Introduction to Electric Circuits

6. 1 Series resonance 129

Since mo = 1/x/(LC), then Q - L/RV'(LC) = ~L/(RV'C), so

e = (1/R)[V(L/C)] (6.5)

Q is the ratio of inductive reactance to resistance and since the unit of both of

these is the same (the ohm), then Q itself is dimensionless.

Q-factors are of the order of 10 in the audio frequency range (up to about

20 kHz), whereas in the radio frequency range they are of the order of 102 and

in the microwave range they can be as high as 10 3. The bigger the Q-factor the

easier it is for the circuit to accept current and power at the resonant frequency

so that, for example, in radio and television receivers a particular station can be

selected and others (which have their own resonant frequency) can be

rejected.

Remember: reducing the resistance in the circuit increases the Q-factor.

Example 6.6

The circuit shown in Fig. 6.11 operates at the resonant frequency of 11.25 kHz.

Determine (1) the Q-factor of the circuit, (2) the capacitance of the capacitor,

and (3) the current in the circuit.

5D 5mr C

It

"YI{)

Figure 6.11

Solution

1 Using Equation (6.4) the Q-factor of the circuit is given by Q = woL/R, so

Q - 27r 11.25 • 10 3 X 5 X 10-3//5 -- 70.68.

2 At the resonant frequency, the capacitive reactance (Xc) is equal to the

inductive reactance (XL). The inductive reactance is

27rf0L - 2rr 11.25 • 10 3 X 5 X 10 -3 -- 353.4 ~. The capacitive reactance of

the capacitor is therefore also 353.4 fl and its capacitance

C- 1/27r 11.25 • 103 • 353.4 - 40 • 10-9F

3 The current is I = V/R - 24/5 - 4.8 A.

Filter applications

Series RLC circuits are commonly used in filter applications and a basic