Question E5A07:
What is the magnitude of the current at the input of a parallel RLC circuit at resonance?  

Choices:  
A: Minimum  
B: Maximum  
C: R/L  
D: L/R    

Correct Answer is  A: Minimum  

At parallel resonance, the circuit’s reactances cancel in admittance, making the input impedance maximum. For a given applied voltage, Ohm’s law then gives the smallest input current.

Why A (Minimum) is correct:
- For a parallel RLC, 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 reactive part cancels:
  $$\omega_0 C - \frac{1}{\omega_0 L} = 0$$
  Hence $$Y_{\text{res}} = \frac{1}{R}$$ is at its minimum magnitude, so
  $$Z_{\text{res}} = \frac{1}{Y_{\text{res}}} = R$$
  is maximum. With fixed applied voltage $V$, the input current is
  $$I_{\text{in}} = \frac{V}{Z}$$
  which is minimum when $Z$ is maximum.
- Note: Large equal-and-opposite inductor and capacitor branch currents circulate internally at resonance ($I_L = I_C$), but they cancel at the input, so the source current is minimal.

Why the other choices are incorrect:
- B: Maximum — True for a series RLC, not parallel. In a series circuit, $Z$ is minimum at resonance, so current is maximum.
- C: $R/L$ — This is the damping factor (units s⁻¹) for an RL transient, not the input current at resonance.
- D: $L/R$ — This has units of seconds (the RL time constant), not a current magnitude.

Useful formulas and rules of thumb:
- Resonant frequency: $$\omega_0 = \frac{1}{\sqrt{LC}}, \quad f_0 = \frac{1}{2\pi\sqrt{LC}}$$
- Parallel RLC at resonance: maximum impedance, minimum source current.
- Series RLC at resonance: minimum impedance, maximum current.
- Branch currents at parallel resonance:
  $$I_L = \frac{V}{\omega_0 L}, \quad I_C = V \omega_0 C, \quad I_L = I_C$$
  Their reactive components cancel at the input, leaving only the resistive current $V/R$.