Effect of a parallel (tank) resonant circuit placed in a filter path: At resonance, what impact does a parallel LC tank have on the downstream or final filter current?

Introduction:
A parallel (tank) resonant circuit exhibits very high impedance at its resonant frequency. When inserted in the signal path or as a shunt in certain filter topologies, this high impedance behavior changes how much current can flow at resonance, which is critical for shaping filter responses and isolating stages.

Given Data / Assumptions:

• Parallel LC tank operated at its resonant frequency
• Ideal components; losses only small unless specified
• Placed so that its impedance influences line current to the next stage

Concept / Approach:

At resonance, a parallel LC has large impedance because the equal and opposite reactive currents in L and C circulate internally, minimizing net current drawn from the line. Consequently, the branch presents a high barrier to current flow, effectively blocking current to the load in that path or sharply reducing it, depending on the exact placement in the filter.

Step-by-Step Solution:

Verification / Alternative check:

Compare with a series resonant circuit, which has minimum impedance at resonance and thus passes maximum current. The parallel case is the complement, confirming the blocking behavior at resonance.

Why Other Options Are Wrong:

• very little: Incorrect; effect is large at resonance due to very high impedance.
• The bandpass frequencies will change.: Placement may affect tuning slightly, but the principal effect at resonance is the high impedance (blocking), not retuning by itself.
• The frequency cutoff will change.: Cutoffs relate to design Q and component values; the primary immediate effect at resonance is current suppression.
• It will automatically double the Q.: Q depends on L, C, and losses; not automatically doubled by topology.

Common Pitfalls:

• Confusing series versus parallel resonance effects on impedance.
• Assuming resonance always means maximum current; that is only for series resonance.

Final Answer:

The impedance will block output.