In one line
Parallel resonance occurs in a coil-capacitor parallel (tank) circuit when the reactive currents of the two branches cancel, making the line current minimum, impedance maximum (\( Z_D = L/CR \)) and power factor unity.
Explained simply
In parallel resonance, energy sloshes back and forth between the coil and the capacitor like water between two connected tanks. The supply only needs to top up the small losses, so very little current is drawn from it, even though a large current circulates inside the loop.
Technically, for a coil (\( R, L \)) in parallel with a capacitor \( C \), resonance occurs when the susceptances cancel: the capacitor current equals the reactive component of the coil current. The circuit then looks like a large pure resistance called the dynamic impedance.
How it works
- Coil current lags; capacitor current leads by 90°. At resonance the coil's reactive component equals the capacitor current.
- This gives \( f_0 = \dfrac{1}{2\pi}\sqrt{\dfrac{1}{LC} - \dfrac{R^2}{L^2}} \). If \( R \) is negligible, \( f_0 = \dfrac{1}{2\pi\sqrt{LC}} \).
- Line current is minimum and in phase with voltage; impedance is maximum \( Z_D = L/(CR) \).
- The circulating current in L and C is \( Q \) times the line current: current magnification.
- Called a rejector circuit because it rejects (offers high impedance to) the resonant frequency. Used in oscillators and radio IF stages.
| Feature | Series | Parallel |
|---|---|---|
| Impedance at \( f_0 \) | minimum (R) | maximum (L/CR) |
| Current at \( f_0 \) | maximum | minimum |
| Magnification | voltage | current |
| Name | acceptor | rejector |
Key formula
\( f_0 = \dfrac{1}{2\pi}\sqrt{\dfrac{1}{LC} - \dfrac{R^2}{L^2}}, \qquad Z_D = \dfrac{L}{CR} \)
\( f_0 \) = resonant frequency (Hz), \( L \) = coil inductance (H), \( R \) = coil resistance (\( \Omega \)), \( C \) = capacitance (F), \( Z_D \) = dynamic impedance (\( \Omega \)).
Quick example
Coil with \( R = 10\,\Omega \), \( L = 0.1 \) H in parallel with \( C = 10\,\mu \)F: \( 1/LC = 10^6 \), \( R^2/L^2 = 10^4 \), \( f_0 = \frac{1}{2\pi}\sqrt{990000} = 158.4 \) Hz, slightly below the series value 159.2 Hz. \( Z_D = 0.1/(10^{-5}\times10) = 1000\,\Omega \).
Common mistakes
- Using the series formula for a tank circuit whose coil resistance is not negligible.
- Saying current is maximum at parallel resonance.
- Mixing up 'acceptor' and 'rejector'.
How MU asks it
Theory only in the MU syllabus: "Explain parallel resonance and derive the resonant frequency and dynamic impedance" (5–8 marks) or "Compare series and parallel resonance" (5 marks). For full marks: draw the circuit and phasor diagram at resonance, derive \( f_0 \), state \( Z_D = L/CR \), sketch impedance and current versus frequency, and give the comparison table.