After learning about series resonance, we will move on to the parallel resonance.

**Contents**show

This kind of resonance occurs when a circuit has an inductor and a capacitor connected together in parallel.

## Parallel Resonance

The parallel RLC circuit in Figure.(1) is the dual of the series RLC circuit.

The admittance is

or

Resonance occurs when the imaginary part of Y is zero,

or

which is the same as the equations for the series resonant circuit. The voltage |**V**| is sketched in Figure.(2) as a function of frequency.

Notice that at resonance, the parallel LC combination acts as an open circuit, so that the entire currents flow through R. Also, the inductor and capacitor current can be much more than the source current at resonance.

We exploit the duality between the series resonance circuit and Figure.(1) by comparing Equation.(2) with Equation.(3). By replacing R, L, and C in the expressions for the series circuit with 1/R, 1/C, and 1/L respectively, we obtain for the parallel circuit

Using Equations.(5) and (7), we can express the half-power frequencies in terms of the quality factor. The result is

Again, for high-Q circuits (Q ≥ 10)

Table.(1) presents a summary of the characteristics of the series and parallel resonant circuits. Besides the series and parallel RLC considered here, other resonant circuits exist.

## Parallel Resonance Examples

Let us review the parallel resonance examples below:

### Parallel Resonance Example 1

In the parallel RLC circuit in Figure.(3), let R = 8 kΩ, L = 0.2 mH, and C = 8 µF. (a) Calculate ω_{0}, Q, and B. (b) Find ω_{1 }and ω_{2}. (c) Determine the power dissipated at ω_{0}, ω_{1}, and ω_{2}.

**Solution:**

(a)

(b) Due to the high value of Q, we can regard this as a high-Q circuit.

Hence,

(c) At ω = ω_{0}, Y = 1/R or Z = R = 8 kΩ. Then

Since the entire current flows through R at resonance, the average power dissipated at ω = ω_{0 }is

or

At ω = ω_{1}, ω_{2},

### Parallel Resonance Example 2

Determine the resonant frequency of the circuit in Figure.(4).

**Solution:**

The input admittance is

At resonance, Im(Y) = 0 and

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