Introduction to Electric Circuits

Example 6.9

6.2 Parallel resonance 135

A coil having an inductance of 200 ~H and a resistance of 50 1) is connected in
parallel with a capacitor having a capacitance of 120 pF to a 100 V supply.
Determine (1) the resonant frequency, (2) the dynamic impedance of the
circuit, (3) the Q-factor of the circuit, and (4) the current in the circuit at
resonance.

Solution

Figure 6.18

m

'00V l

The circuit is shown in Fig. 6.18.

1 From Equation (6.20),

fo = (1/27r)N/[(1/LC) - (R/L)e]}

= (1/27r)V{[(1/(200 x 10-6x 120 x 10-~2)] - (50/200 • 10-6) 2]

= (1/2rr)V(4166 • 101~ 6.25 x 101~

= 1.02 MHz

Note that (R/L): < 1/LC.

2 From Equation (6.23),

Zd = L/CR = 200 • 10-6/(120 • 10 -12 • 50) = 33.3 kf~

3 The Q-factor is given by

o~oL/R = 2rr • 1.02 • 106 • 200 • 10-6/50 - 25.64

4 At resonance, the current is limited by the dynamic impedance so

I = V/Zd = 100/(33.3 • 10 3) = 3 mA

Bandwidth

We have seen that, assuming R ,~ XL, the impedance of the circuit of Fig. 6.16 is

given by Equation (6.22) to be

Z = joJL/[joJCR + (1 - ~o2LC)]

Factorizing the denominator we get

Z- j~oL/{j~oCR[1 + (1/j~CR) - (~LC/j~oCR)]}

= j~oL/{joJCR[1 + (LC/jo~CR)((1/LC)- ~o2)]}