A parallel RLC circuit is driven by an AC current source. As the driving frequency sweeps from low to high, what happens to the voltage across the circuit as it passes through the resonant frequency?
AVoltage drops to a minimum at resonance because the reactive currents cancel, reducing total impedance
BVoltage reaches a maximum at resonance because impedance is maximum at that frequency
CVoltage remains constant because a current source maintains fixed current regardless of impedance
DVoltage reaches a maximum because the capacitor and inductor simultaneously draw maximum current
At parallel resonance, the imaginary parts of the admittance cancel (inductive susceptance and capacitive susceptance sum to zero), leaving only the real admittance 1/R. Total admittance is minimum, so total impedance is maximum. For a current source driving the circuit, V = I × Z, so maximum impedance means maximum voltage. This is the direct opposite of series resonance, where impedance is minimum and current is maximum. The parallel resonant circuit's defining characteristic is its maximum impedance at ω₀ — it looks like a large pure resistor to the source.
Question 2 Multiple Choice
An AM radio tuner uses a parallel LC circuit with a variable capacitor. The engineer wants the tuner to sharply distinguish the 1000 kHz station from the 1010 kHz station. What circuit property is most critical?
ALow resonant frequency, so the LC values are large enough to tune precisely
BHigh Q factor, because narrow bandwidth allows the tuner to select one station while rejecting adjacent ones
CLow Q factor, so the tuner's frequency response is flat and captures the entire AM band equally
DHigh resistance R, because more resistance dissipates unwanted frequencies more effectively
Q = R/(ω₀L) = ω₀RC for a parallel circuit, and bandwidth BW = ω₀/Q. High Q means narrow bandwidth — the impedance peak is sharp, so only frequencies very close to ω₀ receive the high-impedance selectivity that passes the signal. A 10 kHz channel separation in the AM band requires a narrow enough bandwidth to reject the adjacent station; a low-Q circuit with wide bandwidth would pass multiple stations simultaneously. Higher resistance R increases Q (less energy is dissipated per cycle), producing a sharper, more selective peak. This is why tank circuit Q is a primary specification in RF design.
Question 3 True / False
At resonance in a parallel RLC circuit, the circulating current between the inductor and capacitor can be much larger than the current supplied by the external source.
TTrue
FFalse
Answer: True
True, and this is the physical meaning of Q. The inductor current I_L = V/(ω₀L) and capacitor current I_C = Vω₀C are equal in magnitude and opposite in phase at resonance — they circulate internally between L and C each half-cycle. The source only needs to supply the energy lost in the resistance. The ratio of circulating current to source current is exactly Q = R/(ω₀L). A Q of 100 means the internal reactive currents are 100 times larger than the source current. This energy-storage-and-circulation property is why parallel resonant circuits are called 'tank circuits' — they act as reservoirs of oscillating energy.
Question 4 True / False
A series RLC and a parallel RLC circuit with identical L, C, and R values will reach resonance at different frequencies because their energy storage configurations differ.
TTrue
FFalse
Answer: False
False. Both series and parallel RLC circuits resonate at ω₀ = 1/√(LC), regardless of R. The resonant frequency is determined purely by the reactive elements L and C — it is the frequency at which their reactances are equal in magnitude (ω₀L = 1/ω₀C). What differs dramatically between series and parallel resonance is the behavior AT resonance: series resonance gives minimum impedance and maximum current, while parallel resonance gives maximum impedance and minimum current. Same frequency, opposite impedance characteristics.
Question 5 Short Answer
Explain why a parallel resonant circuit (tank circuit) draws minimum current from the source at resonance, even though the inductor and capacitor each carry large reactive currents at that same instant.
Think about your answer, then reveal below.
Model answer: At resonance, the inductor current and capacitor current are equal in magnitude and exactly 180° out of phase (one leads, one lags the voltage by 90°). They cancel each other at the circuit terminals — no net reactive current is visible to the external source. The source only needs to replenish the energy dissipated in the resistance R; all the reactive energy is already circulating internally between L and C each half-cycle. The source current is therefore just V/R (the resistive component), which is minimum when V = I_source × Z is expressed correctly — at resonance Z = R (maximum), so for a given voltage, source current = V/R (minimum). The large internal currents are sustained by the stored energy being continuously exchanged between the magnetic field of the inductor and the electric field of the capacitor.
The tank circuit analogy is apt: like a pendulum swinging between kinetic and potential energy, the LC tank swings energy between magnetic (inductor) and electric (capacitor) forms. Each full cycle, energy flows L→C→L→C with minimal external replenishment needed. The higher Q, the more cycles the energy completes per unit of dissipation, and the smaller the source current relative to the internal currents.