Engineering What is the role of the resistor in a Parallel Resonance Circuit?

AI Thread Summary
In a Parallel Resonance Circuit, the resistor plays a crucial role in determining the circuit's Quality Factor (Q), which affects the sharpness of the resonance peak. The relationship is defined by the formula Q = omega_0 * L / R, where omega_0 is the resonance angular frequency. A higher resistance leads to a lower Q, resulting in a wider resonance bandwidth. Additionally, the resistor influences the leakage current within the circuit. Understanding these effects is essential for optimizing circuit performance.
paul9619
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Hi all.

I have a Parallel Resonant circuit with a Capacitor in Parallel with an Inductor & Resistor in Series together. Capacitor on the left branch and Inductor & Resistor on the right branch.

The question I would like to ask is what effect does the resistor have on the circuit?

I believe it may have something to do with the leakage current? Not really sure though.

Anyone help out??
 
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paul9619 said:
Hi all.

I have a Parallel Resonant circuit with a Capacitor in Parallel with an Inductor & Resistor in Series together. Capacitor on the left branch and Inductor & Resistor on the right branch.

The question I would like to ask is what effect does the resistor have on the circuit?

I believe it may have something to do with the leakage current? Not really sure though.

Anyone help out??
It determines the "Q" (Quality Factor) of the resonance. If omega_0 is the resonance angular frequency,

Q = omega_0 * L / R.

The width of the resonance at the -3dB (half power) points is

delta_omega = omega_0 / Q
 

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