared to the circuit resistance): That describes a series RLC at resonance (minimum impedance). A parallel RLC has maximum impedance at resonance.
- D (High compared to the circuit resistance): Although the impedance is “high” relative to off‑resonant values, at resonance it equals the circuit resistance, not greater than it.

Useful formulas and rules:
- Resonant frequency:
  $$\omega_0 = \frac{1}{\sqrt{LC}}, \quad f_0 = \frac{1}{2\pi\sqrt{LC}}.$$
- Parallel RLC admittance:
  $$Y = \frac{1}{R} + j\left(\omega C - \frac{1}{\omega L}\right).$$
- At resonance:
  $$Z = R \quad \text{(purely resistive, maximum for a parallel tank)}.$$
- Practical rule of thumb: If losses are modeled as a series resistance $R_s$ in the inductor with quality factor $Q = \frac{X_L}{R_s}$, the equivalent parallel resistance at resonance is
  $$R_p \approx Q^2 R_s = \frac{X_L^2}{R_s},$$
  and the tank’s impedance at resonance is approximately $R_p$.