Introduction to Electric Circuits

6. 1 Series resonance 125

9 the phase angle is 4' = 0 and the power factor is cos 4> = 1;

9 the frequency is f~ - 1,/27rk/(LC) and the angular frequency is

~Oo- 1/~(LC).

Example 6.1

Calculate the resonant frequency of a circuit consisting of a coil of inductance
50 mH in series with a capacitor of capacitance 200 nF and a resistor of
resistance 5 f~.

Solution

5fl 50mH 200nF

, II

v

f

Figure 6.4

The circuit is shown in Fig. 6.4. Using Equation (6.2)

fo = a/2~rX/(LC) = 1/2~rX/(50 x 10-3x 200 x 10 -9) - 1/27rX/(10 -8)

= 1/(2Ir • ].0 -4) "-" 104/2"/r- 1.591 • 103 Hz

= 1.591 kHz

Note that this result is independent of the resistance in the circuit, whether it be

that of a separate resistor or that of the coil.

Example 6.2

Determine the value of capacitance to which the variable capacitor C must be

set in order to make the circuit given in Fig. 6.5 resonate at 400 Hz.

5mH C

Figure 6.5

I