Introduction to Electric Circuits

126 Resonance

Solution

Using Equation (6.2) f0 = 1/2~rX/(LC) and rearranging it to make C the subject

we get C = 1/4~fo2L (squaring both sides and then multiplying both sides by

C/fo2). Thus

C = 1/47r 2 4002 X 5 X 10 -3 -- 31.6 x 10-6F

Example 6.3

A resonant series circuit consists of a capacitor having a capacitance of 0.1 ~F

and a coil whose inductive reactance is 60 ~. Calculate the inductance of the

coil.

60 s r O. 1 pF

II

vt9 f

Figure 6.6

Solution

The circuit is shown in Fig. 6.6 and r is the resistance of the coil. Since the circuit

is in a state of resonance we can use Equation (6.1) with f = f0:

2ff0L = 1/2ffoC

X, = /2 oC

fo = 1/2~rXLC (multiplying both sides by fo/XL)

SO

f0 = 1/2zr 60 • 0.1 x 10 .6- 26 526 Hz

and

L = XL/2Zrfo = 60/2zr. • 26 526 = 359.9 • 10-6H

Impedance and current at resonance
The graph of impedance Z = ~/[R2 + (XL- XC) 2] to a base of frequency is
given in Fig. 6.7 and this shows that Z has a minimum value (=R) at the
resonant frequency. Consequently, the circuit current at this frequency will
have its maximum value (= V/R) as shown in Fig. 6.8.
If the resistance is small, the current could be very large and the potential
difference developed across the inductance (IXL) and the capacitance (IXc)
would then be very large (many times bigger than the supply voltage V). Great