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

In the circuit of Fig. 14.24 , R = 2 Ω ,L = 1 mH , and C = 0.4 μF . (a) Find the resonant frequency and the half-power frequencies. (b) Calculate the quality factor and bandwidth. (c) Determine the amplitude of the current at , $ω_{0} , ω_{1} and ω_{2} .$

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(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 $ω = ω_{1} , ω_{2} ,$

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

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