Introduction to Electric Circuits

128 Resonance

Solution

----4

50~ 20mH C

I - II

Figure 6.10

250mV f l 0

The circuit is shown in Fig. 6.10. At resonance, )Co = 1/27r~(LC) so

C- 1/4~foZL = 1]47r 2 (100 • 103) 2 20 • 10 .3= 126.6 X 10-12 F

Therefore

Xc = 1/2rrf0C- 1/27r 100 • 103• 126.6 • 10 -~2- 12 571 f~

At resonance

I = V/R = 250 x 10-3/50 5 mn

The potential difference across the capacitor at resonance is

IX c = 5 X 10 .3 X 12 571 = 62.855 V

The capacitor working voltage must therefore be about 65 V.

Q-factor

The fact that the circuit behaves as a pure resistor and that the power factor is
unity seems to indicate that there is no reactive power in the circuit at
resonance. This is not so: what is happening is that the reactive energy is
continuously being transferred between the capacitor, where it is stored in the
electric field, and the inductor, where it is stored in the magnetic field. Since this
energy transfer is taking place within the circuit, it appears from the outside as
though there is no reactive power. As the resistance of the circuit becomes
smaller so the current becomes larger and the stored energy (12XL = IZXc)
oscillating between the capacitor and the inductor becomes much larger than
the energy (I2R) dissipated in the resistor.
The ratio of 12XL to IZR is called the Q-factor (quality factor) of the coil or
circuit so that
Q = IZx~/(I2R) -XL//R

and since XL = 2"rrf0L = w0L

Q = woL/R (6.4)