Introduction to Electric Circuits

132 Resonance

Dividing both numerator and denominator by IR we have

H = 1,/[1 + j~oL//R + 1/jwCR]

H- 1/{1 + j[wL/R- 1/o~CR]} (6.8)

Now from Equation (6.4), 0 = o)oL//R and multiplying both sides by ~o/~o0 we

get

( OJo/OJo)O - woL// R (6.9)

Also, since woL/R = 1/woCR (because woL - 1/woC at resonance then)

Q = 1/woRC (6.10)

Multiplying both sides by ~o0//w we have

(Wo/w)Q- 1/wRC (6.11)

Substituting from Equations (6.9) and (6.11) into Equation (6.8) we see that

H- 1,,/{1 + jQ[(w/O)o) - (o90/o9)]} (6.12)

The phase angle is given by

tan -~ [(~oL - 1/wC)/R] (6.13)

If H is plotted to a base of frequency we obtain the graph in Fig. 6.14 and the

graph of the phase angle to the same base is given in Fig. 6.15. Note that when

XL = Xc, & = 0; when XL = 0, ~ lies between 0 and -90~ when Xc = 0, 4~ lies

between 0 and +90 ~

Figure 6.14

H '

1 ....

070 ....

0 fl fo f2 f

Figure 6.15

Inductive

Capacitive

+901 ,~

-45 J

-90 I___

"3