Short notes

Series and Parallel Resonance Short Notes

Tune an AM radio receiver and you are mechanically varying the capacitance of a parallel LC tank circuit to bring its resonant frequency into alignment with the 630 kHz carrier of your desired station — at resonance, the tank circuit presents extremely high impedance to the 630 kHz signal while offering near-zero impedance to all other frequencies, effectively selecting one station out of dozens. The Q factor of that tank circuit determines how sharp the selectivity is and how far adjacent stations must be spaced.

EEE, ECE, EI

How it works

Series resonance: at resonant frequency ω₀ = 1/√(LC), inductive reactance X_L = capacitive reactance X_C, so they cancel and total impedance Z = R (minimum, purely resistive). Current is maximum = V/R. Quality factor Q = ω₀L/R = 1/(ω₀CR) = (1/R)√(L/C). Bandwidth B = ω₀/Q = R/L rad/s, or B = f₀/Q in Hz. Half-power frequencies: ω₁ = ω₀ − B/2 and ω₂ = ω₀ + B/2, so f₀ = √(f₁f₂) (geometric mean). Voltage across L or C at resonance = Q×V_source — for high Q circuits this can be very large; Q = 100 gives 100× voltage magnification.

Key points to remember

For a practical parallel RLC circuit at resonance: admittance Y = 1/R (minimum admittance, maximum impedance Z = R_p). Resonant frequency ω₀ = 1/√(LC) for ideal parallel LC. For a practical inductor (resistance r in series with L) in parallel with C, ω₀ = √(1/LC − r²/L²) and dynamic resistance R_d = L/(Cr) at resonance. Q = ω₀L/r for the practical circuit. Series resonance has maximum current (voltage resonance); parallel resonance has minimum current (current resonance). For an RLC series circuit with R = 10Ω, L = 10 mH, C = 10 µF: ω₀ = 1/√(10⁻²×10⁻⁵) = 3162 rad/s, Q = 3162×0.01/10 = 3.16, B = 3162/3.16 = 1000 rad/s.

Exam tip

The examiner always asks you to calculate resonant frequency, Q factor, and bandwidth for a given series or parallel RLC circuit, and sketch the impedance or current vs frequency curve — mark f₀, f₁, f₂, and the −3dB points clearly on the curve.

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