Question 14.7: In the circuit of Fig. 14.24,R = 2Ω,L = 1 mH, and C = 0.4 μF......

(a) The resonant frequency is

ω_{0} = \frac{1}{\sqrt{LC}} = \frac{1}{\sqrt{10^{-3}  ×  0.4  ×  10^{-6}}} = 50  krad/s

■ METHOD 1 The lower half-power frequency is

ω_{1} = – \frac{R}{2L} + \sqrt{(\frac{R}{2L})^{2}  +  \frac{1}{LC}}

= – \frac{2}{2  ×  10^{-3}} + \sqrt{(10^{3})^{2}  + (50  ×  10^{3})^{2}}

= – 1  +  \sqrt{1  +  2500}  krad/s  = 49  krad/s

Similarly, the upper half-power frequency is

ω_{2}  =  1  +  \sqrt{ 1 +  2500}   krad/s =  51  krad/s

(b) The bandwidth is

B =  ω_{2}   –  ω_{1} =  2  krad/s

or

B = \frac{R}{L} = \frac{2}{10^{-3}} = 2  krad/s

The quality factor is

Q = \frac{ω_{0}}{B} = \frac{50}{2} = 25

■ METHOD 2 Alternatively, we could find

Q = \frac{ω_{0}L}{R} = \frac{50  ×  10^{3}  ×  10^{-3}}{2} = 25

From Q,  we find

B = \frac{ω_{0}}{Q} = \frac{50  ×  10^{3}}{25} = 2  krad/s

Since  Q > 10 , this is a high-Q circuit and we can obtain the halfpower frequencies as

ω_{1}  =  ω_{0}  –  \frac{B}{2}  =  50  –  1 =  49  krad/s

ω_{2}  =  ω_{0}  –  \frac{B}{2}  =  50  +  1 =  51  krad/s

as obtained earlier.

(c) At ω  =  ω_{0} ,

I = \frac{V_{m}}{R} = \frac{20}{2} = 10  A

At ω  =  ω_{1} , ω_{2} , 

I = \frac{V_{m}}{\sqrt{2}R} = \frac{10}{\sqrt{2}} = 7.071   A